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ratelena [41]
3 years ago
15

An object has a mass of 2,000kg. what is its weight on earth? show your work​

Physics
2 answers:
artcher [175]3 years ago
6 0

Answer:

F=m*g is the formula and the answer is 19,620 kg

Explanation:

Since the formula is F=m*g and Earth's gravity is 9.81 m/s^2 all you need to do is multiply 2,000 by 9.81

kotegsom [21]3 years ago
4 0

Answer:

20,000\:\mathrm{N}

Explanation:

The weight of an object on Earth is equal to the force of gravity on that object.

We can use the following equation to determine the force of gravity on an object:

F_g=ma_g=mg, where m is the mass of the object and g is acceleration due to gravity (9.81\: \mathrm{m/s}).

Solving, we get:F_g=2,000\cdot9.81=19,620=\fbox{$20,000\:\mathrm{N}$} (one significant figure).

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x and y

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A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball
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Answer:

Part a)

T = 2\sqrt{\frac{R}{3g}}

Part b)

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Part c)

v_y = \sqrt{Rg/3}

Part d)

v = \frac{1}{2}\sqrt{13Rg}

Part e)

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H = \frac{13R}{8}

Part g)

X = \frac{13R}{4}

Explanation:

Initial speed of the launch is given as

initial speed = v_i

angle = \theta_i degree

Now the two components of the velocity

v_x = v_i cos\theta_i

similarly we have

v_y = v_i sin\theta_i

Part a)

Now we know that horizontal range is given as

R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}

maximum height is given as

H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}

so we have

v_i sin\theta = \sqrt{Rg/3}

time of flight is given as

T = \frac{2v_isin\theta_i}{g}

T = \frac{2\sqrt{Rg/3}}{g}

T = 2\sqrt{\frac{R}{3g}}

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

R = v_x T

R = v_x 2\sqrt{\frac{R}{3g}}

v_x = \frac{\sqrt{3Rg}}{2}

Part c)

Initial vertical velocity is given as

v_y = v_i sin\theta_i

v_i sin\theta = \sqrt{Rg/3}

Part d)

Initial speed is given as

v = \sqrt{v_x^2 + v_y^2}

so we will have

v = \sqrt{Rg/3 + 3Rg/4}

v = \frac{1}{2}\sqrt{13Rg}

Part e)

Angle of projection is given as

tan\theta_i = \frac{v_y}{v_x}

tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}

\theta_i = 33.7 degree

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

H = \frac{v^2}{2g}

H = \frac{13R}{8}

Part g)

For maximum range the angle should be 45 degree

so maximum range is

X = \frac{v^2}{g}

X = \frac{13R}{4}

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3 years ago
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