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UkoKoshka [18]
3 years ago
10

An object accelerates at a rate of 8 m/s² over a displacement of 9 m. If the object started with a velocity of 5 m/s, what is it

s final velocity?
Physics
1 answer:
Crank3 years ago
8 0

Answer:

13 m/s

Explanation:

The computation of the final velocity is shown below;

We know that

v^2 = u^2 + 2as

Here v represent the final velocity

u denotes the initial velocity

a denotes the acceleration

And, s denotes the displacement

So, the final velocity is  

v^2 = (5)^2 + 2(8)(9)

= sqrt(25 + 144)

= sqrt(169)

= 13 m/s

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D. Strong nuclear forces hold the nucleus of an atom together. Weak nuclear forces are involved when certain types of atoms break down.

Explanation:

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2 years ago
What happens when an object is dropped?
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When an object is dropped, tossed, or kicked, as long as it is not laying on the ground, it accelerates downward, because of the force of gravity acting on it.
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What process contributes most to acid precipitation?
kramer

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7 0
2 years ago
Two children are riding on a merry-go-round that is rotating with a constant angular speed. Abbie is one meter from the center o
klio [65]

Answer:

The acceleration of Abbie is half of the Zak's.

Explanation:

The centripetal acceleration of an object on a circular path is given by :

a=r\omega^2

Two children are riding on a merry-go-round that is rotating with a constant angular speed. Let r_1 is distance of Abbie from the merry-go-round and r_2 is distance of Zak's from the merry-go-round. Acceleration of Abbie is :

a_1=r_1\omega^2 ...... (1)

r_1=1\ m

Acceleration of Zak's is :

a_2=r_2\omega^2 .......(2)

r_2=2\ m

Dividing equation (1) and (2) we get :

\dfrac{a_1}{a_2}=\dfrac{r_1}{r_2}\\\\\dfrac{a_1}{a_2}=\dfrac{1}{2}\\\\a_1=\dfrac{a_2}{2}

So, the acceleration of Abbie is half of the Zak's.

7 0
3 years ago
A sphere 5 cm in diameter has a small scratch on its surface. When the scratch is viewed through the glass from a position direc
Molodets [167]

Answer

given,

diameter of sphere = 5 cm

radius of sphere = -2.5 cm

refractive index for glass n₁ = 1.5

refractive index for air n₂ = 1

magnification of the glass = ?

now,

\dfrac{n_1}{S} +\dfrac{n_2}{S'}= \dfrac{n_2-n_1}{R}

\dfrac{1.5}{5} +\dfrac{1}{S'}= \dfrac{1-1.5}{-2.5}

\dfrac{1.5}{5} +\dfrac{1}{S'}= \dfrac{0.5}{-2.5}

                  S' = - 10 cm

magnification

 m = \dfrac{-n_1S'}{n_2S}

 m = \dfrac{-1.5\times -10}{1\times 5}

 m = + 3

3 0
3 years ago
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