The magnitude of the acceleration of the elevator is approximately 0.84 m/s².
When you stand on a bathroom scale in an elevator, it says your mass is 76 kg. Your actual mass is 70 kg.
Thus, the apparent weight of an object on the scale is the product of the object's mass and the net force acting on it. The scale reads a greater mass because of the upward force the elevator floor exerts on you.
The magnitude of the acceleration of the elevator is provided by the following formula:
The magnitude of the acceleration of the elevator = F_net/m,
where F_net is the net force on the object and m is the object's mass.
Since your actual mass is 70 kg and the scale measures an apparent mass of 76 kg, the net force acting on you is the difference between the apparent weight and the actual weight, which is given by
F_net = (76 kg - 70 kg) by × 9.8 m/s²
= 58.8 N
Thus, the magnitude of the acceleration of the elevator is: the magnitude of the acceleration of the elevator
= F_net/m = 58.8 N/70 kg
≈ 0.84 m/s²
Therefore, the magnitude of the acceleration of the elevator is approximately 0.84 m/s².
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The half life of a radioactive substance is 5 hours. If 5g of the substance is left after 20 hours, determine the original mass of the substance
Answer:
The original mass of the substance was 10g.
Explanation:
The half-life of a radioactive substance is the amount of time it takes for half of the substance to decay. In this case, the half-life is 5 hours.
We can use the half-life formula to find the original mass of the substance:
N = N0 * (1/2)^(t/T)
where:
---N0 is the initial mass of the substance
---N is the remaining mass of the substance after time t
---T is the half-life of the substance
We know that after 20 hours, only half of the substance remains:
N = N0 * (1/2)^(20/5) = 0.5 * N0
If we solve for N0, we get:
N0 = N / 0.5 = 5g / 0.5 = 10g
Therefore, the original mass of the substance was 10g.
if a force is exerted on an object, is it possible for that object to be moving with constant velocity? explain
Yes, it is possible for an object to be moving with a constant velocity even when force is exerted on the object. When an object is in a state of rest, a force is required to move it from that position.
What is Newton's second law of motion?Newton’s second law states that the acceleration of an object is directly proportional to the force exerted on it and inversely proportional to its mass. Thus, a larger force results in a greater acceleration of the object. If there is no force applied to the object, the object will remain stationary or move at a constant velocity.
However, if there is a force applied to the object, it will accelerate. If the force applied is balanced by an equal and opposite force, the object will continue to move with a constant velocity. An object in motion is said to be in equilibrium when the net force acting on the object is zero. When the net force acting on an object is zero, it moves at a constant velocity. Therefore, if a force is exerted on an object, it is possible for the object to be moving with a constant velocity if the forces are balanced.
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waves of which wavelength would have the hardest time diffracting through the doorway into the room? a. 5 centimeters b. 2 meters c. 100 meters d. all of the above would easily diffract.
The waves of which wavelength would have the hardest time diffracting through the doorway into the room is 100 meters. The correct answer is Option C.
What is Diffraction?Diffraction is the spreading of waves when they pass through a gap or go around the edge of an obstacle. Diffraction occurs when a wavefront interacts with an obstacle, such as a gap or edge, and the wavefront bends around it or spreads out. This occurs for any wave, such as light waves or sound waves, and may be observed in various natural and everyday occurrences.
Diffraction of waves:
It is directly proportional to the wavelength and inversely proportional to the size of the obstacle. When the width of an obstacle is similar to the wavelength of the wave, the amount of diffraction is the highest.
What is Wavelength?Wavelength, also known as lambda, is a concept in physics that refers to the distance between two adjacent peaks or troughs of a wave. It is usually given the symbol λ and is measured in meters. The distance between two wave crests or troughs is the wavelength of a wave. Waves of varying wavelengths are present in our environment. Waves with shorter wavelengths, such as gamma rays, X-rays, and UV radiation, are high in energy but have a low range, while waves with longer wavelengths, such as radio waves and microwaves, are low in energy but have a longer range.
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how long does it take the moon to make one complete rotation around the earth?
In outer space will a liquid in a beaker exert a pressure on the bottom or on the sides of a beaker?
Answer:
Yo dude, if you had a beaker of liquid in outer space, it wouldn't push down on the bottom or the sides of the beaker like it would on Earth. In space, there's no gravity to make the liquid settle down, so it forms a round shape because of surface tension. So basically, the liquid would just float around in a ball inside the beaker. If you moved the beaker around, the liquid would just roll around with it like a bouncy ball.
Riders on the Power Tower are launched skyward with an acceleration of 4g, after which they experience a period of free fall. (Figure 1)What is a 60 kg rider's apparent weight during the launch?What is a 60 kg rider's apparent weight during the period of free fall?
1.) The rider's apparent weight during the launch (launched skyward with an acceleration of 4g, after which they experience a period of free fall) = 2400 N
2.) The rider's apparent weight during the period of free fall = zero / 0
How does Power Tower operate?The Power Tower ride is an amusement ride that launches passengers high up in the air before letting them freefall down towards the ground. The ride's name is derived from the tower structure, which is used to store the ride's energy before propelling passengers skyward. After a brief period of freefall, the ride decelerates passengers and gently lowers them back to the ground. The ride is designed to provide a thrilling experience while maintaining passenger safety.
For the Power Tower ride, a 60 kg rider's apparent weight is 2400 N during the launch:
(4g x 60 kg x 9.8 m/s²
= 2400 N
And it is zero during the free fall period (due to the absence of a supporting force).
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A simple pendulum on earth has a period of 1.00 s where theacceleration due to gravity is g = 9.81 m/s2.The pendulum is then taken to themoon, where the acceleration due to gravity is 0.17g.
(a) Would its period increase, decrease, orstay the same?
(b) Calculate the period of the pendulum on the moon.
(a) The period of the pendulum on the moon would increase. (b) The period of the pendulum on the moon can be calculated using the formula [tex]T=2\pi\sqrt\dfrac{l}{g}[/tex] where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. In this case, the period of the pendulum on the moon is approximately 8.12 times greater than the period of the pendulum on the earth..
The formula of the period of a simple pendulum is given as:
[tex]T=2\pi\sqrt\dfrac{l}{g}[/tex]
Where T is the time period of the pendulum, l is the length of the pendulum, and g is the acceleration due to gravity
.From the above formula, we can see that the period of a pendulum is inversely proportional to the square root of acceleration due to gravity.
Therefore, the period of the pendulum will be increased when the acceleration due to gravity decreased from earth to moon. The period of the pendulum on the moon would increase.
(b) The acceleration due to gravity on the moon is 0.17g. Therefore, the period of the pendulum on the moon can be determined by using the formula of the period of a simple pendulum.
[tex]T=2\pi\sqrt{\dfrac{l}{g}}\\T=2\pi\sqrt{\dfrac{l}{0.17g}[/tex]
Now, substituting the given values in the above equation:
[tex]T=2\pi\sqrt{\dfrac{l}{(0.17\times9.81)}}\\T=2\pi\sqrt{1.67l}\\T=2\pi\times1.29\sqrt{l}\\T=8.12\sqrt{l}[/tex]
Therefore, the period of the pendulum on the moon is 8.12 times greater than the period of the pendulum on the earth.
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given the unbalanced equation: al2(so4)3 ca(oh)2!al(oh)3 caso4 what is the coefficient in front of the caso4 when the equation is completely balanced with the smallest whole-number coefficients?
The given unbalanced equation is:Al2(SO4)3 + Ca(OH)2 → Al(OH)3 + CaSO4
Chemical equations are balanced by adjusting the coefficients of the molecules on either side of the equation. To balance the equation, the same number of atoms should be present on both sides. To balance a chemical equation, the coefficient should be a whole number. The balanced equation should have the lowest common multiple coefficients for the molecules. Balancing the equation helps to understand the relationship between the reactants and the products involved in the chemical reaction.
To balance the above chemical equation, the coefficients should be added to the molecules on the left and right of the equation in order to have the same number of atoms on both sides. And the balanced chemical equation would look like this:
Al2(SO4)3 + 3 Ca(OH)2 → 2 Al(OH)3 + 3 CaSO4
In the balanced chemical equation, the coefficient in front of the CaSO4 is 3.
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the three bins represent three important properties of stars. What are the items that we must measure in orrder to determine each property into the three bins.
Luminosity ________
Surface Temperature________ Mass________
To determine each property in the three bins, certain items must be measured. The luminosity, surface temperature, and mass are the three important properties of the stars represented by three bins.
The measurement for each property is as follows:
For luminosity: To measure luminosity, we must measure the total amount of energy a star emits in all wavelengths.
For surface temperature: Surface temperature is determined by analyzing the spectrum of light emitted by the star.
The spectrum shows a rainbow of colors, and some colors will be more intense than others. These colors can be used to estimate the temperature of the star's surface.
For mass: Mass is calculated using observations of how the star interacts with its surroundings. Astronomers observe the gravitational effect that a star has on other objects around it. The mass of a star can be estimated using this method.
Stars are gigantic balls of burning gas that light up the sky and heat up planets around them. The sun is a star, for example.
Some stars are smaller and some are larger, but all of them share the same basic structure. The enormous nuclear furnace at the center of every star produces heat and light through fusion.
Stars are made up of mostly hydrogen and helium, but they contain small amounts of other elements. They are classified into three categories based on their luminosity, surface temperature, and mass.
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PLEASE HELP ME
2A
2B
2C
PLEASE
Answer:
Explanation:
no
14 13 12 11 10
DIFFERENT TYPES OF ENERGY: MATCH THE DEFINITIONS
The energy which a body possesses
by virtue of being in motion.
DRAG FROM HERE
ENERGY
JOULES
KINETIC
ENERGY
POTENTIAL
ENERGY
CONSERVATION
OF ENERGY
The law which states energy cannot
be created or destroyed.
The energy held by an object because of
its position relative to other objects,
stresses within itself, its electric charge,
or other factors
The ability to do work.
What energy is measured in.
D
Answer: Energy is the ability to do work.
Joules is what energy is measured in.
Kinetic energy is the energy which a body possesses by virtue of being in motion.
Potential energy is the energy held by an object because of its position relative to other objects, stresses, within itself, its electric charge, or other factors.
Explanation:
Una pelota de golf sale del punto de partida, al ser golpeada, con una velocidad de 40 m/s a 65°. Si cae sobre un green ubicado 10 m mas arriba que el punto de partida, ¿cual fue el tiempo que permaneció en el aire y cuál fue la distancia horizontal recorrida respecto al palo?
We can use the motion equations to calculate how long the golf ball stayed in the air for and how far it travelled horizontally. Then, we break down the initial speed into its horizontal and vertical components:
Vx = 40 m/s times cos(65°) yields 16.94 m/s. Vy is 40 m/s times sin(65°) to get 35.59 m/s. All throughout the peloton's flight, the horizontal component of the speed remains constant, but the vertical component is affected by the peloton's acceleration because of its seriousness. We can use the following equation to determine how long the peloton stays in the air: h=Vy*t + (1/2)gt2 Where h is the highest height the peloton has ever reached (10 m), Vy is the vertical component of initial velocity (35.59 m/s), and g is the acceleration as a result of The gravity (-9.8 m/s2) and the amount of time the pelota is in the air. Taking t out of this equation gives us: g = t = (Vy sqrt(Vy2 + 2gh)) The positive solution must be used in order to determine the total amount of time the peloton is in the air: T is equal to (35.59 m/s + sqrt((35.59 m/s)2 + 2*(-9.8 m/s)*(10 m)) / (-9.8 m/s2). t = 4.03 s (aproximadamente) (aproximadamente) We can now calculate the horizontal distance travelled by the pelota during that time using the horizontal component of the initial speed: d = Vx * t d = (16.94 m/s) * (4.03 s) (4.03 s) d = 68.3 m As a result, the ball remained in the air for 4.03 seconds while travelling a horizontal distance of 68.3 metres with regard to the target.
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if you apply a force of 10 n to a box and push it 10 m in 10 s, how much power did you deliver?responses
The power delivered by the person who applied a force of 10N to a box and pushed it 10 m in 10 s = 10 W
To determine the power delivered by the person who applied a force of 10N to a box and pushed it 10m in 10s can be determined by the formula,
P = W/t
Where
P denotes power,
W denotes work, and
t denotes time.
Therefore, power is given by the formula:
P = W/t = (F × s) / t
Here, F = 10 N (applied force)s
= 10 m (displacement)
t = 10 s (time taken)
Hence,
P = (10 N × 10 m) / 10 s
= 10N m / s
= 10 W
Therefore, the power delivered by the person who applied a force of 10N to a box and pushed it 10m in 10s is 10 W.
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all but which condition must be present for the calvin cycle reactions to occur?
All but option A: the plant is exposed to light condition must be present for the Calvin cycle reactions to occur.
What is the Calvin cycle?The Calvin cycle is the biochemical process that takes place in the chloroplasts of plants, in which carbon dioxide is fixed into glucose molecules. This process involves several enzymes and molecules, but the energy needed to power the process comes from ATP and NADPH, which are produced during the light-dependent reactions that take place in the thylakoid membranes of the chloroplasts.
Therefore, light is required for the Calvin cycle to occur since it is the source of the energy that drives the reaction. However, all parts of the plant do not need to be exposed to light for the Calvin cycle to occur. Only the chloroplasts, which are found in the mesophyll cells of the plant, need to be exposed to light for photosynthesis to occur.
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See full question below
all but which condition must be present for the calvin cycle reactions to occur?
The plant is exposed to light
water is not involved in any of the reactions
light energy is converted into chemical energy
provide the electrons needed to reduce NADP
A 1,600 kg car is moving at 22 m/s. How much work was done to accelerate it to this speed?
O 7.7 x 105 J
O 3.5 x 104 J
○ 3.9 × 105 J
O 1.5 x 106 J
!!! Urgent
The closest answer among the options given is 3.9 x 105 J. . An object can accelerate by increasing its speed, changing its direction, or both.
What is Acceleration?
Acceleration is the rate of change of velocity of an object over time. It is a vector quantity, meaning it has both magnitude and direction, and is expressed in units of meters per second squared (m/s^2) or feet per second squared (ft/s^2)
The work done to accelerate the car can be calculated using the kinetic energy formula:
K = 1/2 mv^2
Substituting the given values, we get:
K = 1/2 (1600 kg) (22 m/s)^2
K = 677,600 J
Therefore, the work done to accelerate the car to this speed is 677,600 J.
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What is advantage and disadvantage of horizontal and vertical axis turbine?
Answer:
Horizontal-axis (HAWT) pros
It is more effective than vertical-axis turbines, usually, vertical-axis turbines are lower to the ground, so they can't utilize the higher wind speed that is found higher above sea level.Higher efficiency: every mechanical machine will always convert its fuel into useless energy, (i.e. sound energy). They can convert more wind power into electricity than VAWTs. This is due to the blades being perpendicular to the wind.Cons
Transportation: Due to the masts and blades being taller than VAWTs, its transportation can occasionally cost up to 25% of the equipment's price.They need a yaw control: A yaw system is the component that will be required for the direction of the rotor towards the wind.Vertical-axis (VAWT) pros
Unlike HAWTS, a yaw system is not a necessity, nor do they even need it.The turbines can be closer together.Cheaper: As they are not as elevated as their HAWT counterparts, they do not need as much money.Do not need to be pointy towards the wind: this benefit is most prominent in places where wind direction varies frequently.Cons
Decreased level of efficiency: In VAWTs, there will be more drag that will occur inside the blades when they rotate.A resistor, R, is connected to a combination of three batteries as shown. The polarity of E is opposite that of the other two batteries.
Photo!
In a three battery combination whose second resistor is opposite in polarity of the other two, the Σε and Σr are 24 V and 14 Ω respectively.
How to calculate resistor and voltage?Begin by finding the equivalent voltage and resistance of the circuit:
Since ε₂ is in the opposite direction, subtract it from the other two voltages:
∑ε = ε₁ + ε₃ - ε₂ = 12 V + 18 V - 6 V = 24 V
To find the equivalent resistance, first combine the 6 Ω and 12 Ω resistors in parallel:
1/Rp = 1/6 Ω + 1/12 Ω = 1/4 Ω
Rp = 4 Ω
Then combine the 4 Ω resistor in series with the 10 Ω resistor:
R = 4 Ω + 10 Ω = 14 Ω
Therefore, ∑ε = 24 V and ∑r = 14 Ω.
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A block on a horizontal surface is placed in contact with a light spring with spring constant k, as shown in Figure 1. When the block is moved to the left so that the spring is compressed a distance d from its equilibrium length, the potential energy stored in the spring-block system is Em . When a second block of mass 2m is placed on the same surface and the spring is compressed a distance 2d, as shown in Figure 2, how much potential energy is stored in the spring compared to the original potential energy Em ? All frictional forces are considered to be negligible.
The required potential energy stored in the spring-block system, when the second block is placed on the surface and the spring is compressed by twice the distance, is four times the original potential energy Em.
Let's denote the original potential energy when the spring is compressed by distance d as Em. When the spring is compressed, it exerts a restoring force given by Hooke's Law:
F = -kx,
Where F is the restoring force, k is the spring constant, and x is the displacement from the equilibrium position.
When the spring is compressed by distance d, the potential energy stored in the system is given by:
[tex]E_m = \dfrac{1}{2} kd^2[/tex]
Now, let's consider the situation when the second block of mass 2m is placed on the surface, and the spring is compressed by a distance 2d. Since the spring is compressed by twice the distance, the displacement is 2d. In this case, the potential energy stored in the system can be calculated as:
[tex]E_2 = \dfrac{1}{2} k((2d)^2) \\E_2= 4\times \dfrac{1}{2}k(d^2) \\E_2= 4E_m[/tex]
Therefore, the potential energy stored is four times the original potential energy Em.
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How long will be required for a car to go from a speed of 20. 0 m/s to a speed of 25. 0 m/s if the acceleration is 3. 0 m/s2?
It will take approximately 1.67 seconds for the car to go from a speed of 20.0 m/s to a speed of 25.0 m/s if the acceleration is 3.0 m/s².
[tex]v = u + at[/tex]
Given:
u = 20.0 m/s (initial velocity)
v = 25.0 m/s (final velocity)
a = 3.0 m/s² (acceleration)
Rearranging the equation, we get:
[tex]t= \frac{(v - u)}{a}[/tex]
Substituting the given values:
t = [tex]\frac{(25.0 m/s - 20.0 m/s)}{3.0 m/s^{2} }[/tex]
t = 1.67 s
Speed is the measure of the rate at which an object moves. It is defined as the distance an object travels per unit of time. Speed is a scalar quantity and is expressed in units of distance per unit of time, such as meters per second (m/s) or kilometers per hour (km/h). It is important to note that speed does not tell us anything about the direction of motion or any changes in direction. This is where velocity comes in, which is a vector quantity that includes both speed and direction.
Speed can be either average or instantaneous. Average speed is the total distance traveled divided by the total time taken, while instantaneous speed is the speed at a particular instant in time. Speed can also be described in terms of velocity, which includes direction as well as magnitude.
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Listed following are characteristics that can identify a planet as either terrestrial or jovian. Match these to the appropriate category. Consider only the planets of our own solar system.
Terrestrial planets are those within our Solar System composed of silicate rocks or metals, while Jovian planets are gas giants composed mainly of hydrogen and helium.
Terrestrial Planets:
- Small size
- High density
- Rocky or metallic composition
- Solid surfaces
Jovian Planets:
- Large size
- Low density
- Gas and liquid composition
- No solid surfaces
The four terrestrial planets in our Solar System are Mercury, Venus, Earth, and Mars. They have small sizes and high densities, and are composed mainly of silicate rocks or metals, with solid surfaces.
The four jovian planets are Jupiter, Saturn, Uranus, and Neptune. They have large sizes and low densities, and are composed mainly of hydrogen and helium gas and liquid. They do not have solid surfaces.
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Complete question:
identify a planet as either terrestrial or jovian. Match the planet to the appropriate category. Consider only the planets of our own solar system.
If five vectors were added tail-to-tip and they ended up where they started from, what would be the magnitude and direction of R?
When five vectors are added tail-to-tip and they end up where they started from, the magnitude and direction of R are zero.
What are vectors?
A vector is a quantity that has a magnitude (or length) and direction. It is typically represented by an arrow, with the length of the arrow representing the magnitude of the vector and the arrowhead indicating the direction in which the vector points. Vectors can be added or subtracted to create new vectors.
When vectors are added, they must be placed head-to-tail or tail-to-tip. In other words, the tail of one vector must be placed at the head of the previous vector to create a chain. The resulting vector is called the resultant vector and can be found by drawing a straight line from the tail of the first vector to the head of the final vector.
The magnitude of a vector is the length of the arrow that represents the vector. It is typically represented by the letter "R". So when five vectors are added tail-to-tip and they end up where they started from, the magnitude and direction of R are zero.
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a 200 ohm resistor is connected to a 12 v car battery. if the resistor is then removed and connected to a battery with 6 v or potential difference, what happens to the current passing through the resistor?
When a resistor is connected to a battery with a lower voltage, the amount of current flowing through it would decrease.
What does a potential difference of 6 volts mean?R is the resistance applied through the 6 volt cell. We obtain by replacing values. As a result, the potential difference across the 6V cell is 8.4 V. The voltage is the difference in potential between two sites in an electric field. In an electric field, the current is the movement of charges between two sites.
According to Ohm's Law
I = V/R
The current flowing through a 200 ohm resistor attached to a 12 V automotive battery is:
I = V/R
= 12 V / 200 ohms
= 0.06 A
Applying Ohm's Law once more, the new current would be:
I = V/R
= 6 V / 200 ohms
= 0.03 A
Therefore, When a resistor is connected to a battery with a lower voltage, the amount of current flowing through it would decrease.
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Complete the following ---
Area of piston A: 0.01m^2
Force applied on piston A: 6N
Pressure in the liquid: ?
Area of piston B: 0.1cm^2
Force produced by piston B: ?
(note: there's a difference of m^2 and cm^2 in the areas so it's difficult for me...)
Answer: The force produced by piston B is 0.006 N.
Explanation:
No problem! We can convert the area of piston B to square meters to make the units consistent:
Area of piston B = 0.1 cm^2 = 0.1 x (0.01 m/cm)^2 = 0.00001 m^2
Now we can use the formula:
pressure = force/area
For piston A, we have:
pressure = 6 N / 0.01 m^2 = 600 Pa
So the pressure in the liquid is 600 Pa.
To find the force produced by piston B, we rearrange the formula:
force = pressure x area
Using the pressure we just found and the area of piston B, we get:
force = 600 Pa x 0.00001 m^2 = 0.006 N
So the force produced by piston B is 0.006 N.
Which of the following best approximates the percentages of sand, clay, and silt in a silty loam? Use the soil texture table below to answer.(picture is at the bottom)Public DomainSand 10Clay 25Silt 65Sand 70Clay 10Silt 20Sand 20Clay 60Silt 20Sand 30Clay 10Silt 60'
The correct option iD. Sand 20% Clay 20% Silt 60% best approximates the percentages of sand, clay, and silt in a silty loam.
Soil texture is the roughness or softness of soil or soil particles. Soil texture can either be smooth/soft or rough soil texture. The soil texture table helps to determine the percentages of sand, clay, and silt in a silty loam. Among the given options, the best approximation for the percentages of sand, clay, and silt in a silty loam is 20% sand, 60% silt, and 20% clay. Therefore, the correct option for the question is option D. Sand 20% Clay 20% Silt 60%So, this is the answer to your question.
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At a major league baseball game, a pitcher delivers a 45 m/s (100.7 mph) fastball to the first player at bat, who bunts (meets the pitch with a loosely held stationary bat) so that the ball leaves the bat at only 5 m/s (11.2 mph) directly back towards the pitcher. The second player at bat also receives a 45 m/s fastball from the pitcher, but he swings his bat hard and sends the ball in a fast line drive directly back towards the pitcher at 50 m/s (111.8 mph). The mass of a standard baseball is 0.145 kg.
Calculate the impulse delivered to the baseball by the baseball bat for the first player (who bunts the ball). Assume the initial pitch is in the positive x-direction, and the ball moves in the negative x-direction after it strikes the bat.
Calculate the impulse delivered to the baseball by the baseball bat for the second player (who hits the fast line drive). Assume the initial pitch is in the positive x-direction, and the ball moves in the negative x-direction after it strikes the bat.
Calculate the magnitude of the work done by the baseball bat on the baseball for the first player (who bunts the ball). Report your answer as a positive number for positive work done on the ball or a negative number for negative work done on the ball.
Calculate the work done by the baseball bat on the baseball for the second player (who hits the fast line drive). Report your answer as a positive number for positive work done on the ball or a negative number for negative work done on the ball.
1) The impulse delivered to the baseball by the baseball bat is 40 kg-m/s.
2) The impulse delivered to the baseball by the baseball bat is 5 kg-m/s.
3) The magnitude of the work done by the baseball bat on the baseball for the first player is 1800 Joules.
4) The work done by the baseball bat on the baseball for the second player is 225 Joules.
The impulse delivered to the baseball by the baseball bat for the first player (who bunts the ball) can be calculated by subtracting the final velocity of the ball (5 m/s) from the initial velocity of the ball (45 m/s). The impulse delivered to the baseball by the baseball bat is 40 kg-m/s.
The impulse delivered to the baseball by the baseball bat for the second player (who hits the fast line drive) can be calculated by subtracting the final velocity of the ball (50 m/s) from the initial velocity of the ball (45 m/s). The impulse delivered to the baseball by the baseball bat is 5 kg-m/s.
The magnitude of the work done by the baseball bat on the baseball for the first player (who bunts the ball) can be calculated by multiplying the impulse (40 kg-m/s) by the initial velocity of the ball (45 m/s). The magnitude of the work done by the baseball bat on the baseball for the first player is 1800 Joules.
The work done by the baseball bat on the baseball for the second player (who hits the fast line drive) can be calculated by multiplying the impulse (5 kg-m/s) by the initial velocity of the ball (45 m/s). The work done by the baseball bat on the baseball for the second player is 225 Joules.
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the ball is initially accelerated downward by the gravitational force. when it reaches the floor, its quickly changes in direction, and the ball heads back upward.
When the ball is initially dropped, it is accelerated downward by gravitational force.
When the ball is released, it has potential energy that is transformed into kinetic energy as it accelerates downwards under the gravitational force.
At the moment the ball hits the ground, the kinetic energy is converted into elastic potential energy due to the compression of the ball's material.
As a result of this compression, the ball's motion is reversed, and the elastic potential energy is converted back into kinetic energy, which causes the ball to rise again.
This process of energy transformation continues until the ball reaches its maximum height, where its kinetic energy has been transformed back into potential energy.
Overall, the gravitational force plays a critical role in this process by providing the initial acceleration that allows the ball to fall toward the ground. Without this force, the ball would remain stationary in the air, unable to move in any direction.
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in the bohr model of the hydrogen atom, an electron in the lowest energy state moves at a speed of 2.19 106 m/s in a circular path having a radius of 5.29 1011 m. what is the effective current associated with this orbiting electron?
The effective current associated with an electron in the lowest energy state of the hydrogen atom in the Bohr model is 6.63×10-7 A.
To calculate the effective current, we can use the formula:
I = qv/T
where I is the effective current, q is the charge of the electron, v is its velocity, and T is the time period of its circular orbit.
In the lowest energy state of the hydrogen atom, the electron is in a circular orbit with a radius of 5.29×1011 m and a speed of 2.19×106 m/s. The time period of the orbit can be calculated using the formula for centripetal acceleration:
a = v^2/r
F = ma = (mv^2)/r
F = kQq/r^2
mv^2/r = kQq/r^2
T = 2pir/v
where F is the electrostatic force between the electron and the proton in the nucleus, k is Coulomb's constant, Q is the charge of the nucleus, q is the charge of the electron, m is the mass of the electron, and r is the radius of the orbit.
Substituting the given values, we get:
T = 2pi(5.29×10^(-11) m)/(2.19×10^6 m/s) = 2.42×10^(-16) s
Using the charge of an electron, q = -1.6×10^(-19) C, and the velocity calculated above, we get:
I = qv/T = (-1.6×10^(-19) C)(2.19×10^6 m/s)/(2.42×10^(-16) s) = -6.63×10^(-7) A
The negative sign indicates that the effective current is in the opposite direction of the electron's motion.
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because the direction of earth's motion around the sun continually changes during the year, the apparent position of a star in the sky moves in a small loop, known as the aberration of starlight. in order to better understand this phenomenon, it is sometimes helpful to use visual analogies. in these visual analogies, the car is analogous to the earth, and the rainfall is analogous to starlight. determine which visual analogies correspond to the following scenarios: a) the earth moving around the sun and interacting with light from a distant star b) a person on the moving earth observing the light from a distant star c) a person on a motionless earth observing the light from a distant star items (4 images) (drag and drop into the appropriate area below)
The appropriate visual analogies that correspond to the given scenarios are as follows:
A) The car traveling in a circle and the rain falling from the sky - this analogy corresponds to the Earth moving around the Sun and interacting with light from a distant star.
B) The car traveling in a straight line and the rain falling from the sky - this analogy corresponds to a person on the moving Earth observing the light from a distant star.
C) The car is stationary and the rain falls from the sky - this analogy corresponds to a person on a motionless Earth observing the light from a distant star.
What is a star?
As we know that the direction of the earth's motion around the sun continually changes during the year, and the apparent position of a star in the sky moves in a small loop, known as the aberration of starlight. Hence, the visual analogies that correspond to the given scenarios are as follows:'=
a) The Earth moving around the Sun and interacting with light from a distant star is analogous to the first picture, where the car is moving and it is raining. This visual analogy explains that when the Earth moves around the Sun and interacts with light from a distant star, it results in a small loop of light in the sky.
b) A person on the moving Earth observing the light from a distant star is analogous to the second picture, where a person is sitting inside the moving car and looking at the rain. This visual analogy explains that when a person is on the moving Earth and observes the light from a distant star, it creates an illusion in the sky.
c) A person on a motionless Earth observing the light from a distant star is analogous to the third picture, where a person is standing outside the car and looking at the rain. This visual analogy explains that when a person is on a motionless Earth and observes the light from a distant star, it appears as if the star is moving in a small loop in the sky.
Therefore, the appropriate visual analogies that correspond to the given scenarios are as follows: Image 1: The Earth moving around the Sun and interacting with light from a distant star image 2: A person on the moving Earth observing the light from a distant star image 3: A person on a motionless Earth observing the light from a distant star.
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two identical point charges q 7.20 x 10 6 c are fixed at diagonally opposite corners of a square with sides of length 0.480 m a negative test charge q0 2.40 x 10 8 c with a mass of 6.60 x 10 8 kg is released from rest at an empty corner of the square determine the speed of the test charge when it reaches the center of the square use conservation of energy
Therefore, the speed of the test charge when it reaches the center of the square is 50.14 m/s
To find the speed of the test charge when it reaches the center of the square, use conservation of energy. The initial potential energy (PE) of the test charge is 0 as it is released from rest. The total energy (E) of the test charge is conserved as it moves towards the center of the square. Therefore, the initial kinetic energy (KE) of the test charge must equal the PE of the test charge when it reaches the center of the square.
The PE of the test charge at the center of the square is the sum of the electrostatic potential energy between the test charge and the two point charges, which is given by:
PE = (q0)(q1)/(4πɛ0L) + (q0)(q2)/(4πɛ0L)
Where L is the length of the side of the square and q1 and q2 are the charges of the two point charges.
We can then calculate the initial Kinetic energy of the test charge using the formula:
KE = 1/2mv2
Where m is the mass of the test charge and v is the speed of the test charge. We can equate the PE and KE to find the speed of the test charge:
KE = PE
1/2mv2 = (q0)(q1)/(4πɛ0L) + (q0)(q2)/(4πɛ0L)
v2 = 2(q0)(q1)/(m4πɛ0L) + 2(q0)(q2)/(m4πɛ0L)
v = √(2(q0)(q1)/(m4πɛ0L) + 2(q0)(q2)/(m4πɛ0L))
Substituting the values given in the question, we get:
v = √(2(2.40 x 108 C)(7.20 x 106 C)/(6.60 x 108 kg x 4π x 8.85 x 10-12 C2/Nm2 x 0.480 m) + 2(2.40 x 108 C)(7.20 x 106 C)/(6.60 x 108 kg x 4π x 8.85 x 10-12 C2/Nm2 x 0.480 m))
v = 50.14 m/s
Therefore, the speed of the test charge when it reaches the center of the square is 50.14 m/s.
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Small aperture phone aperture stop doubly-telecentric full control over the light collection uniform across the entire field of view. furthermore it would allow resolution to be uniform across the same field of view.
True or False
The given statement "Small aperture phone aperture stop doubly-telecentric full control over the light collection uniform across the entire field of view. Furthermore it would allow resolution to be uniform across the same field of view" is False.
Aperture refers to the opening in a camera lens that regulates the amount of light that enters a camera. The camera aperture is determined by the size of the lens opening, which may be adjusted using a ring on the lens. The smaller the opening, the larger the f-number, which reduces the amount of light entering the camera.
Resolution refers to the ability of a lens to capture an image in detail. The resolution of an image is measured in pixels, and it is directly proportional to the number of pixels in the image. The resolution of an image is determined by the size of the camera sensor, the pixel density, and the quality of the lens.
Doubly telecentric lenses are those that have a telecentric entrance pupil, a telecentric exit pupil, or both. The telecentric entrance pupil implies that the aperture stop is located at the entrance pupil's location. The telecentric exit pupil implies that the pupil is located at the aperture stop's location. The doubly telecentric means that both the entrance and exit pupils are at the same location. As a result, the lens is able to provide sharp images with uniform resolution across the entire field of view.
The given statement, "Small aperture phone aperture stop doubly-telecentric full control over the light collection uniform across the entire field of view. Furthermore it would allow resolution to be uniform across the same field of view" is False.
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