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3 years ago

A 100 kg object and a 10 kg object are dropped simultaneously in a vacuum. Which of the following statements is true?

2 answers:

7 0

In a vacuum, all objects fall at the same rate. Meaning that the 100 kg ball will fall at the same speed as the 10 kg ball. Assuming that both objects share the same starting acceleration, they will keep that acceleration until the fall is stopped.
In other words, your answer is the first one, Both objects will accelerate at 9.8 m/s

Zepler [3.9K]3 years ago

3 0

Answer:

Both objects will accelerate at 9.8 m/s².

Explanation:

According to given condition, a 100 kg object and a 10 kg object are dropped simultaneously in a vacuum. If there is no air resistance, when two objects of same or different masses will reach the ground at same time. The rate of descent does not depend on the amount of matter contained inside the object.

In the vacuum, no air resistance is present. Both of the objects will accelerate at 9.8 m/s² i.e. under the action of gravity.

So, the correct option is (a). Hence, this is the required solution.

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3 years ago

Linea de tiempo sobre la aportaciones griegas a la astronomia y astrologia

rusak2 [61]

Answer:

1. 100 CE

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8. 585 BCE

Thales of Miletus lived during now.

Explanation:

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3 years ago

What is the acceleration of a block on a ramp inclined 35o to the horizontal if µk = 0.4?

lina2011 [118]

We can solve the problem by applying Newton's second law, which states that the resultant of the forces acting on an object is equal to the product between its mass and its acceleration:
$\sum F = ma$

We should consider two different directions: the direction perpendicular to the inclined plane and the direction parallel to it. Let's write the equations of the forces along the two directions, decomposing the weight of the object (mg):

$mg \sin \theta - \mu_K N = ma$ (parallel direction) (1)
$mg \cos \theta - N =0$ (perpendicular direction) (2)
where
$\theta=35^{\circ}$ is the angle of the inclined plane, N is the normal reaction of the plane, $\mu_K N$ is the frictional force, with $\mu_K=0.4$ being the coefficient of friction.

From eq.(2), we find
$N=mg \cos \theta$
and if we substitute into eq.(1), we can find the acceleration of the block:
$mg \sin \theta - \mu_k mg \cos \theta = ma$
from which
$a=g(\sin \theta - \mu_K \cos \theta)=(9.81 m/s^2)(\sin 35^{\circ} - 0.4 \cos 35^{\circ})=2.41 m/s^2$

7 0

3 years ago