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Katen [24]
3 years ago
15

Newton’s laws do not apply to small objects?

Physics
1 answer:
Soloha48 [4]3 years ago
7 0

Answer:

Yes Newton's laws apply to small objects

EX: Newton s first law

when body at rest always want to be at rest

or body at motion always want to be at motion

unles an external force acts upon it

for example a eraser on the table will be at rest

if so e apply some force then it comes motion

so, Newton s law apply to small object s

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Five hundred joules of heat are added to a closed system. The initial internal energy of the system is 87 J, and the final inter
Aleonysh [2.5K]

We can solve the problem by using the first law of thermodynamics:

\Delta U= Q-W

where

\Delta U is the variation of internal energy of the system

Q is the heat added to the system

W is the work done by the system

In this problem, the variation of internal energy of the system is

\Delta U=U_f-U_i=134 J-87 J=47 J

While the heat added to the system is

Q=500 J

therefore, the work done by the system is

W=Q-\Delta U=500 J-47 J=453 J

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3 years ago
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Energy transformation in wound spring of a toy car?<br>give your own answrr​
BaLLatris [955]

Answer:

The work done in winding the spring gets stored in the wound up spring in the form of elastic potential energy (i.e potential energy due to change in shape). ... During this process, the potential energy stored in it gets converted to kinetic energy. This turns the wheels of the toy car.

Explanation:

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2 years ago
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Two children ride side-by-side on a carousel. Their paths are shown in the image below.
Sonbull [250]

Answer:

The child represented by a star on the outside path.

Explanation:

5 0
3 years ago
A simple pendulum has a period of 3.45 second, when the length of the pendulum is shortened by 1.0m, the period is 2.81 second c
den301095 [7]

Answer:

Original length = 2.97 m

Explanation:

Let the original length of the pendulum be 'L' m

Given:

Acceleration due to gravity (g) = 9.8 m/s²

Original time period of the pendulum (T) = 3.45 s

Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.

New length of the pendulum is, L_1=L-1

New time period of the pendulum is, T_1=2.81\ s

We know that, the time period of a simple pendulum of length 'L' is given as:

T=2\pi\sqrt{\frac{L}{g}}-------------- (1)

So, for the new length, the time period is given as:

T_1=2\pi\sqrt{\frac{L_1}{g}}------------ (2)

Squaring both the equations and then dividing them, we get:

\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1

Now, plug in the given values and calculate 'L'. This gives,

L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m

Therefore, the original length of the simple pendulum is 2.97 m

4 0
2 years ago
A 60 kg student in a rowboat on a still lake decides to dive off the back of the boat. The studen'ts horizontal aceleration is 2
TiliK225 [7]
As per the third law of Newton, the force exerted by the boat over the student is equal in magnitude to the force that the student exerted on the boat.

So, calculate the force on the student using the second law of Newton, Force = mass * acceleration.

Force on the student = 60 kg * 2.0 m/s^2 = 120 N.

=> horizontal force exerted by the student on the boat = 120 N

Answer: option d. 120 N. toward the back of the boat.

Of course it is toward the back because that is where the student jumped from..
4 0
3 years ago
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