1 year ago

To overcome an object's inertia, it must be acted upon by __________. A. gravity B. energy C. force D. acceleration

2 answers:

7 0

In order to overcome an object’s inertia (resistance to change), it must be acted upon by an unbalanced force, so the answer to the problem is letter C.

Jlenok [28]1 year ago

4 0

According to Newton's first law, the law of inertia, an object's velocity will not change unless it is acted upon by an outside force. This means that objects at rest tend to stay at rest, and objects in motion tend to stay in motion.

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A spring-powered dart gun is unstretched and has a spring constant 16.0 N/m. The spring is compressed by 8.0 cm and a 5.0 gram p

stepladder [879]

Answer:

Explanation:

Given that,

Spring constant = 16N/m

Extension of spring

x = 8cm = 0.08m

Mass

m = 5g =5/1000 = 0.005 kg

The ball will leave with a speed that makes its kinetic energy equal to the potential energy of the compressed spring.

So, Using conservation of energy

Energy in spring is converted to kinectic energy

So, Ux = K.E

Ux = ½ kx²

Then,

Ux = ½ × 16 × 0.08m²

Ux = 0.64 J

Since, K.E = Ux

K.E = 0.64 J

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1 year ago

A student, along with her backpack on the floor next to her, are in an elevator that is accelerating upward with acceleration a.

Anna007 [38]

Answer:

$\mu_k = \frac{2(vt - L)}{(g + a) t^2}$

Explanation:

As we know that backpack is kicked on the rough floor with speed "v"

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$N = mg + ma$

now the magnitude of kinetic friction on the block is given as

$F_f = \mu N$

$F_f = \mu (mg + ma)$

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$a = - \frac{F_f}{m}$

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now we know that bag hits the opposite wall at L distance away in time t

so we have

$d = v t + \frac{1}{2}at^2$

$L = vt - \frac{1}{2}(\mu_k)(g + a) t^2$

$\mu_k = \frac{2(vt - L)}{(g + a) t^2}$

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1 year ago

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2 years ago

Anna applies a force of 19.5 newtons to push a book placed on a table. If the normal force of the book is 51.7 newtons, what is

GarryVolchara [31]

that would be given by

[email protected]

@ representing coefficient of kinetic friction.

thus 19.5/51.7 = 0.377

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1 year ago

The additional energy in a beaker of hot water compared to an otherwise identical beaker of room temperature water is _ _ _ _ _

Fynjy0 [20]

<span>The additional energy in a beaker of hot water compared to an otherwise identical beaker of room temperature water is thermal energy.
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2 years ago

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