A mass is suspended from each end of a rod of unequal lengths. The rod is balanced horizontally by adjusting an unknown mass. Two cases are shown.
When a mass is suspended from each end of a rod of unequal lengths, the rod will not remain horizontal unless an unknown mass is suspended at a specific point on the rod. This point can be determined by balancing the rod horizontally. The position of the unknown mass can be calculated using the principle of moments, which states that the sum of the moments acting on an object must be equal to zero for it to be in equilibrium. In this case, the moments due to the masses on each end of the rod and the unknown mass must be equal and opposite. The position of the unknown mass can then be calculated using the formula m = (l1/l2) * M, where m is the unknown mass, l1 and l2 are the lengths of the rod on either side of the pivot point, and M is the mass on one end of the rod.
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A kettle transfers 6,000j of energy electrically. 1,500j of this is wasted. what is the efficiency of this kettle?
Answer:75 percent
Explanation:so in order tro fin d thge efficiency i used the forumla ,efficency=useful output energy/input energyx100%,in order to use this formula i needed the output,which i found by subtracting the input energy with wasted energy,that gave me the output,and after founding the output,i put that into the formula,
output energy=input energy - wasted energy
output energy=6000j-1500j
output energy=4500
put that into the formula
efficiency =output energy/input enrgy x100%
efficiency=4500/6000 multiplied by 100%
efficiency=0.75x100%
efficiency=75%
What will be the pressure exerted by the object if 5,000 N of force
area of 200 cm²?
Answer:
[tex]250000 \frac{n}{m {}^{2} } [/tex]
Explanation:
pressure is the force applied perpendicular to the surface area of an object per unit area.
for the question,
Force (F) = 5000N
[tex]Area = 200cm {}^{2} = 0.02m {}^{2} \\ Pressure = \frac{force}{area} \\ Pressure = \frac{5000}{0.02} \\ Pressure = 250000 \frac{n}{m {}^{2} } [/tex]
A proton exits the cyclotron 1. 0 ms after starting its spiral trajectory in the center of the cyclotron. How many orbits does the proton complete during this 1. 0 ms ?.
The diameter of the largest orbit just before the protons exit the cyclotron is 39 cm.
The number of orbits completed by the proton during this 1.0 ms is 14000 revolutions.
The kinetic energy for the protons can be computed by using the formula:
K.E = 1/2mv²mv² = 2 K.E
v = sqrt( 2*KE / M)
the kinetic energy of the medical isotopes = 6.5 MeV
substituting the values,
v = sqrt( 2* 6.5* 1.6* 10^-13 / 1.67* 10^-27 )
v = 3.53 × 10⁷ m/s
The radius of the orbit can be estimated by using the formula:
mv² / R = qvB
r = q*v / mv²
r = ( 1.67* 10^-27 * 3.53 × 10⁷ ) / ( 1.9* 1.6* 10^-19 )
r = 0.19415 m
Since diameter (D) = 2r,
D= 2(0.19415 m)
D= 0.39 m
D≅ 39 cm
The time period to complete a revolution around the spiral trajectory is:
T = 2πr / v
T = 2*3.14* 0.1941 / 3.53*10^7
T = 0.7 × 10⁻⁷ s
Finally, the number of orbits that the proton does to complete the revolution in 1 ms is:
n = t / T
n = 10^-3 / (0.7*10^-7)
n = 14285.71
n ≅ 14000 revolutions
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Your question is incomplete, but most probably full question was:
A medical cyclotron used in the production of medical isotopes accelerates protons to 6.5 MeV. The magnetic field in the cyclotron is 1.9 T.
(a) What is the diameter of the largest orbit, just before the protons exit the cyclotron? Express your answer with the appropriate units. d = 57 cm Previous
(b) A proton exits the cyclotron 1.0 ms after starting its spiral trajectory in the center of the cyclotron. How many orbits does the proton complete during this 1.0 ms?
Your best friend weighs 81.5 kg and is a rugby player. In one of his games, he slides to a stop in a phenomenal manner. The coefficient of kinetic friction between the player and the ground is 0.70. His speed at the start of the slide is 8.23 m/s. I a) Calculate his acceleration during the slide. b) How long (in time) does he slide until he stops?
A. The acceleration during the slide is 6.86 m/s²
B. The time taken to slide until he stops is 1.2 s
How to determine the force of frictionMass (m) = 81.5 KgCoefficient of friction (μ) = 0.7Acceleration due to gravity (g) = 9.8 m/s²Normal reaction (N) = mg = 81.5 × 9.8 = 798.7 NFrictional force (F) =?F = μN
F = 0.7 × 798.7
F = 559.09 N
A. How to determine the accelerationMass (m) = 81.5 KgFrictional force (F) = 559.09 NAcceleration (a) =?a = F / m
a = 559.09 / 81.5
a = 6.86 m/s²
B. How to determine the time Initial velocity (u) = 8.23 m/sFinal velocity (v) = 0 m/sDecceleration (a) = -6.86 m/s²Time (t) =?a = (v – u) / t
t = (v – u) / a
t = (0 – 8.23) / -6.86
t = 1.2 s
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As the electric potential near the ground increases during a thunderstorm, a positive charge current can move up pointed objects, such as masts of ships, producing a luminous halo or glow known as ____.
Answer:
St. Elmo's Fire - well known to old time sailors
what is kinematics ?
explain ~
[tex] \\ \\ [/tex]
tysm! :)
Answer:
Kinematics is the study of the motion of mechanical points, bodies and systems without consideration of their associated physical properties and the forces acting on them. The study is often referred to as the geometry of motion, and it models these motions mathematically using algebra
Answer:
the branch of mechanics concerned with the motion of objects without reference to the forces which cause the motion is kinematics
A 10 kg box initially at rest is pulled with a 50 N horizontal force for 4 m across a level surface. The force of friction acting on the box is a constant 20 N. How much work is done by the 50 N force?
The work is done by the 50 N force is 200 J.
Work done by the 50 N forceWork done = Fd
where;
F is Force applied d is the displacement of the objectWork done = 50 N x 4 m = 200 J
Thus, the work is done by the 50 N force is 200 J.
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The pressure due to the atmosphere is very large, why are we not crushed?
Answer:
It is because of the blood pressure balances with our atmospheric pressure
Which type of galaxy is likely to contain only o-spectral type stars?
What force is large enough to change the direction of a planet?
A. Gravity
B. Friction
C.Tension
D. Air Resistance
Answer:
Gravity
Explanation:
The more massive the object is, the higher the gravity
Hence, a massive star, like our sun, has enough gravity to pull on planets and keeping them in orbit
Earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.
a) Determine the Earth’s average orbital speed expressed in kilometers per hours.
b) Based on the information given in this question, calculate the approximate mass of the Sun.
The Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x [tex]10^{24}[/tex] Kg
Relationship between Linear and angular speedLinear speed is the product of angular speed and the maximum displacement of the particle. That is,
V = Wr
Where
V = Linear speedW = Angular speedr = RadiusGiven that the earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.
a) To determine the Earth’s average orbital speed, we will make use of the below formula to calculate angular speed
W = 2[tex]\pi[/tex]/T
W = (2 x 3.143) / (365.26 x 24)
W = 6.283 / 876624
W = 7.2 x [tex]10^{-4}[/tex] Rad/hr
The Earth’s average orbital speed V = Wr
V = 7.2 x [tex]10^{-4}[/tex] x 149.6 x [tex]10^{6}[/tex]
V = 107225.5 kilometers per hours.
b) Based on the information given in this question, to calculate the approximate mass of the Sun, we will use Kepler's 3rd law
M = (4[tex]\pi ^{2}[/tex][tex]r^{3}[/tex]) / G[tex]T^{2}[/tex]
M = (4 x 9.8696 x 3.35 x [tex]10^{24}[/tex]) / (6.67 x [tex]10^{-11}[/tex] x 7.68 x [tex]10^{11}[/tex])
M = 1.32 x [tex]10^{26}[/tex] / 51.226
M = 2.58 x [tex]10^{24}[/tex] Kg
Therefore, the Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x [tex]10^{24}[/tex] Kg
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Two 5000-kg passenger cars roll without friction (one at 1 m/s, the other at 2 m/s) toward one another on a level track. They collide, couple, and roll away together with a combined momentum of
The combined momentum of the passengers is 5000 kgm/s.
Combined momentum of the passengerThe combined momentum of the passengers is calculated as follows;
P = mv1 + mv2
where;
m is mass of the passengersv1 is velocity of the first passengerv2 is velocity of the second passengerP = m(v1 + v2)
P = 5000(-1 + 2)
P = 5000 kgm/s
Thus, the combined momentum of the passengers is 5000 kgm/s.
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in the photoelectric effect a photon with an energy of 5.3 * 10^-19 J strikes an electron in a metal
The velocity of the photo electron is 6.11 × 105 ms
Given :
Supplied energy, Es = 5.3 x 10-19 J
Minimum energy of the electron to escape from the metal, E. = 3.6 x 10-19 J
To Find :
Velocity of photoelectron
Solution : The energy supplied by the photon is equal to the sum of minimum escape energy and the kinetic energy of the escaping electron. So
5.3 × 10^-19 J = 3.6 × 10^-19 J + K
K = 5.3 x 10^-19 - 3.6 x 10^-19
K = 1.7 × 10^-19 J
The formula of kinetic energy is given by:
K = 1/2 mv^2
v = √2K/m
= √2 x 1.7 x 10^-19
√ 9.1 x 10^-31
v = 6.11 x 10^5 m/s
So, the velocity of the photo electron is 6.11 x 10^5 m/s
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How much power will be required to force a current of 4.13 amps to flow through a conductor whose resistance is 113 ohms? Use two decimals for your answer. Round your answer to two decimals.
The power required to force the current of 4.13 A to flow through the conductor is 1927.43 watts
What is power?This is defined as the rate in which energy is consumed. Electrical power is expressed mathematically as:
Power (P) = square current (I²)× resistancet (R)
P = I²R
How to determine the powerCurrent (I) = 4.13 AResistance (R) = 113 ohmsPower (P) =?P = I²R
P = 4.13² × 113
P = 1927.43 watts
Thus, the power required is 1927.43 watts
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An engine has an input of heat energy of 10,750 J and does 2,420 J of work. Which of the following is the heat loss?
Group of answer choices
a. 13,200 J
b. 8,330 J
c. 4.44 J
c. 0.225 J
B. The heat loss by the heat engine is 8,330 J.
Heat loss by the heat engine
The heat loss by the heat engine is calculated as follows;
H = E - W
where;
H is heat lossE is input energyW is work doneH = 10,750 J - 2,420 J
H = 8,330 J
Thus, the heat loss by the heat engine is 8,330 J.
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The image above shows a crash test dummie's head traveling at -1.005 m/s while striking a headrest from a car traveling 4.524 m/s with a mass of 2005.6 kg. If the crash test dummie's head head bounces off the headrest with a 9.965 m/s and the car continues traveling at 4.487 m/s, calculate the mass of the crash test dummie's head?
Answer:
From the calculation, the mass of the dummies head is 1647.44 Kg
What is the mass of the crash test dummies head?We know that the momentum after collision is equal to the momentum before collision.
Mass of the headrest = 2005.6 kg
Initial velocity of the head rest = 4.524 m/s
Final velocity of the head rest = 4.487 m/s
Mass of the dummy = m
Initial velocity of the dummy = -1.005 m/s
Final velocity of the dummy = 9.965 m/s
Then;
(m * -1.005) + (2005.6 * 4.524 ) = (2005.6 * 4.487) + (m * 9.965)
-1.005m + 9073.33 = 8999.13 + 9.965m
9073.33 - 8999.13 = 9.965m + 1.005m
18072.46 = 10.97m
m = 18072.46/ 10.97
m = 1647.44 Kg
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A 15.2 kg mass has a gravitational potential energy of -342 J. How high from the ground is it? Group of answer choices GPE cannot be negative 2.3 m 22.5 m 530 m
A. The height above the ground reached by the mass is 2.3 m.
Height traveled by the mass
P.E = mgh
where;
m is mass h is height reached by the objectg is acceleration due to gravityh = P.E/mg
h = (342)/(15.2 x 9.8)
h = 2.3 m
Thus, the height above the ground reached by the mass is 2.3 m.
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question is down below
The vertical components of velocity is 10.35 m/s and the horizontal component of velocity is 38.6 m/s
What are the components of velocity?We know that velocity is a vector quantity, a vector often can be resolved into its components. The vertical components is V sinθ while the horizontal component is vcosθ.
Hence;
Vertical component = 40 m/s sin 15 degrees = 10.35 m/s
Horizontal component = 40 cos 15 degrees = 38.6 m/s
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1. A 500 g piece of silver at 250°C is submerged in 1000 g of water at 5°C to be cooled. Determine the
final temperature of the silver and water. Given Cwater = 4180 J/kg°C and Csilver = 240 J/kg°C.
Taking into account the definition of calorimetry, the final temperature of the silver and water is 11.84 °C.
CalorimetryCalorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).
So, the equation that allows to calculate heat exchanges is:
Q = c× m× ΔT
where:
Q is the heat exchanged by a body of mass m. c is the specific heat substance. ΔT is the temperature variation.Final temperature of the silver and waterIn this case, you know:
For silver:Mass of silver= 500 g= 0.5 kg (being 1000 g= 1 kg)Initial temperature of silver= 250 °CFinal temperature of silver= ?Specific heat of silver= 240 [tex]\frac{J}{kgC}[/tex] For water:Mass of water = 1000 g= 1 kgInitial temperature of water= 5 ºCFinal temperature of water= ?Specific heat of water = 4180 [tex]\frac{J}{kgC}[/tex]Replacing in the expression to calculate heat exchanges:
For silver: Qsilver= 240 [tex]\frac{J}{kgC}[/tex]× 0.5 kg× (Final temperature of silver - 250 C)
For water: Qwater= 4180 [tex]\frac{J}{kgC}[/tex] × 1 kg× (Final temperature of water - 5 C)
If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.
Then, the heat that the silver gives up will be equal to the heat that the water receives. Therefore:
- Qsilver = + Qwater
And the final temperature of the silver is equal to the temperature of the water (Final temperature of silver= Final temperature of water= Final temperature). Then:
- 240 [tex]\frac{J}{kgC}[/tex]× 0.5 kg× (Final temperature - 250 C)= 4180 [tex]\frac{J}{kgC}[/tex] × 1 kg× (Final temperature - 5 C)
Solving:
- 120 [tex]\frac{J}{C}[/tex]× (Final temperature - 250 C)= 4180 [tex]\frac{J}{C}[/tex]× (Final temperature - 5 C)
120 [tex]\frac{J}{C}[/tex]× (250 C - Final temperature) = 4180 [tex]\frac{J}{C}[/tex]× (Final temperature - 5 C)
120 [tex]\frac{J}{C}[/tex]×250 C - 120 [tex]\frac{J}{C}[/tex]×Final temperature = 4180 [tex]\frac{J}{C}[/tex]×Final temperature - 4180 [tex]\frac{J}{C}[/tex]×5 C
30,000 J - 120 [tex]\frac{J}{C}[/tex]×Final temperature = 4180 [tex]\frac{J}{C}[/tex]×Final temperature - 20,900 J
50,900 J= 4300 [tex]\frac{J}{C}[/tex]×Final temperature
50,900 J÷ 4300 [tex]\frac{J}{C}[/tex]= Final temperature
11.84 °C= Final temperature
Finally, the final temperature of the silver and water is 11.84 °C.
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What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 1.50×10^4v ?
Black smokers are hot volcanic vents that emit smoke deep in the ocean floor. Many of them teem with exotic creatures, and some biologists think that life on earth may have begun around such vents. The vents range in depth from about 1500 m to 3200 m below the surface. What is the gauge pressure at a 2452-m deep vent, assuming that the density of water does not vary
If the density of water does not vary and the vents range in depth from about 1500 m to 3200 m below the surface, then the gauge pressure at a 2452-m deep vent is 224.268 atm.
Calculation:
Step-1:
It is given that the vents range in depth from about 1500 m to 3200 m below the surface. If we are assuming that the density of water does not vary. Then it is required to calculate the gauge pressure at a 2452-m deep vent.
The gauge pressure at a particular depth of ocean water is calculated as:
[tex]$$P=\rho g h$$[/tex]
Here [tex]\rho[/tex] is the density of water, P is the required pressure, h is the depth of water, and g is the gravitational acceleration.
Step-2:
Now we are substituting the values to calculate the pressure at the depth of 2452-m.
[tex]$$\\\begin{aligned}\\P&=\rho gh\\&=1030 (\text{ kg/m}^3)\times 9.8 (\text{ m/s}^2)\times 2452 \text{ m}\\&=24.75\times 10^6 \text{ Pa}\times\frac{1 \text{ atm}}{10.1325 \times10^4 \text{ Pa}}\\&=224.268 \text{ atm}\\\end{aligned}\\$$[/tex]
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When operating an aircraft at cabin pressure altitudes above 12,500 feet msl up to and including 14,000 feet msl, supplemental oxygen shall be used during?
Supplemental oxygen shall be used in that flight time in excess of 30 minutes at those altitudes.
Most plane cabins are pressurized to 8,000 feet above sea level, an altitude that lowers the amount of oxygen inside the blood by approximately 4 percentage points, researchers say.
The higher the altitude, the less oxygen there is in the air and the lower the overall air pressure is. If flights were not pressurized, passengers could be prone to various physiological illnesses. Because of this, federal policies require that all commercial flights over 8,000 feet be pressurized.
The flight crew must use supplemental oxygen for the entire duration of flight operations above a cabin pressure altitude of 14,000 feet MSL.
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A mountain skier has begun descending the 30o slope. if the coefficient of kinetic friction is 0.10, what is her acceleration?
The acceleration of the body is found to be -4.05 m/s^2
What is the acceleration?We define the acceleration as the rate of change of the velocity with time. Hence, we can write that;
a = (μcosθ - sinθ)g
a = acceleration
μ = coefficient of kinetic friction
θ = angle of inclination
g = acceleration due to gravity
Thus;
a = (0.10cos 30 - sin 30) 9.8
a = -4.05 m/s^2
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20. Q: How long will it take for an apple falling from a 29.4m-tall tree to hit the ground?
A: 1.56 s
B: 2.04 s
C: 2.45 s
D: 3.72 s
Answer:
2.45s
Explanation:
is explanation needed too?
John is talking to his friend about painting a fence. He states that if he uses a roller brush he will cover the fence faster than if he uses a bristle brush. This is an example of a a. Law b. Belief c. Theory d. Hypothesis
Answer:
b
Explanation:
a pulley is assumed massless and frictionless and rotates freely about its axle. the blocks have masses m1=40g and a block m2= 20g, and block m1 is pulled to the right by a horizontal force magnitude f=0.03n. find the magnitude of acceleration of a block m2 and the tension in the cord.....?
The magnitude of acceleration of a block m₂ is 0.05 m/s² and the tension in the cord is 0.01 N.
Given:
mass of block 1, m₁ = 40 gm = 40×10⁻³ kg
mass of block 2, m₂ = 20 gm = 20×10⁻³ kg
Applied force, F = 0.03 N
Calculation:
Consider the free-body diagram of the system as shown below. Using Newton's second law of motion we get:
F = ma
where F is the applied force
m is the total mass of the system
a is the acceleration of block 2 (as it is pulled by horizontal force)
From the above equation we get:
0.03 N = (m₁+m₂) a
⇒ a = (0.03 N) / (m₁+m₂)
⇒ a = (0.03 N) / (40×10⁻³ kg + 20×10⁻³ kg)
⇒ a = 0.5 m/s²
Now, from the free-body diagram of block 2 as shown in figure 3, we get:
Balancing the forces along the horizontal:
∑Fₓ = 0
∴ T = m₂ a
where T is tension in the string
a is the acceleration of block 2
Applying values in the above equation we get:
T = (20×10⁻³ kg) × (0.5 m/s²)
= 0.01 N
Therefore, the acceleration of block 2 due to the applied horizontal force is 0.5 m/s² and the tension in the cord is 0.01 N.
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A fireworks shell is accelerated from rest to a velocity of 60.0 m/s over a distance of 0.200 m. How long did the acceleration last
The acceleration lasted for 0.007s.
To find the answer, we need to know about the Newton's equation of motion.
What's the Newton's equation of motion that relates initial velocity, final velocity, acceleration, time and distance?Newton's equations are
V²-U²= 2aSV²-U²= 2aSV= U+atV= final velocity, U= initial velocity, a= acceleration, t= time and S = distance
What's the acceleration, if the fireshell is accelerated from rest to a velocity of 60.0 m/s over a distance of 0.200 m?Here, U= 0m/s, V= 60.0 m/s, S= 0.200 m60²-0= 2a×0.2= 0.4a
a= 3600/0.4= 9000m/s²What's the time taken by the fireshell to achieve 60m/s with 9000m/s² acceleration from rest?Here, U= 0m/s, V = 60.0 m/s, a= 9000m/s²60= 0+9000tt= 60/9000= 0.007 sThus, we can conclude that the acceleration of the fireshell lasted for 0.007s.
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A velocity-time graph is shown below:
The average acceleration in the first 10 seconds of the journey is 2 m/s²; option B.
What is acceleration?Acceleration is the rate of change of velocity with time.
Acceleration = change in velocity/timeFrom the graph;
Acceleration in first seconds = 20 - 0/5 = 4 m/s²
Acceleration in second 5 seconds = 0
Average acceleration = 4 + 0/2
Average acceleration = 2 m/s²
In conclusion, average acceleration is the average of the final and initial acceleration.
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How is a rainbow formed?
Answer:
easy
Explanation:
it when the sunlight hits a rain droplets ,some of the light is reflected .The electromagnetic spectrum is made of light with many different wavelengths and each is reflected at a different angle . Thus spectrum is seperated producing a rainbow
Based on illustrations of magnetic field lines, where could an object be placed so it would not experience a magnetic force
Halfway between the like poles of two magnets, because the field lines bend away and do not enter this area.
How does a magnetic field diagram show where the field is strongest?
The magnetic field lines do not ever cross. The lines include arrowheads to indicate the direction of the force exerted by a magnetic north pole. The closer the lines are to the poles, the stronger the magnetic field (thus the magnetic field from a bar magnet is highest closest to the poles).Where is magnetic field the strongest and weakest on a magnet?
The bar magnet's magnetic field is strongest at its core and weakest between its two poles. The magnetic field lines are densest immediately outside the bar magnet and least dense in the core.Which two locations on the magnet would have the greatest attractive forces?
Inside the magnet itself, the field lines run from the south pole to the north pole. The magnetic field is strongest in areas of greatest density of magnetic field lines, or areas of the greatest magnetic flux density.Learn more about magnetic field
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