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3 years ago

When a force is applied to an object for an amount of time, it is known as __________.

Physics

2 answers:

stiks02 [169]3 years ago

8 0

Answer:

Impulse

Explanation:

I took the test and got it correct.

Hope this helps! :)

NikAS [45]3 years ago

7 0

The correct answer to the question is : D) Impulse

EXPLANATION:

Before going to answer this question, first we have to understand impulse.

Impulse of a body is defined as change in momentum or the product of force with time.

Mathematically impulse = F × t = m ( v - u ).

Here, v is the final velocity

u is the initial velocity

F is the force acting on the body for time t.

Hence, the perfect answer of this question is impulse m i.e the force multiplied with time is known as impulse.

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Your car's speedometer works in much the same way as its odometer, except that it converts the angular speed of the wheels to a

brilliants [131]

Answer:

Speed will be 30810 rpm

Explanation:

We have given diameter of the tire d = 24 inch

So radius $r=\frac{d}{2}=\frac{24}{2}=12imch$

We have given linear velocity v = 35 mph

We know that linear velocity is given by $v=\omega r$

$35=\omega \times 12$

$\omega =\frac{35}{12}\times \frac{63360}{60}=3080rad/min$

As we know that 1 mile = 63360 inch and 1 hour = 60 min

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3 years ago

What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.

valkas [14]

<h2>Answer: 56.718 min</h2>

Explanation:

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

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

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

Where;

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

$M=6.39(10)^{23}kg$ is the mass of Mars

$a=3.4(10)^{6}m$ is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}$ (2)

$T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}$ (3)

$T=\sqrt{11581157.44 s^{2}}$ (4)

Finally:

$T=3403.1099s=56.718min$ This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0

3 years ago

A 60-kg rollerblader rolls 10 m down a 30? incline. When she reaches the level floor at the bottom, she applies the brakes. The

VARVARA [1.3K]

Answer:

s = 20 m

Explanation:

given,

mass of the roller blader = 60 Kg

length = 10 m

inclines at = 30°

coefficient of friction = 0.25

using conservation of energy

$\dfrac{1}{2}mu^2 = m g d sin \theta$

$u^2 = 2 g d sin30^0$

$u= \sqrt{2\times 9.8 \times 10 sin30^0}$

u = 9.89 m/s

Using second law of motion

ma =μ mg

a = μ g

a = 0.25 x 9.8

a = 2.45 m/s²

Using third equation of motion ,

v² - u² = 2 a s

0² - 9.89² = 2 x 2.45 x s

s = 20 m

the distance moved before stopping is 20 m

3 0

3 years ago

not all objects have a volume that is measured easily. If you were to determine the mass, volume, and density of your textbook,

Maslowich

Text book: We can measure the mass of the text book easily by weighing machine, to measure the volume we need to measure the length, width, and height of the text book by the ruler, by multiplying these dimension we can get the volume of the text book, and by dividing the mass of the book with its volume we can get the density of the book.

Milk Container: We can measure the mass of the milk container easily by weighing machine, now (assuming the milk container is cylindrical in shape) we need to measure its height, and and diameter and by the formula (π*r^2*h) we can measure its volume, and and by dividing the mass with its volume we can get the density of the milk container.

Air filled balloon: we can measure the mass of the air filled balloon by weighing it weight machine, we know that the density of air is 28.97 kg/m^3, by dividing the mass of the balloon with the denisty of air we can get the volume of the balloon.

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3 years ago

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