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3 years ago

5

The acceleration due to gravity on the moon is 1.6 m/s2. What is the gravitational potential energy of a 1200-kg lander resting

on top of a 350-m hill?

2 answers:

Monica [59]3 years ago

8 0

Gravitational potential energy<span> is </span>energy<span> an object possesses because of its position in a </span>gravitational<span> field. </span><span>The equation for </span>gravitational potential energy<span> is GPE = mgh.

GPE = 1200(1.6)(350) = 672000 J

Hope this answers the question. Have a nice day.</span>

Gala2k [10]3 years ago

8 0

the second box would be 6..........

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san4es73 [151]

Answer:

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2 years ago

Which method is used to deliver heat in most central heating systems?

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3 years ago

Read 2 more answers

A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve

Gemiola [76]

Answer:

(a) v = 5.42m/s

(b) vo = 4.64m/s

(c) a = 2874.28m/s^2

(d) Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

$v=\sqrt{2gh}$ (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

$v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}$

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

$h_{max}=\frac{v_o^2}{2g}$ (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

$v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}$

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}$

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}$

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

$\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m$

THe compression of the ball when it strikes the floor is 5.11*10^-3m

4 0

3 years ago

A 18.0-kg rock is sliding on a rough, horizontal surface at 7.10 m/s and eventually stops due to friction. the coefficient of ki

Bond [772]

A = .3*g = 2.94 m/s²

<span>t = v/a = 9/2.94 = 3.061 sec </span>

<span>W = E/t = ½mv²/t = ½*40*9²/3.061 = 529.2 watts</span>

4 0

3 years ago

A bird is about 6.26.2 in. long, with a thin, dark bill and a wide, white wing stripe. If the bird can fly 9292 mi with the w

Trava [24]

Answer:

$209$ mph

Explanation:

$V$ = Speed of bird in still air

$v$ = Speed of wind = 44 mph

Consider the motion of the bird with the wind

$D_{1}$ = distance traveled with the wind = 9292 mi

$t_{1}$ = time taken to travel the distance with wind

Time taken to travel the distance with wind is given as

$t_{1} = \frac{D_{1}}{V + v}$

$t_{1} = \frac{9292}{V + 44}$ eq-1

Consider the motion of the bird with the wind

$D_{2}$ = distance traveled against the wind = 6060 mi

$t_{2}$ = time taken to travel the distance against wind

Time taken to travel the distance against wind is given as

$t_{2} = \frac{D_{2}}{V + v}$

$t_{2} = \frac{6060}{V - 44}$ eq-2

As per the question,

Time taken with the wind = Time taken against the wind

$t_{1} = t_{2}$

$\frac{9292}{V + 44} = \frac{6060}{V - 44}$

$(9292) (V - 44) = (6060) (V + 44)$

$9292V - 408848 = 6060V + 266640$

$3232V = 675488$

$V = 209$ mph

5 0

3 years ago