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mart [117]
3 years ago
13

Gravity on Earth is 9.8 m/s^2, and gravity on Jupiter is 23.1 m/s^2. So, if the mass of a rock is 70 kilograms, it's weight on E

arth is____newtons and its weight on Jupiter is____newtons?
Physics
2 answers:
Levart [38]3 years ago
3 0
~686newtons on earth and
~1617 newtons on jupiter
the formula is weight = gravitational acceleration * mass of the object
prohojiy [21]3 years ago
3 0

Answer:

686 N, 1617 N

Explanation:

Gravity on earth, ge = 9.8 m/s^2

Gravity on Jupiter, gj = 23.1 m/s^2

m = 70 kg

Weight on earth, We = m x ge = 70 x 9.8 = 686 N

Weight on Jupiter, Wj = m x gj = 70 x 23.1 = 1617 N

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How tightly does mass need to be compacted in order to become a black hole??? (2 words)
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is it infinite density?

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A steel tank of weight 600 lb is to be accelerated straight upward at a rate of 1.5 ft/sec2. Knowing the magnitude of the force
VikaD [51]

Answer:

a) the values of the angle α is 45.5°

b) the required magnitude of the vertical force, F is 41 lb

Explanation:

Applying the free equilibrium equation along x-direction

from the diagram

we say

∑Fₓ = 0

Pcosα - 425cos30° = 0

525cosα - 368.06 = 0

cosα = 368.06/525

cosα = 0.701

α = cos⁻¹ (0.701)

α = 45.5°

Also Applying the force equation of motion along y-direction

∑Fₓ = ma

Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)

525sin45.5° + F + 212.5 - 600 = 27.95

374.46 + F + 212.5 - 600 = 27.95

F - 13.04 = 27.95

F = 27.95 + 13.04

F = 40.99 ≈ 41 lb

8 0
3 years ago
Protons and neutrons grouped in a specific pattern
alexgriva [62]
Answer b protons and electrons
5 0
3 years ago
a body weights 28N at a height of 3200km from the earth surface.What will be the gravitational force on that body if its lies on
alekssr [168]

Answer:

The object would weight 63 N on the Earth surface

Explanation:

We can use the general expression for the gravitational force between two objects to solve this problem, considering that in both cases, the mass of the Earth is the same. Notice as well that we know the gravitational force (weight) of the object at 3200 km from the Earth surface, which is (3200 + 6400 = 9600 km) from the center of the Earth:

F_G=G\,\frac{M_E\,m}{d^2} \\28\,\,N=G\,\frac{M_E\,m}{9600000^2}

Now, if the body is on the surface of the Earth, its weight (w) would be:

F_G=G\,\frac{M_E\,m}{d^2} \\w=G\,\frac{M_E\,m}{6400000^2}

Now we can divide term by term the two equations above, to cancel out common factors and end up with a simple proportion:

\frac{w}{28} =\frac{9600000^2}{6400000^2} \\\frac{w}{28} =\frac{9}{4} \\\\ \\w=\frac{9\,*\,28}{4}\,\,\,N\\w=63\,\,N \\

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3 years ago
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