Explanation:
1.
We use the equation
h = [tex]\frac{gt^2}{2}[/tex], where
h is the height traveled,
g is the acceleration due to gravity and
t is the time taken to reach height h.
We can now calculate t to be
[tex]\sqrt{\frac{2*1.2 m}{9.81 m/s^2} }[/tex]
= 0.495 s
Let v be the initial velocity of the player.
The player deaccelarates from v m/s to 0 m/s in 0.495 s at the rate of 9.81 m/s^2.
v = 9.81 m/s^2 x 0.495 s = 4.85 m/s
2.
The player takes 0.3 s to increase his velocity from 0 m/s to 4.85 m/s. So his average accelaration is
4.85 m/s / 0.3 s = 16.2 m/s^2
A boat moves up and down as water waves pass under the boat. If the amplitude of the wave gets bigger, then
A)
the boat will rise up higher.
B)
the boat will not rise up as high.
C)
the boat will go up and down more often.
D)
the boat will continue to move the same way.
Answer: The Boat will rise
Explanation: Because high amplitude means high in heights.
The (kinetic, radiant) theory explains the motion of particles in matter.
A)Kinetic
B)Radiant
Answer:
radiant
Explanation:
Answer:
kinetic
Explanation:
this is because the kinetic theory is when energy is used in motion.
The other name of eureka can
At an air show, a stunt pilot performs a vertical loop-the-loop in a circle of radius 3.63 x 103 m. During this performance the pilot whose weight is 676 N, maintains a constant speed of 2.25 x 102 m/s. At what speed, in m/s, will the pilot experience weightlessness
Answer:
189 m/s
Explanation:
The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.
So, F = W
mv²/r = mg
v² = gr
v = √gr where v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m
So, v = √gr
v = √(9.8 m/s² × 3.63 × 10³ m)
v = √(35.574 × 10³ m²/s²)
v = √(3.5574 × 10⁴ m²/s²)
v = 1.89 × 10² m/s
v = 189 m/s
A motorcycle passes over the top of a hill that has a radius of curvature of 100 m. The mass of the motorcycle plus rider is 300 kg. The motorcycle is moving at a speed of 30 m/s. The surface exerts a normal force of magnitude FN on the motorcycle.The motorcycle passes over the top of the hill again but now is moving at a speed of 33 m/s. How does the new normal force exerted on the motorcycle compare to FN
Answer:
[tex]F_N>F_N'[/tex]
Explanation:
From the question we are told that
Radius of curvature[tex]r=100m[/tex]
Mass [tex]M=300kg[/tex]
initial Speed of Motorcycle [tex]V_1=30m/s[/tex]
Final Speed of Motorcycle [tex]V_2=33m/s[/tex]
Generally the equation Force at initial speed is mathematically given as
[tex]F_N=mg-\frac{mv^2}{R}[/tex]
[tex]F_N=300*9.8-\frac{(300*30)^2}{100}[/tex]
[tex]F_N=240N[/tex]
Generally the equation Force at Final speed is mathematically given as
[tex]F_N'=mg-\frac{mv'^2}{R}[/tex]
[tex]F_N'=300*9.8-\frac{(300*33)^2}{100}[/tex]
[tex]F_N'=-327N[/tex]
Therefore
[tex]F_N>F_N'[/tex]
Following are the calculation to the given question:
Solution:
Using formula:
[tex]\to mg - F^{'}_N=\frac{mv^{2}}{R} \\[/tex]
Calculating the Initial value:
[tex]\to F_N = mg - \frac{mv^2}{R}\\\\[/tex]
[tex]= 300 \times (9.8 - \frac{(30)^2}{100}) \\\\= 300 \times (9.8 - \frac{900}{100}) \\\\= 300 \times (9.8 - 9) \\\\= 300 \times (0.8) \\\\ = 240\ N \\\\[/tex]
Calculating the Final value:
[tex]\to F^{'}_{N}= 300 (9.8 -\frac{33^2}{100})\\\\[/tex]
[tex]= 300 (9.8 -\frac{1089}{100})\\\\= 300 (9.8 - 10.89)\\\\= 300 (- 1.09)\\\\=-327[/tex]
Therefore, the answer is "the new normal force is less than [tex]F_{N}[/tex]" or [tex]\bold{F^{'}_{N}< F_{N}}[/tex].
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A truck travels on a straight road at a velocity of 17 meters per second. Over 20
seconds, it accelerates uniformly to 27 meters per second. What distance did the truck
travel during this acceleration?
Answer:
Distance, S = 440 meters.
Explanation:
Given the following data;
Initial velocity, u = 17m/s
Time, t = 20 seconds
Final velocity, v = 27m/s
To find the distance;
First of all, we would determine the acceleration of the truck.
Acceleration = (v-u)/t
Substituting the given values into the equation, we have;
Acceleration = (27 - 17)/20
Acceleration = 10/20
Acceleration = 0.5m/s²
Now, we would use the second equation of motion to find the distance traveled.
S = ut + ½at²
S = 17*20 + ½*0.5*20²
S = 340 + 0.25*400
S = 340 + 100
S = 440m
The equations of motion can be used to obtain the distance covered as 440 m.
We have to use of the equations that are used for uniformly accelerated motion in solving the problem. The chosen equation must be;
v^2 = u^2 + 2as and v = u + at
v = final velocity
u = initial velocity
a = acceleration
s = distance
To obtain the acceleration;
27 = 17 + 20(a)
27 - 17 = 20a
a = 0.5 ms-2
Now, to obtain the distance;
v^2 = u^2 + 2as
v^2 - u^2/as = s
s = (27)^2 - (17)^2/2(0.5)
s = 440 m
Learn more about the acceleration: https://brainly.com/question/12134554
A virtual image larger than the object can be produced by
1) concave mitor 2) convex mutor 3)plane mirror
concave lens
O1
Answer:
concave mirror
Explanation:
Why This is Correct Because, concave mirrors only form a virtual image when a object is larger than the other. when the object is produced between the focus, object and the image object, it becomes a virtual image.
Answer:
The answer is concave mirror
what is the definition of power ? what are the units of power?
Answer:
power is the rate at which energy is transferred or converted
the unit of power is Watts
The bond order for a single covalent bond is.
A. two
B. four
C. one
D. three
Answer:
I think it should be C, which is one
3. What is the acceleration of a cart with a F = 80 N and m = 32 kg? *
2.0 m/s^2
O 2.5 m/s^2
2.8 m/s^2
2.7 m/s^2
Answer:
2.5 m/s^2
Explanation:
Given the following data;
Force = 80N
Mass = 32kg
To find acceleration;
Force is given by the multiplication of mass and acceleration.
Mathematically, Force is;
[tex] F = ma[/tex]
Where;
F represents force.
m represents the mass of an object.
a represents acceleration.
Making acceleration (a) the subject, we have;
[tex]Acceleration (a) = \frac{F}{m}[/tex]
Substituting into the equation;
[tex]Acceleration (a) = \frac{80}{32}[/tex]
Acceleration = 2.5m/s²
A sports car of mass m has the same kinetic energy as an SUV with a mass 3m as each is driven along the same road. Which vehicle, if either, has the larger momentum and what is the difference in their momenta, if any
Answer:
Explanation:
Kinetic energy ( KE ) = 1/2 m v²
= m²v² / 2 m = p² / 2m where p is momentum
KE = p² / 2m
p² = 2m KE
KE is constant
p is proportional to mass
So car having higher mass will have higher momentum .
p₁ = √ ( 2 m x KE )
p₂ = √ ( 6 m x KE )
p₂ - p₁ = √ ( 6 m x KE ) - √ ( 2 m x KE )
= √KE m ( √6 - √2 )
Kinetic energy ( K.E )
[tex]= \frac{1}{2} m v^2\\\\= \frac{m^2 v^2}{2 m} \\\\= \frac{p^2}{2m}[/tex]
where p is momentum
[tex]K.E =\frac{p^2}{2m}\\\\p^2 = 2m. KE[/tex]
KE is constant
p is proportional to mass
So car having higher mass will have higher momentum .
[tex]p_1 =\sqrt{(2m*K.E)}\\\\p_2 = \sqrt{(6m*K.E)} \\\\p_2 - p_1 = \sqrt{(6m*K.E)} -\sqrt{(2M*K.E} \\\\p_2 - p_1 = \sqrt{K.E m(\sqrt{6}-\sqrt{2}) }[/tex]
The difference is shown above.
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Police driving with a velocity of 50 m/s decide to chase a speeder who is 3 km ahead and moving at 55 m/s. The police car accelerates at 2 m/s2. Instantly the speeder becomes aware that he is being chased and starts to accelerate at 1 m/s2. How much time (in s) passes until the police catch the speeder
Answer:
The time that passes until the police catch the speeder is 82.6204 seconds.
Explanation:
A body performs a uniformly accelerated rectilinear motion or uniformly varied rectilinear motion when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases its modulus in a uniform way.
The position is calculated by the expression:
x = x0 + v0*t + 1/2*a*t²
where:
x0 is the initial position. v0 is the initial velocity. a is the acceleration. t is the time interval in which the motion is studied.First, let’s look at the police car’s equations of motion. In this case:
x0= 0 v0= 50 m/s a= 2 m/s²So: x = 50 m/s*t + 1/2*2 m/s²*t²
Now for the speeder’s car’s equations of motion you know:
x0= 3 km= 3,000 m v0= 55 m/s a= 1 m/s²So: x = 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²
When the police catch the speeder they are both in the same position. So:
50 m/s*t + 1/2*2 m/s²*t²= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²
Solving:
0= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t² - 50 m/s*t - 1/2*2 m/s²*t²
0= 3,000 + 55 *t + 1/2*t² - 50*t - 1*t²
0= 3,000 + 55 *t - 50*t - 1*t² + 1/2*t²
0= 3,000 + 5*t - 1/2*t²
Applying the quadratic formula:
[tex]x1,x2=\frac{-5+-\sqrt{5^{2}-4*(-\frac{1}{2})*3000 } }{2*(-\frac{1}{2} )}[/tex]
x1= -72.6209
and x2= 82.6209
Since you are calculating the value of a time and it cannot be negative, then the time that passes until the police catch the speeder is 82.6204 seconds.
The discharge of a pump is 3 m above the inlet. Water enters at a pressure of 138 kPa and leaves at a pressure of 1380 kPa. The specific volume of the water is 0.001 m3/kg. If there is no heat transfer and no change in kinetic or internal energy, what is the work per unit mass
Answer:
The answer is "[tex]1.271 \ \frac{KJ}{kg}\\[/tex]"
Explanation:
[tex]\Delta e_{mech} =\frac{P_2-P_1}{P} + \frac{v_{2}^2-v_{1}^2}{2}+g(z_2-z_1)\\\\\Delta e_{mech} =\frac{ 1380 -138 \times 1000 }{1000} + 0+g(3-0)\\\\P = \frac{1}{v}= \frac{1}{0.001} = 1000 \frac{kg}{m} \\\\ \Delta e_{mech} =1242 +9.81(3)= 1271.43 \frac{J}{kg} \\\\\text{work per unit pass}= 1.271 \ \frac{KJ}{kg}\\[/tex]
diffraction of light is the __________ of light as it passes through the Edges of a barrier or a slit.
1, reflections
2, refraction
3,bending
4,absorbing
Answer:
3 bending
Explanation:
One of the fastest pitches ever thrown in Major League Baseball was by Aroldis Chapman and had a velocity of 105.1 miles/hour. How many seconds did it take this pitch to travel the 60 feet and 6 inches from the pitcher's mound to home plate
Answer:
t = 0.39 s
Explanation:
Assuming that the ball is launched horizontally, once in the air, if we neglect the resistance of the air, the ball moves at a constant speed, equal to the initial velocity, in this case, 105.1 mi/hr.In order to find time in seconds, it is advisable to convert the speed in mi/hr to m/s, as follows:[tex]105.1 mi/hr * (\frac{1hr}{3600s})*\frac{1609m}{1mi} = 47.0 m/s (1)[/tex]
In the same way, it's advisable to convert 60' 6'' (60.5') to m, as follows:[tex]60.5 ft * \frac{0.3058m}{1ft} = 18.5 m (2)[/tex]
Applying the definition of average velocity, we can find the time traveled by the ball from pitcher's mound to home plate, as follows:[tex]t = \frac{18.5m}{47m/s} = 0.39 s (3)[/tex]
Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece's weight. Assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a 12.0 mg piece of tape held 1.00 cm above another.
Answer:
36.1 nC
Explanation:
The electrostatic force F = kqq'/r² since q = q', F = kq²/r² where q = charge on tape, r = distance between tapes = 1.00 cm = 1 × 10⁻² m and k = 9 × 10⁹ Nm²/C².
Given that F = weight of 12.0 mg piece of tape, F = mg where m = mass of tape = 12.0 mg = 12 × 10⁻³ kg and g = acceleration due to gravity = 9.8 m/s²
So, kq²/r² = mg
q²/r² = mg/k
q² = mgr²/k
taking square-root of both sides,
q = √(mg/k)r
So, substituting the values of the variables into the equation, we have
q = √(mg/k)r
q = √(12 × 10⁻³ kg × 9.8 m/s²/ 9 × 10⁹ Nm²/C²)1 × 10⁻² m
q = √(117.6 × 10⁻³ kgm/s²/9 × 10⁹ Nm²/C²)1 × 10⁻² m
q = √(13.07 × 10⁻¹² C²/m²)1 × 10⁻² m
q = 3.61 × 10⁻⁶ C/m × 1 × 10⁻² m
q = 3.61 × 10⁻⁸ C
q = 36.1 × 10⁻⁹ C
q = 36.1 nC
An electric field is represented by the force on which particle?
a neutron
a dipole
a negative test charge
a positive test charge
Answer: a positive test charge
Explanation:
An electric field is a vector measure. Its direction is determined as the direction that a positive test charge would experience a force when it is placed in the field.
A positive charge has a higher potential and a negative charge is lower potential.
Thus, an electric field is represented by the force on a positive test charge.
Hence, the correct option is "a positive test charge".
Answer:
D. a positive test charge.
Explanation:
Edge 2021
Conductivities are often measured by comparing the resistance of a cell filled with the sample to its resistance when filled with some standard solution,such as aqueous potassium chloride. The conductivity of water is 76 mS m^(-1) at 25 C and the conductivity of 0.100 mol dm^(-3) KCl (aq) is 1.1639 S m^(-1) A cell has a resistance of 33.21ohm when filled with 0.100 mol dm^(-3) KCl (aq) and 300.0 ohm when filled with 0.100 mol dm CHaCOOH (aq). What is the molar conductivity of acetic acid at that concentration and temperature?
Answer:
1200 Sm^2mol^-1
Explanation:
Given data :
conductivity of water ( kwater ) = 76 mS m^-1 = 0.076 Sm^-1
conductivity of kcl (aq)( Kkcl ) = 1.1639 Sm^-1
Kkcl = 1.1639 - 0.076 = 1.0879 Sm^-1
Resistance = 33.21 Ω
where conductivity can be expressed as = [tex]\frac{Cell constant}{Resistance }[/tex]
hence cell constant = conductivity * Resistance
= 1.0879 * 33.21 = 36.13m^-1
conductivity of CH3COOH ( kCH3COOH ) = 36.13 / 300
= 0.120 Sm^-1
Determine the molar conductivity of acetic acid
= ( kCH3COOH * 1000 ) / C
C = 0.1 mol dm
= (0.120 * 1000) / 0.1 = 1200 Sm^2mol^-1
Elizabeth has always believed that people's thoughts can help heal them. She wants to help people use positive thinking to positively affect their
illnesses. What type of psychology would be MOST appropriate for Elizabeth to study?
Answer: Family
Explanation:
physical properties for soap
Answer:
there are germs in your hand and soap has chemicals that kill germs
Answer:
dino nuggies
Explanation:
cause whynot
A bald eagle is flying to the left with a speed of 34 meters
per second when a gust of wind blows back against the
eagle causing it to slow down with a constant acceleration
of a magnitude 8 meters per second squared.
What will the speed of the bald eagle be after the wind has
blown for 3 seconds?
Answer:
the speed after 3 seconds is 10 m/s
Explanation:
The computation of the speed is shown below:
As we know that
V = U + at
Here,
U = 34 m/s
a = - 8 m/s²
t = 3 Sec
V = velocity after 3 sec
V = 34 + (-8)3
= 34 - 24
V = 10 m/s
Hence, the speed after 3 seconds is 10 m/s
Can I get help on this question I’m scared to get it wrong .
acceleration greater than 1,000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.63 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.3 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. hardwood floor magnitude m/s2 hardwood floor duration ms carpeted floor magnitude m/s2 carpeted floor duration ms g
The primary reason for the path of motion of an object being a smooth curve is: Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a the third derivative of parabolas is always zero. b inertia. c tangent direction unit vectors change continuously. d calculus must have continuous derivatives to apply correctly.
Answer:
the correct answer is d
Explanation:
The laws of mechanics are related
F = m a
the acceleration of the body is given by the kinematics
a = [tex]\frac{dv}{dt}[/tex]
v = [tex]\frac{dx}{dt}[/tex]
substituting
a = \frac{d2x}{dt^2}
F = m [tex]\frac{d^2x}{dt^2}[/tex]
Therefore, in order to obtain the force (interaction of a body), continuous curves are needed and derivable from the position and the speed, for which all change in the trajectory of a body must be smooth where smooth is understood to have until the second derived.
Consequently the correct answer is d
Jack D. Ripper flipped out after missing a Must-Do-It question for the third time on his Minds On Physics assignment. Outraged by the futility of his efforts, he flings a 4.0-gram pencil across the room. The pencil lodges into a 221.0-gram Sponge Bob doll which is at rest on a countertop. Once in motion, the pencil/doll combination slide a distance of 11.9 cm across the countertop before stopping. The coefficient of friction between the doll and the countertop is 0.325. Determine the speed at which the pencil is moving prior to striking Sponge Bob.
Answer:
Explanation:
Let the velocity after the 4 gram pencil strikes is v .
kinetic energy of the combination = 1/2 m v²
= .5 x ( 4 + 221 ) x 10⁻³ x v² = work done by friction
friction force acting on the combination = 225 x10⁻³x .325 x 9.8 = .7166 N
work done by friction
= .7166 x .119 = .085 J
.5 x 225 x 10⁻³ v² = .085
v² = .085 / .1125 = .7555
v = .8692 m = 86.92 cm /s
Velocity of combination after collision = 86.92 cm /s
Let velocity of pencil before collision be V
Applying law of conservation of momentum at the time of collision ,
4 x V = 225 x 86.92
V = 4889.25 cm / s
= 48.9 m /s .
What is sin(77)?
Α. 0.77
Β. 4.33
Ο Ο Ο
C. 0.22
Ο D. 0.97
Answer:
0.97 id the correct answer
Explanation:
What genetic test would you get if there was a specific genetic disease in your family
A student weighing 5.4 × 102 newtons takes 15 seconds to run up a hill. The top of the hill is 10 meters vertically above her starting point. What power does the student develop during her run?
Answer:
P = 360 Watts
Explanation:
Given that,
The weight of a student, [tex]F=5.4\times 10^2\ N[/tex]
It takes 15 seconds to run up a hill.
The top of the hill is 10 meters vertically above her starting point.
We need to find the power develop during her run. We know that te power developed is given by :
[tex]P=\dfrac{W}{t}\\\\P=\dfrac{mgh}{t}\\\\P=\dfrac{5.4\times 10^2\times 10}{15}\\\\P=360\ W[/tex]
So, the power develop during her run is 360 W.
rocket fuel contains 50j of energy in its chemical store. the rocket has a mass of 1kg. what is the maximum speed the rocket could reach
____made up of glucose and fructose and found in plants.
1.Xylose
2.Maltose
3.Lactose
4.Sucrose
Answer:
1.Lactose
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