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jeka57 [31]
3 years ago
12

Which is the best example of Newton's Second Law of Motion? (1 point) Select one: a. A student has on roller skates and she deci

des to push against the railing. She immediately begins to roll backwards. b. A small, lightweight ball and a large, heavy ball are dropped off of a roof. They both strike the ground at the same time. c. A baseball player hits a baseball that is pitched to him. The ball immediately soars back in the direction of the pitcher. d. You are riding in a car that makes a quick right turn. You immediately slide across the back seat.
Physics
1 answer:
slava [35]3 years ago
4 0
D, I believe is the correct answer. Hope I answered your question, have a good day.<span />
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When you push with a horizontal 15-N force on a book that slides at constant
Marat540 [252]

Answer:

f = 15 N

Explanation:

It is given that, when you push with a horizontal 15-N force on a book that slides at constant  velocity, a frictional force also acts on it. Frictional force is an opposing force. The magnitude of applied force and frictional forces are same. So, the force of friction on the book is equal to 15 N.

8 0
3 years ago
Match the traits to the ogransims
k0ka [10]
Need more than that to answer this question
4 0
3 years ago
Suppose a 60-kg boy and a 41-kg girl use a massless rope in a tug-of-war on an icy, resistance-free surface. If the acceleration
Digiron [165]

Answer:

Approximately 2.05\; {\rm m\cdot s^{-2}}.

Explanation:

The net force on the girl would be:

\begin{aligned}m(\text{girl}) \, a(\text{girl}) &= 41\; {\rm kg} \times 3.0\; {\rm m\cdot s^{-2}} \\ &= 123.0\; {\rm N} \end{aligned}.

Under the assumptions, the net force on this girl would be equal to the tension force in the rope. All other forces on the girl would be balanced.

In other words, the tension force that the rope exerted on the girl would be 123.0\; {\rm N}. The girl would exert a reaction force on the rope at the same magnitude (123.0\; {\rm N}\!) in the opposite direction. This force would translate to a 123.0\; {\rm N}\!\! force on the boy towards the girl.

Under similar assumptions, the net force on the boy would also be 123.0\; {\rm N}. Since the mass of the boy is m(\text{boy}) = 60\; {\rm kg}, the acceleration of the boy would be:

\begin{aligned}a(\text{boy}) &= \frac{(\text{net force})}{m(\text{boy})} \\ &= \frac{123.0\; {\rm N}}{60\; {\rm kg}} \\ &= 2.05\; {\rm m\cdot s^{-2}}\end{aligned}.

3 0
2 years ago
A proton moves with a speed of 0. 985c. (a) calculate its rest energy
mojhsa [17]

The rest energy of the proton will be 938 MeV. The rest energy is related to the rest mass of the proton.

<h3>What is rest energy?</h3>

The rest energy is the rest mass times the square of the speed of light of a particle at rest in an inertial frame of reference.

The rest energy is found as;

\rm E_R= mC^2\\\\  E_R=(1.67 \times 10^{-27})\times 3 \times 10^8)^2\\\\ E_R= 9.375 \times 10^8 / eV

Convert into the mega electron volt;

\rm E_R= 938 \  MeV

Hence, the rest energy of the proton will be 938 MeV

To learn more about the rest energy, refer to the link;

brainly.com/question/15047418

#SPJ4

4 0
2 years ago
Prove that, s=u+v/2<br>please I need help ​
Hoochie [10]

Answer:

Actually the formulae is S=(u+v/2)t

.TO explain,from average velocity=distance /time

and average velocity is u+v/2

hence using the formular

Average velocity=distance/time

distance will be=average velocity ×time

so S=(u+v/2)t

5 0
3 years ago
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