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3 years ago

Starting from rest, the Road Runner accelerates at 3 m/s for ten seconds. What is the final velocity of the Road Runner?

Physics

2 answers:

Vanyuwa [196]3 years ago

8 0

(3 m/s²) x (10 s) = <em>30 m/s</em>

His speed at the end of 10 sec is 30 m/s. We can't describe his velocity, because we have no information about the <em>direction</em> he's moving.

Charra [1.4K]3 years ago

6 0

Answer

The answer to this question is $30 ms^{-1}$

Explanation

As we know that accelartion is the rate of change of velocity. So, it can be write as

$a = (V_f -V_i) /t$

where

$V_f$ is the final velocity

$V_i$ is the initial velocity

t is the time

a is the accelartion

as we konw

$a = 3 ms^{-2}$

t = 10 s

From rest So,

$V_i$ = 0

$V_f$ = ?

Putting values

$3 = (V_f - 0)/10$

$3 * 10 = V_f$

$30 = V_f$

$V_f = 30 ms^{-1}$

So, the right answe is $30 ms^{-1}$

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A 500-g lump of clay is dropped onto a 1-kg cart moving at 60 cm/s. The clay is moving downward at 30 cm/s just before l dith t

viktelen [127]

Answer:

The speed of the cart and clay after the collision is 50 cm/s .

Explanation:

Given :

Mass of lump , m = 500 g = 0.5 kg .

Velocity of lump , v = 30 cm/s .

Mass of cart , M = 1 kg .

Velocity of cart , V = 60 cm/s .

We know by conservation of momentum :

$mv+MV=(m+M)v'$

Here , $v'$ is the speed of the cart and clay after the collision .

Putting all value in above equation .

We get :

$0.5\times 30+1\times 60=(0.5+1)\times v'\\\\v'=\dfrac{15+60}{1.5}\\\\v'=50\ cm/s$

Hence , this is the required solution .

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3 years ago

Starting from Newton’s law of universal gravitation, show how to find the speed of the moon in its orbit from the earth-moon dis

WARRIOR [948]

Answer: 1010.92 m/s

Explanation:

According to Newton's law of universal gravitation:

$F=G\frac{Mm}{r^{2}}$ (1)

Where:

$F$ is the gravitational force between Earth and Moon

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24} kg$ is the mass of the Earth

$m=7.349(10)^{22} kg$ is the mass of the Moon

$r=3.9(10)^{8} m$ is the distance between the Earth and Moon

Asuming the orbit of the Moon around the Earth is a circular orbit, the Earth exerts a centripetal force on the moon, which is equal to $F$ :

$F=m.a_{C}$ (2)

Where $a_{C}$ is the centripetal acceleration given by:

$a_{C}=\frac{V^{2}}{r}$ (3)

Being $V$ the orbital velocity of the moon

Making (1)=(2):

$m.a_{C}=G\frac{Mm}{r^{2}}$ (4)

Simplifying:

$a_{C}=G\frac{M}{r^{2}}$ (5)

Making (5)=(3):

$\frac{V^{2}}{r}=G\frac{M}{r^{2}}$ (6)

Finding $V$ :

$V=\sqrt{\frac{GM}{r}}$ (7)

$V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24} kg)}{3.9(10)^{8} m}}$ (8)

Finally:

$V=1010.92 m/s$

5 0

2 years ago

What is the process of evaporation through plant leaves called

Marianna [84]

<h2>Answer: Transpiration </h2>

Vegetal transpiration is the loss of water in the form of vapor, in the plant through its different parts, especially its leaves.

In this process, soil water is absorbed by the roots of the plant and transported in liquid form to the leaves to be converted into water vapor, while a part is used in photosynthesis. That is why vegetal transpiration is considered a vital function in the photosynthesis process.

This is possible because the leaves have small pores that allow water to escape into the atmosphere in the form of vapor and absorb carbon dioxide. Then, most of the water in the plants is used in the process of transpiration and only a small percentage is retained in liquid state and used for its growth and storage.

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3 years ago

A sky-diver jumps from a stationary balloon. His initial downwards acceleration is 10m/s².

Tatiana [17]

Explanation:

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Gina is driving her car to work, but she’s stopped at a red light. When the light turns green, she presses the gas pedal and acc

rusak2 [61]

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3 years ago

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