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Aleksandr [31]
3 years ago
13

A 6 kg weight is lifted off the ground to a height that gives it 70.56 j of gravitational potential energy. what is its height?

acceleration due to gravity is g = 9.8 m/s2.
Physics
2 answers:
Levart [38]3 years ago
8 0

Answer:

Height, h = 1.2 meters

Explanation:

It is given that,

Mass of an object, m = 6 kg

Gravitational potential energy, E = 70.56 J

It is lifted off a ground to a height h. We need to find its height. It can be determined using the expression for the potential energy i.e.

E = m g h

g = acceleration due to gravity

h=\dfrac{E}{mg}

h=\dfrac{70.56\ J}{6\ kg\times 9.8\ m/s^2}

h = 1.2 meters

Hence, it is lifted off the ground to a height of 1.2 meters.

Marina CMI [18]3 years ago
5 0
GPE= 70.56 J -------------------> GPE= mgh-------------> X= height
70.56 = 6(kg) * 9.8(m/s/s) * X
70.56 = 58.8X
70.56/58.8= 58.8X/58.8
X= 1.2
The height is 1.2 feet or meters (whatever unit you are using in this problem)

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Answer:

9.4\cdot 10^{10} m

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

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T_e=365 d is the orbital period the Earth

Solving the equation for r_o, we find

r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m

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3 years ago
Suppose the coefficient of kinetic friction between mA and the plane in the figure(Figure 1) is μk = 0.15, and that mA=mB=2.7kg.
luda_lava [24]
A ) 
T = mB g + mB a
T + mA a - mA g sin 35° = (Mi) mA g cos 35°
------------------------------------------------------------
T = 2.7 · 9.81  + 2.7 a
T = 26.487 + 2.7 a
26.487 + 2.7 a + 2.7 a - 2.7 · 9.81 · 0.574 = 0.15 · 2.7 · 9.81 · 0.819
5.4 a + 26.487 - 15.2023 = 3.2539
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a = 1.487 ≈ 1.5 m/s²
B )
T = 2,7 · 9.81 = 26.487 
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On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = go at the north pole
ohaa [14]
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I attached a diagram that shows how this force aligns with the force of gravity.
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