Ask question

Ask question

All categories

3 years ago

11

Impulse is defined as the average force acting on an object multiplied by the time the force acts. If we let I represent impulse

, F the average force, and t the time, is I = F/t a correct way of expressing this definition? Explain

1 answer:

Dmitrij [34]3 years ago

7 0

Answer:

Impulse, $J=F\times t$

Explanation:

The impulse applied on an object is defined as the force acting on it for a short duration of time. It is given by :

$J=F\times t$ ...........(1)

Here

F is the applied force

t is the time taken

In this case the given expression, $I=\dfrac{F}{t}$ is incorrect. The correct expression is given in equation (1). Hence, this is the required solution.

You might be interested in

The mass of the earth is 5.98 1024 kg, the mass of the moon is 7.36 1022 kg, and the distance between the centers of the earth a

Setler [38]

Answer:

x_{cm} = 4.644 10⁶ m

Explanation:

The center of mass is given by the equation

$x_{cm}$ = 1 / $M_{total}$ ∑ $x_{i} m_{i}$

Where M_{total} is the total masses of the system, $x_{i}$ is the distance between the particles and $m_{i}$ is the masses of each body

Let's apply this equation to our problem

M = Me + m

M = 5.98 10²⁴ + 7.36 10²²

M = 605.36 10²² kg

Let's locate a reference system located in the center of the Earth

Let's calculate

x_{cm} = 1 / 605.36 10²² [Me 0 + 7.36 10²² 3.82 10⁸]

x_{cm} = 4.644 10⁶ m

4 0

3 years ago

Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W

jenyasd209 [6]

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

$T^{2}=\frac{4\pi^{2}}{GM}r^{3}$ (1)

Where;

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24}kg$ is the mass of the Earth

$r$ is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period $T_{X}$ is:

$T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}$ (2)

Where $r_{X}=1.2(10)^{6}m$

$T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}$ (3)

$T_{X}=413.712 s$ (4)

For satellite Y, the orbital period $T_{Y}$ is:

$T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}$ (5)

Where $r_{Y}=1.9(10)^{5}m$

$T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}$ (6)

$T_{Y}=26.064 s$ (7)

This means $T_{X}>T_{Y}$

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

$V_{X}=\sqrt{\frac{GM}{r_{X}}}$ (8)

$V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}$

$V_{X}=18224.783 m/s$ (9)

<u>For Satellite Y:</u>

$V_{Y}=\sqrt{\frac{GM}{r_{Y}}}$ (10)

$V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}$

$V_{Y}= 45801.13 m/s$ (11)

This means $V_{Y}>V_{X}$

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

4 0

3 years ago

Read 2 more answers

What is causing the boat to move toward the shore?

ivann1987 [24]

Answer:c

Explanation:because when u push the oar back u go forward

4 0

3 years ago

Read 2 more answers

Is a measure of waves that passes a point in a given amount of time

GREYUIT [131]

FREQUENCY is the number of complete waves that pass a given point in a certain amount of time.

Good luck :)

8 0

3 years ago

The period of an ocean wave is 5 seconds. What is the wave's frequency?

Oksana_A [137]

Answer:

f=1/5= 0.2 Hz

..,...........

7 0

3 years ago