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kicyunya [14]
3 years ago
7

A student pushes on a crate with the force of 100 Newtons directed to the right. What force does the crate exert on the student?

Note: It has to be done with the Newton's 2nd Law of Motion.
Physics
1 answer:
ale4655 [162]3 years ago
6 0

Answer: The Force exerted on the student by crate will be 100 N.

Explanation: As per the given Question student is trying to push the crate in right direction with the force of 100N.

And we know that, Newton's third law states that every action has equal and opposite reaction.

So from the Newton's 3rd law of motion it is very clear the student must be experiencing the same amount of force which he is applying.

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What must you know to find the amount of work done on an object
Cerrena [4.2K]

Answer:

The work is calculated by multiplying the force by the amount of movement of an object (W = F * d). A force of 10 newtons, that moves an object 3 meters, does 30 n-m of work. A newton-meter is the same thing as a joule, so the units for work are the same as those for energy – joules.

Explanation:

5 0
3 years ago
Read 2 more answers
A baseball hits a car, breaking its window and triggering its alarm which sounds at a frequency of 1210 Hz. What frequency (in H
IRINA_888 [86]

Answer:

The frequency of sound heard by the boy is 1181 Hz.

Explanation:

Given that,

Frequency of sound from alarm  f_{0} = 1210\ Hz

Speed = -8.25 m/s

Negative sign show the boy riding away from the car

Speed of sound = 343

We need to calculate the heard frequency

Using formula of frequency

f = f_{0}(\dfrac{v+v_{0}}{v-v_{s}})

Where, f_{0} = frequency of source

v_{0} = speed of observer

v_{s} = speed of source

v = speed of sound

Put the value into the formula

f=1210\times\dfrac{343+(-8.25)}{343-0}

here, source is at rest

f=1180.8\ Hz

f=1181\ Hz

Hence, The frequency of sound heard by the boy is 1181 Hz.

8 0
3 years ago
Which of the following is used in Sonata II's prepared piano?
Delvig [45]
It is C. Copper wire.
8 0
3 years ago
If μs is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. calculate this
fiasKO [112]

weight = mg acts downwards <span>
normal force = N acts upwards. 
and force F acts at an angle θ below the horizontal. 
(Let us assume that the woman pushes from the left, so F is acted towards the right, which is below the horizontal) 
so that, Frictional force, f=us*N acts towards the left 

Now we balance the forces along x and y directions: 
y direction: N = mg + F sinΘ 
x direction: us * N = F cosΘ 

We let the value of µs be equal to a value such that any F will not be able to move the crate. Then, if we increase F by an amount F', then the force pushing the crate towards the right also increases by F' cosΘ. Additionally, the frictional force f must raise by exactly this amount. 
Since f can’t exceed us*N, so the normal force must increase by F' cosΘ/us. 
Also, from the y direction equation, the normal force exceeds by F' sin Θ. 

<span>These two values must be the same, therefore:
<span>us = cot θ</span></span></span>

4 0
3 years ago
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