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Serhud [2]
3 years ago
10

Consider the same 70kg/686N student on the surface on another planet from the table above: Jupiter. Tompared to the gravitationa

l force or weight on Earth. How would we expect the student's weight to change? Explain. A) Because the mass of Jupiter is about 300 times greater than Earth's we would expect the student's weight to be 205,800N. B) Both the mass and the diameter of Jupiter is greater than Earth's. Considering both, the student's Earth weight would be multiplied by 3, or 2058N. Because Jupiter is considered a gas giant, rather than a rocky solid planet, there is no gravity and we could not calculate the student's new weight. D) Because the distance to Jupiter is about 500 million km farther from the Sun than Earth, we can expect the weight of the student to be 686N/5 or 137.2N​
Physics
1 answer:
Yuliya22 [10]3 years ago
6 0

Answer:

Dont mind me

Explanation:

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An object with a mass of 5kg accelerates at 2m/s2. How much force in Newtons(N) is needed to cause this to happen??
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Answer: The formula of Newtons second law of motion is F=MA so therefore it would be written like this Force = Mass X Acceleration

F = 5 x 2

F = 10 N

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3 years ago
Two horses on opposite sides of a narrow stream are pulling a barge up the stream. Each hors pulls with a force of 720N. The rop
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Fx = Fcos(∅)
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8 0
3 years ago
What is the wavelength of violet light that has a frequency of 7.5x10^14
Crank
Λ= V/f 
<span>but change it to represent the speed of light, c </span>
<span>λ= c/f </span>
<span>c = 3.00 x 10^8 m/s </span>
<span>Plug in your given info and solve for λ(wavelength) </span>
<span>λ= 3.00 x 10^8 m/s / 7.5 x 10^14 Hz
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6 0
3 years ago
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A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo
Talja [164]

Answer:

T = 16.72 N

Explanation:

When the object is swung in a circular path, and in a vertical plane, there are two forces external to the object acting on it at any time: the gravity (which is always downward) and the tension in the string (which always points towards the center of the circle).

At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac =  v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

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3 years ago
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0.049

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