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Advocard [28]
3 years ago
15

Instructions:Select the correct answer. A meteor is approaching Earth. Which statement about its motion is true? The meteor trav

els upward relative to Earth. The meteor accelerates. The meteor is out of Earth’s gravitational force field. The meteor falls with a constant velocity. NextReset
Physics
2 answers:
xz_007 [3.2K]3 years ago
5 0

Answer:

The meteor accelerates

Explanation:

Let's see the options:

- the meteor travels downward relative to Earth,  that is, from the sky to the Earth.

- The meteor is affected by Earth’s gravitational force field, which attracts it to Earth.

- As a consequence of the previous item, it suffers acceleration (Earth gravitational acceleration), and then, its velocity is not constant.

Lynna [10]3 years ago
3 0
<span> The meteor accelerates after it is brought into earths force field 

</span>
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What is the acceleration of a 10 kg mass pushed by a 5 N force?
yanalaym [24]
g-\ gravitational \ acceleration \\ g= \frac{G}{m} = \frac{5N}{10kg}=  \\ g=0,5 \frac{N}{kg}
8 0
3 years ago
Read 2 more answers
An electron is initially moving at 1.4 x 107 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N
algol13

Answer:

K.E = 15.57 x 10⁻¹⁷ J

Explanation:

First, we find the acceleration of the electron by using the formula of electric field:

E = F/q

F = Eq

but, from Newton's 2nd Law:

F = ma

Comparing both equations, we get:

ma = Eq

a = Eq/m

where,

E = electric field intensity = 120 N/C

q = charge of electron = 1.6 x 10⁻¹⁹ C

m = Mass of electron = 9.1 x 10⁻³¹ kg

Therefore,

a = (120 N/C)(1.6 x 10⁻¹⁹ C)/(9.1 x 10⁻³¹ kg)

a = 2.11 x 10¹³ m/s²

Now, we need to find the final velocity of the electron. Using 3rd equation of motion:

2as = Vf² - Vi²

where,

Vf = Final Velocity = ?

Vi = Initial Velocity = 1.4 x 10⁷ m/s

s = distance = 3.5 m

Therefore,

(2)(2.11 x 10¹³ m/s²)(3.5 m) = Vf² - (1.4 x 10⁷)²

Vf = √(1.477 x 10¹⁴ m²/s² + 1.96 x 10¹⁴ m²/s²)

Vf = 1.85 x 10⁷ m/s

Now, we find the kinetic energy of electron at the end of the motion:

K.E = (0.5)(m)(Vf)²

K.E = (0.5)(9.1 x 10⁻³¹ kg)(1.85 x 10⁷ m/s)²

<u>K.E = 15.57 x 10⁻¹⁷ J</u>

4 0
3 years ago
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6 0
3 years ago
. Consider a fully extended arm that is rotating about the shoulder such that with a shoulder-to-hand length of 30 cm. If the ar
Rainbow [258]

Answer:

13.309 m/s²

Explanation:

Length from shoulder to hand, l = 30 cm = 0.3 m

initial velocity, u = 1 m/s

final velocity, v = 2.5 m/s

time, t = 3 s

Let the tangential acceleration is a.

by using first equation of motion

v = u + at

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a = 0.5 m/s²

Let the centripetal acceleration is a'.

a' = v'²/l

a' = 2 x 2 / 0.3

a' = 13.3 m/s²

The tangential acceleration and the centripetal acceleration are both perpendicular to each other. So, the net acceleration is given by

A=\sqrt{a^{2}+a'^{2}}

A=\sqrt{0.5^{2}+13.3^{2}}

A = 13.309 m/s²

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3 years ago
Describe in terms of kinetic and potential energy what happens if an apple falls from a tree and comes to rest on the ground( wr
Lunna [17]

Answer:

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as the apple falls from the tree it loses its potential energy and gains kinetic energy due to the movement of the apple. Its kinetic energy will be given by KE= 1/2mv²  where m is the mass of the apple and v is the speed with which the apple falls.

As the apple falls the height or the distance reduces and PE becomes reduces. But it gains Kinetic energy due to its speed.

But when the apple falls to the ground and comes to rest its kinetic energy is converted to potential energy.

thus the total energy remains the same. it changes from one form to the other but remains unaltered.

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