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3 years ago

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As an object in motion becomes heavier, its kinetic energy _____. A. increases exponentially B. decreases exponentially C. incre

ases proportionally D. decreases proportionally

2 answers:

diamong [38]3 years ago

4 0

Answer:

USE SOCRACTIC IT WOULD REALLY HELP

Aleksandr-060686 [28]3 years ago

3 0

The answer is c. as an object is in motion speeds up or “becomes heavier”, it’s kinetic energy increases proportionally: double the velocity and you quadruple the kinetic energy. this is why a tiny bullet traveling at high speed does so much more damage than a huge truck bumping into something at 1 mph. so the answer is c

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What is the difference between speed and velocity?

ElenaW [278]

B - speed and direction are combined in another quantity, called velocity. It can be thought of as its speed in a particular direction. Speed is the corresponding scalar quantity, because it does not have a direction.

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2 years ago

Identifying Advantages of Parallel Circuits

melisa1 [442]

Answer: If one bulb goes out the other bulbs stay lit.

If there is a break in one branch of the circuit, current can still flow through the other branches.

Explanation:

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2 years ago

Read 2 more answers

A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h

Yuki888 [10]

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

$v_{1}=\dfrac{3.0\times10^{3}}{3600}$

$v_{1}=0.833\ m/s$

The force F₁is constant acceleration is also a constant.

$F_{1}=ma_{1}$

We need to calculate the acceleration

Using formula of acceleration

$a_{1}=\dfrac{v}{t}$

$a_{1}=\dfrac{0.833}{10}$

$a_{1}=0.083\ m/s^2$

Similarly,

$F_{2}=ma_{2}$

For total force,

$F_{3}=F_{2}+F_{1}$

$ma_{3}=ma_{2}+ma_{1}$

The speed of second tugboat is

$v=\dfrac{11\times10^{3}}{3600}$

$v=3.05\ m/s$

We need to calculate total acceleration

$a_{3}=\dfrac{v}{t}$

$a_{3}=\dfrac{3.05}{10}$

$a_{3}=0.305\ m/s^2$

We need to calculate the acceleration a₂

$0.305=a_{2}+0.083$

$a_{2}=0.305-0.083$

$a_{2}=0.222\ m/s^2$

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

$\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}$

$\dfrac{F_{1}}{F_{2}}=3.7$

$F_{1}=3.7F_{2}$

Hence, The magnitude of F₁ is 3.7 times of F₂

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2 years ago

A 0.140-kg baseball is dropped from rest. It has a speed of 1.20 m/s just before it hits the ground, and it rebounds with an upw

irga5000 [103]

Answer:

22 N upward

Explanation:

From the question,

Applying newton's second law of motion

F = m(v-u)/t....................... Equation 1

Where F = Average force exerted by the ground on the ball, m = mass of the baseball, v = final velocity, u = initial velocity, t = time of contact

Note: Let upward be negative and downward be positive

Given: m = 0.14 kg, v = -1.00 m/s, u = 1.2 m/s, t = 0.014 s

Substitute into equation 1

F = 0.14(-1-1.2)/0.014

F = 0.14(-2.2)/0.014

F = 10(-2.2)

F = -22 N

Note the negative sign shows that the force act upward

6 0

3 years ago

What’s the answer ?????

Mice21 [21]

Answer:

<h2>Refer the attachment for answer and explanation please</h2>

Explanation:

This might surely help you ☺️❤️

7 0

2 years ago