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LiRa [457]
4 years ago
6

A bag of sugar weighs 5.00 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration is one-sixt

h that on Earth? Repeat for Jupiter, where g is 2.64 times that on Earth. Find the mass of the bag of sugar in kilograms at each of the three locations.
Physics
1 answer:
gizmo_the_mogwai [7]4 years ago
3 0

Answer:

Earth: 22.246 N

Moon: 3.71 N

Jupiter: 58.72 N

Explanation:

The mass of an object will remain constant in any location, its weight however, can fluctuate depending on its location. For example, a golf ball will weigh less on the moon, but its mass will not be different if it was on earth.

To calculate anything, we need to convert to standard measurements.

5.00 lbs = 2.27 kg

On earth, gravity is measured to be 9.8 m/s², so the weight in Newtons on Earth would be: (2.27 kg) x (9.8 m/s²) = 22.246 N

Repeated on the moon where gravity is (9.8 m/s²) x (1/6) = 1.633 m/s², so the weight in Newtons on the moon would be: (2.27 kg) x (1.633 m/s²) = 3.71 N

Repeated on Jupiter where gravity is (9.8 m/s²) x (2.64) = 25.87 m/s², so the wight in Newtons on Jupiter would be: (2.27 kg) x (25.87 m/s²) = 58.72 N

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3 years ago
If there is a net force on an object in the direction it is already moving, what has to happen to the speed of the object?
baherus [9]
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Either the object may speed up or slow down.
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4 years ago
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Explanation:

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3 years ago
A solid cylinder of uniform density of 0.85 g/cm3 floats in a glass of water tinted light blue by food coloring.
SVEN [57.7K]

Each side has to have at least 44 horses

F61160 N. This is further explained below.

<h3>What is the force?</h3>

Generally, We are only interested in the component that operates horizontally since the vertical components all cancel each other out. The pressure difference works on the hemisphere to generate a normal force all over the surface, but we are only concerned with that force's horizontal component. This may be determined by supposing the hemispheres to be two flat circular plates of the same radius as the hemispheres that have been forced together.

Therefore, force is equal to pressure multiplied by area, which is

F= (970 -15 )( * (0.45 m)2)

F=60754 N for each side.

Therefore, each side has to have at least 44 horses

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2 years ago
At highway speeds, a particular automobile is capable of an acceleration of about 2.0 m/s2 . Part A At this rate, how long does
Natasha_Volkova [10]

Answer:

Time taken, t = 4.86 seconds

Explanation:

Given that,

Acceleration of a particular automobile, a=2\ m/s^2

Initial speed of the automobile, u = 75 km/h = 20.83 m/s

Final speed of the automobile, v = 110 km/h = 30.55 m/s

We need to find the time taken to accelerate from u to v. Let t is the time taken. It can be calculate as :

t=\dfrac{v-u}{a}

t=\dfrac{30.55-20.83}{2}

t = 4.86 seconds

So, the time taken by the automobile is 4.86 seconds. Hence, this is the required solution.

5 0
3 years ago
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