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3 years ago

A graph of velocity versus time (velocity displayed in m/s and time in seconds) displays a

Physics

1 answer:

xxMikexx [17]3 years ago

5 0

Answer: acceleration is 2m/s^2

initial velocity is 4m/s

Explanation:

Since slope is rise/ run in this graph it will be change in velocity/ change and time. Therefore the slope of the graph will tell you the acceleration is 2m/s^2. The y intercepts is the velocity then time is 0, therefore the initial velocity is 4 m/s

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If the distance between two objects decreased, what would happen to the force of gravity between them?

evablogger [386]

Answer: A- It would increase

Explanation:

According to the law of universal gravitation:

$F=G\frac{m_{1}m_{2}}{r^2}$

Where:

$F$ is the module of the attraction force exerted between both objects

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both objects

$r$ is the distance between both objects

As we can see, the gravity force is directly proportional to the mass of the bodies or objects and inversely proportional to the square of the distance that separates them.

In other words:

<h2>If we decrease the distance between both objects, the gravitational force between them will increase. </h2>

6 0

3 years ago

Read 2 more answers

Two satellites are in circular orbits around a planet that has radius 9.00x10^6 m. One satellite has mass 53.0 kg , orbital radi

Andreas93 [3]

Answer:

The orbital speed of this second satellite is 5195.16 m/s.

Explanation:

Given that,

Orbital radius of first satellite $r_{1}= 8.20\times10^{7}$

Orbital radius of second satellite $r_{2}=7.00\times10^{7}\ m$

Mass of first satellite $m_{1}=53.0\ kg$

Mass of second satellite $m_{2}=54.0\ kg$

Orbital speed of first satellite = 4800 m/s

We need to calculate the orbital speed of this second satellite

Using formula of orbital speed

$v=\sqrt{\dfrac{GM}{r}}$

From this relation,

$v_{1}\propto\dfrac{1}{\sqrt{r}}$

Now, $\dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}$

$v_{2}=v_{1}\times\sqrt{\dfrac{r_{1}}{r_{2}}}$

Put the value into the formula

$v_{2}=4800\times\sqrt{\dfrac{ 8.20\times10^{7}}{7.00\times10^{7}}}$

$v_{2}=5195.16\ m/s$

Hence, The orbital speed of this second satellite is 5195.16 m/s.

4 0

3 years ago

A flywheel of mass 182 kg has an effective radius of 0.62 m (assume the mass is concentrated along a circumference located at th

musickatia [10]

Answer:

A)5524J,

B) 29.2Nm

Explanation:

This question can be treated using work- energy theorem

Work= change in Kinectic energy

W= Δ KE

Work= difference between the final Kinectic energy and intial Kinectic energy.

We know that

Kinectic energy= 1/2 mv^2 .............eqn(1)

This can be written in term of angular velocity, as

KE= 1/2 I

4 0

2 years ago

A box slides down a frictionless incline, gaining speed. The work done by the normal force n is _______.

jeka57 [31]

The work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

<h3>What is normal force?</h3>

The force of contact is called the normal force. When the two surfaces are in contact with each other, then the normal force acts.

This force is applied by the solid bodies on each other in order to prevent the passing through each other.

A box slides down a frictionless incline, gaining speed. For this box, the value of work done by normal force has to be found out. Let's analyze the given condition.

The body is gaining the speed, which means there is a change in kinetic energy.
The change in kinetic energy is equal to the work done.
The friction force is the product of coefficient of the friction and normal force.
The friction force for the given case is zero. Thus, the normal force must be equal to the zero.

Thus, the work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

Learn more about the normal force here;

brainly.com/question/10941832

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2 years ago

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The current through a heater is 12 A when it is plugged into a

jasenka [17]

Answer:

A. 10Ω

Explanation:

7 0

3 years ago