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3 years ago

The ???? ,of an object is a vector that is directed toward the center of the Earth.

1 answer:

5 0

The Weight is a vector whose magnitude is the product of the mass m of the object and the magnitude of the local gravitational acceleration. Its always directed toward the center of the Earth.

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You have just landed on Planet X. You take out a ball of mass 101 g , release it from rest from a height of 16.0 m and measure t

IRINA_888 [86]

Answer:

0.3817 N

Explanation:

Remark

One thing is certain: the ball has a mass of 101 grams wherever it is in the universe. That is not true of the force. The force on the moon is a whole lot less than it is on earth, and maybe planet x as well.

Givens

m = 101 g

vi = 0 That's what at rest means.

t = 2.91 s

d = 16 m

F= ?

Formulas

d = vi*t + 1/2*a * t^2

Force = m * a

Solution

16 = 0 + 1/2 a * 2.91^2

16 = 4.234 a Divide by 4.234

16/4.234 = a

a = 3.779

F = m * a

a = 3.779

m = 101 g = 1 kg / 1000 grams

m = 0.101 kg

F = 0.101 * 3.779

F = 0.3817N

8 0

3 years ago

Answer the question below please I give brainliest

Basile [38]

Answer: The first one

Explanation: I think it's the first one because it says what is the "least" gravitational potential energy story between the prairie dog and Earth that said resting in its borrow is using less energy

7 0

3 years ago

Read 2 more answers

I cant solve this problem, and our teacher said that this would be in the test we'll have tomorrow, can someone help me?

Ad libitum [116K]

Answer:

d = 11.1 m

Explanation:

Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:

$\frac{1}{2} m {v}^{2} = mgh$

Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:

$\sin(15) = \frac{h}{d} \\ or \: h = d \sin(15)$

Plugging this into the energy conservation equation and cancelling m, we get

${v}^{2} = 2gd \sin(15)$

Solving for d,

$d = \frac{ {v}^{2} }{2g \sin(15) } = \frac{ {(7.5 \: \frac{m}{s}) }^{2} }{2(9.8 \: \frac{m}{ {s}^{2} })(0.259)} \\ = 11.1 \: m$

3 0

3 years ago

a 2,000-kilogram railroad car moving at 8m/s to the right collides with a 6,000-kilogram railroad car moving at 3m/s to the west

astra-53 [7]

A freight car of mass 20,000 kg moves along a frictionless level railroad track ... After the push the skateboarder II moves with a velocity of 2 m/s to ... After the collision the cars stick to each other and ... diver jumps with a velocity of 3 m/s in opposite ... A 10 kg object moves at a constant velocity 2 m/s to the right and collides

3 0

3 years ago

a bullet moving with a velocity of 100m/s pierce a block of wood and moves out with a velocityof 10 m/s.if the thickness of the

erma4kov [3.2K]

The emerging velocity of the bullet is 71 m/s.

The bullet of mass m moving with a velocity u has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed v when it emerges from the block.

If the block exerts a resistive force F on the bullet and the thickness of the block is x then, the work done by the resistive force is given by,

$W=Fx$