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3 years ago

According to Aristotle where do moving objects tend to reach

Physics

2 answers:

Mekhanik [1.2K]3 years ago

3 0

They seek their natural resting places

astraxan [27]3 years ago

3 0

I believe a resting place (stopping point).

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8. Radioactive isotopes, differentiation of Earth’s iron core, gravity-driven compression,and meteoric bombardment all caused ea

Olegator [25]

Answer:

option C

Explanation:

the correct answer is option C.

All the activity such as Radioactive isotopes, differentiation of Earth’s iron core, gravity-driven compression,and meteoric bombardment will make the earth hotter than what it is at present.

Radioactive isotope whose nuclei are unstable dissipate excess energy.

Differentiation of iron core means formation of into layer which includes iron.

meteorite bombardment will affect the atmosphere of the earth which protect harmful radiation.

All these activity will make earth hotter

6 0

3 years ago

An outfielder throws a baseball with an initial speed of 81.8mi/h. Just before an infielder catches the ball at the same level,

agasfer [191]

Answer:

-20.158ft-lb

Explanation:

Check the attached files for the explanation.

3 0

3 years ago

A car stands from rest and attains a speed of 20 m/s in 10 s. What distance will it cover ?

sashaice [31]

Answer:

100m

Explanation:

100m

s=ut+1/2at^2

s= unknown, u=0, a=2, t=10

s=0*10+1/2(2)(10)^2

s=1/2(2)(100)

s=1(100)

displacement = 100 meters

4 0

3 years ago

An astronaut's pack weighs 17.5 N when she is on earth, but only 3.24 N when she is on the surface of an asteroid. What is the a

Nimfa-mama [501]

Answer:

a= 1.81 m/s²

Explanation:

Ratio of the weights on both surfaces can be calculated as,

$\frac{W_{a} }{W_{E} }$ = $\frac{ma}{mg}$

Here;

Wa is the weight on the asteroid, $W_{E}$ is weight on earth,

m is mass of pack, a is acceleration due to an asteroid and

g is acceleration due to gravity.

Rearrange above equation for a,

a=g ( $\frac{Wa}{W_{E} }$ )

Substitute 3.24 N for Wa, 17.5 N for $W_{E}$ ,

9.81 m/s² for g in the above equation,

a= 9.81( $\frac{3.24}{17.5}$ )

a= 1.81 m/s²

7 0

4 years ago

A 0.140-kg baseball is dropped from rest from a height of 1.8 m above the ground. It rebounds to a height of 1.4 m. What change

aliina [53]

Answer:

$\Delta p=-1.56\ kg-m/s$

Explanation:

It is given that,

Mass of the baseball, m = 0.14 kg

It is dropped form a height of 1.8 m above the ground. Let u is the velocity when it hits the ground. Using the conservation of energy as :

$u=\sqrt{2gh}$

h = 1.8 m

$u=\sqrt{2\times 9.8\times 1.8}$

u = 5.93 m/s

Let v is the speed of the ball when it rebounds. Again using the conservation of energy to find it :

$v=\sqrt{2gh'}$

h' = 1.4 m

$v=-\sqrt{2\times 9.8\times 1.4}$

v = -5.23 m/s

The change in the momentum of the ball is given by :

$\Delta p=m(v-u)$

$\Delta p=0.14(-5.23-5.93)$

$\Delta p=-1.56\ kg-m/s$

So, the change in the ball's momentum occurs when the ball hits the ground is 1.56 kg-m/s. Hence, this is the required solution.

4 0

4 years ago