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3 years ago

The mass of the earth is 5.96/10kg^24. The radius of the earth is approximately 6.37x10^6 calculate the force of gravity

Physics

1 answer:

n200080 [17]3 years ago

3 0

<h2>Answer:g=9.79 $ms^{-2}$ ,A object of mass $m$ at the surface of earth experiences a force $mg$ </h2>

Explanation:

Let $M$ be the mass of earth.

Let $R$ be the radius of earth.

Let $G$ be the universal gravitational constant.

Given,

$M=5.96\times 10^{24}Kg$

$R=6.37\times 10^{6}m$

$G=6.67259 \times 10^{-11}$ $Nm^{2}Kg^{-2}$

Let $g$ be the acceleration due to gravity.

Then, $g=\dfrac{GM}{R^{2}}$

$g=\frac{6.67259 \times 10^{-11}\times 5.96\times 10^{24}}{(6.37\times 10^{6})^{2}}$

$g=9.79ms^{-2}$

A object of mass $m$ at the surface of earth experiences a force $mg$

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A 0.500 kg bullet is fired from a gun at 25.0 m/s, how much kinetic energy does it have?

slamgirl [31]

Considering the definition of kinetic energy, the bullet has a kinetic energy of 156.25 J.

<h3>Kinetic energy</h3>

Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and in a rest position, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless there is a change in speed or the body returns to its rest state by applying a force to it.

The kinetic energy is represented by the following expression:

Ec= ½ mv²

Where:

Ec is the kinetic energy, which is measured in Joules (J).
m is the mass measured in kilograms (kg).
v is the speed measured in meters over seconds (m/s).

<h3>Kinetic energy of a bullet</h3>

In this case, you know:

m= 0.500 kg
v= 25 m/s

Replacing in the definition of kinetic energy:

Ec= ½ ×0.500 kg× (25 m/s)²

Solving:

Finally, the bullet has a kinetic energy of 156.25 J.

Learn more about kinetic energy:

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5 0

1 year ago

A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.0 s after being hit. Then 2.5 s afte

vaieri [72.5K]

Answer:

Part a)

$H = 44.1 m$

Part b)

$y = 13.48 m$

Part c)

$d = 8.86 m$

Explanation:

Part a)

As we know that ball will reach at maximum height at

t = 3 s

now we will have

$t = \frac{v sin\theta}{g}$

now we have

$3 = \frac{vsin\theta}{9.8}$

$v sin\theta = 29.4 m/s$

Now maximum height above ground is given as

$H = \frac{v^2sin^2\theta}{2g}$

$H = \frac{29.4^2}{2(9.8)}$

$H = 44.1 m$

Part b)

Height of the fence is given as

$y = (vsin\theta) t - \frac{1}{2}gt^2$

$y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)$

$y = 13.48 m$

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

$x = v_x t$

$97.5 = v_x (5.5)$

$v_x = 17.72 m/s$

now total time of flight is given as

$T = 3 + 3 = 6 s$

so range is given as

$R = v_x T$

$R = (17.72)(6)$

$R = 106.4 m$

so the distance from the fence is given as

$d = 106.4 - 97.5$

$d = 8.86 m$

7 0

3 years ago

What happens to matter during chemical changes?

Harrizon [31]

The object that goes through chemical change, changes completely to where you can not change it back to its original form. Physical change you can undo

7 0

2 years ago

At the race track, one race car starts its engine with a resulting intensity level of 104.0 dB at point P. Then 6 more cars star

dem82 [27]

To solve the problem it is necessary to apply the concepts related to sound intensity. The most common approach to sound intensity measurement is to use the decibel scale:

$\beta (dB) = 10log_{10}(\frac{I}{I_0})$

Where,

$I_0 = 1*10^{-12}$ is a reference intensity. It is the lowest or threshold intensity of sound a person with normal hearing can perceive at a frequency of 1000 Hz.