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velikii [3]
3 years ago
15

A car (mass of 830 kg) is sitting on a car lift in a shop (neglect the mass of the lift itself). While the car is being lifted u

p, it is speeding up with 3.8 m/s2. What is the magnitude of the lifting force?
Physics
1 answer:
OLEGan [10]3 years ago
3 0
There are two force acting on an object that is being lifted. (1) the weight of the car, (2) the upward force. The difference of these force should be equal to the product of the mass and the acceleration. (This is the content of Newton's 2nd Law of Motion). If we let the lifting force be F,
                       F - (830)(9.8) = (830)(3.8)
The value of F from the equation is 11288 N. 
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Hypsometer (upper fixed point​
aivan3 [116]

Answer:a device for calibrating thermometers at the boiling point of water at a known height above sea level or for estimating height above sea level by finding the temperature at which water boils.

Explanation:

3 0
3 years ago
The net force acting on a 5 kg discus is 50 n. what is the acceleration of the discus
Lyrx [107]
Answer: 10 m/s^2

Explanation:

1) The second law of Newton gives the definition and formula to calculate the net force:

Net force acting on an object = mass * acceleration.

2) From that, when you know the net force acting of the object and its mass, you can solve for the acceleration:

acceleration = Net force / mass

acceleration = 50 N / 5 kg = 10 m/s^2, which is the answer.
8 0
3 years ago
A muon is a short-lived particle. It is found experimentally that muonsmoving at speed 0.9ccan travel about 1500 meters between
Monica [59]

Answer:

a)t=5.5\mu s

b)\Delta t'=12.5\mu s

Explanation:

a) Let's use the constant velocity equation:

v=\frac{\Delta x}{\Delta t}

  • v is the speed of the muon. 0.9*c
  • c is the speed of light 3*10⁸ m/s

t=\frac{\Delta x}{v}=\frac{1500}{0.9*(3*10^{8})}

t=5.5\mu s

b) Here we need to use Lorentz factor because the speed of the muon is relativistic. Hence the time in the rest frame is the product of the Lorentz factor times the time in the inertial frame.

\Delta t'=\gamma\Delta t

\gamma =\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

v is the speed of muon (0.9c)

Therefore the time in the rest frame will be:

\Delta t'=\frac{1}{\sqrt{1-\frac{(0.9c)^{2}}{c^{2}}}}\Delta t

\Delta t'=\frac{1}{\sqrt{1-0.9^{2}}}\Delta t

\Delta t'=\frac{1}{0.44}\Delta t

No we use the value of Δt calculated in a)

\Delta t'=2.27*5.5=12.5\mu s

I hope it helps you!

     

 

8 0
3 years ago
Sam and his Dad went grocery shopping. As they went up and down the aisle, adding items to their cart, something changed. The ca
Amanda [17]
B)
The speed of the cart changed because it stopped. 

Hope I could help!
-Marshy
8 0
3 years ago
Read 2 more answers
Check b and c plz :DD
Inga [223]
Both of your answers are correct
4 0
2 years ago
Read 2 more answers
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