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2 years ago

3. An astronaut travels to a far-away moon with a 1.5 m long pendulum. She finds it takes

1 answer:

5 0

Have you tried to look this up on Quizlet. I just had a similar question on one on my quizzes & the whole test was on Quizlet

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If you went to a planet that had the twice the radius as Earth, but the same mass, a 1 kg pineapple would have a weight of

kicyunya [14]

Use the law of universal gravitation, which says the force of gravitation between two bodies of mass m₁ and m₂ a distance r apart is

F = G m₁ m₂ / r²

where G = 6.67 x 10⁻¹¹ N m²/kg².

The Earth has a radius of about 6371 km = 6.371 x 10⁶ m (large enough for a pineapple on the surface of the earth to have an effective distance from the center of the Earth to be equal to this radius), and a mass of about 5.97 x 10²⁴ kg, so the force of gravitation between the pineapple and the Earth is

F = (6.67 x 10⁻¹¹ N m²/kg²) (1 kg) (5.97 x 10²⁴ kg) / (6.371 x 10⁶ m)²

F ≈ 9.81 N

Notice that this is roughly equal to the weight of the pineapple on Earth, (1 kg)g, where g = 9.80 m/s² is the magnitude of the acceleration due to gravity, so that [force of gravity] = [weight] on any given planet.

This means that on this new planet with twice the radius of Earth, the pineapple would have a weight of

F = G m₁ m₂ / (2r)² = 1/4 G m₁ m₂ / r²

i.e. 1/4 of the weight on Earth, which would be about 2.45 N.

7 0

3 years ago

Pls help my science teacher is a horrible teacher...

Anettt [7]

First of all, looks like your teacher is indeed pretty horrible. Secondly, the constraints to consider would be proper weight distribution, methods to minimize excessive motion of the building structure, and quantities such as volume and density, which would help in determining the optimal structure. Keeping the frequency of oscillation for a building low in case of an earthquake or natural disaster would also be a priority.

4 0

3 years ago

If the amplitude of a wave is increased the frequency of the wave will

Travka [436]

Answer:

Remain the same

Explanation:

There is no relationship between amplitude frequency.

3 0

3 years ago

What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from

ElenaW [278]

When is at the end of the runway the velocity of the plane is given by the equation $vf^{2}=0+2*a*s$ where s=1800 m is the runway length. Thus
$vf^{2}=2*5*1800=18000 (m/s)^{2}$
$vf =134.164 (m/s)$

At half runway the velocity of the plane is
$v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}
$
$v= \sqrt{9000}=94.87 ( \frac{m}{s})$

Therefore at midpoint of runway the percentage of takeoff velocity is
‰ $P= \frac{v}{vf}= \frac{94.87}{134.164}=0.707$

6 0

3 years ago

A parachutist of mass 56.0 kg jumps out of a balloon at a height of 1400 m and lands on the ground with a speed of 5.10 m/s. How

Vlad [161]

Answer:

769,048.28Joules

Explanation:

A parachutist of mass 56.0 kg jumps out of a balloon at a height of 1400 m and lands on the ground with a speed of 5.10 m/s. How much energy was lost to air friction during this bump

The energy lost due to friction is expressed using the formula;

Energy lost = Potential Energy + Kinetic Energy

Energy lost = mgh + 1/2mv²

m is the mass

g is the acceleration due to gravity

h is the height

v is the speed

Substitute the given values into the formula;

Energy lost = 56(9.8)(1400) + 1/2(56)(5.10)²

Energy lost = 768,320 + 728.28

Energy lost = 769,048.28Joules

Hence the amount of energy that was lost to air friction during this jump is 769,048.28Joules

6 0

2 years ago