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kkurt [141]
3 years ago
6

An object is moving in the absence of a net force. Which of the following best describes the object's motion? -please help?

Physics
2 answers:
Gelneren [198K]3 years ago
7 0
You can solve this by using Newton's First Law or Newton's Second Law.

1) Newton's First Law or Inertia Law states that in the abscense of a net force acting over an object, this will not chage its state of movement, i.e it will  remain at rest (if it is no moving) or with uniform movement (if the object is moving).

2) Newton's Second Law: Net force = mass * acceleration => acceleration = net force / mass = 0 / mass = 0.

Zero accelerations means rest or uniform movement.

First Law is implicit in Second Law.
anzhelika [568]3 years ago
7 0
There are two cases in which no net force is being applied on an object:
1) The object is stationary
2) The object is not accelerating; that is, the magnitude or direction of its velocity is not changing.
Because this question requires the object to be in motion, the velocity of this object must be constant but non-zero.
Therefore, the correct option would be the one where the velocity is constant and in a straight line. (circular motion means the velocity's direction is constantly changing, thus the velocity is not constant)
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On August 10, 1972, a large meteorite skipped across the atmosphere above the western United States and western Canada, much lik
Anon25 [30]

a) 4.62\cdot 10^{14} J

b) 0.110 megatons

c) 8.46 bombs

Explanation:

a)

The energy lost by the meteorite is equal to the difference between its final kinetic energy and its initial kinetic energy:

\Delta K=K_f-K_i

Which can be rewritten as:

\Delta K=\frac{1}{2}mv^2-\frac{1}{2}mu^2

where:

m=3.2\cdot 10^6 kg is the mass of the meteorite

v=0 is the final speed of the meteorite

u=17 km/s = 17,000 m/s is the initial speed of the meteorite

Substituting the values into the equation, we found the loss in energy of the meteorite:

\Delta K=0-\frac{1}{2}(3.2\cdot 10^6)(17000)^2=-4.62\cdot 10^{14} J

So, the energy lost by the meteorite is 4.62\cdot 10^{14} J

b)

The energy equivalent to 1 megaton of TNT is

E_{TNT}=4.2\cdot 10^{15} J

Here the energy lost by the meteorite is

E=4.62\cdot 10^{14} J

Therefore, in order to write the energy lost by the meteorite as a multiple of the energy of 1 megaton of TNT, we have to divide the energy lost by the meteorite by the energy equivalent to 1 TNT; we find:

\frac{E}{E_{TNT}}=\frac{4.62\cdot 10^{14}}{4.2\cdot 10^{15}}=0.110

So, the energy lost by the meteorite corresponds to 0.110 megatons.

c)

The energy of one atomic bomb explosion in Hiroshima is equal to

E'=13 kt (13 kilotons)

which corresponds to

E'=0.013 Mt (0.013 megatons)

Here the energy of the meteorite is equal to

E=0.110 Mt (0.110 megatons)

Therefore, we can find how many Hiroshima bombs are equivalent to teh meteorite impact by using the following rules of three:

\frac{1 bomb}{0.013 Mt}=\frac{x bombs}{0.110 Mt}\\x=\frac{1\cdot 0.110}{0.013}=8.46

So, 8.46 bombs.

5 0
3 years ago
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