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3 years ago

A car traveling at 90 m/s can stop in a distance of 110 m. What is the magnitude of the cars acceleration as it slows down?

Physics

1 answer:

LenKa [72]3 years ago

8 0

Please find attached photograph for your answer. Hope it helps! Please do comment

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Indicate the substance that is NOT an element.

Volgvan

Water is Not an element

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2 years ago

Why doesn't the motor work?

exis [7]

Answer:

Explanation: its weird

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3 years ago

A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught

meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

$y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}$ (Eq. 1)

Where:

$y_{o}$ - Initial height of the softball, measured in meters.

$y$ - Final height of the softball, measured in meters.

$v_{o}$ - Initial velocity of the softball, measured in meters per second.

$t$ - Time, measured in seconds.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $y = y_{o}$ , $t = 3.56\,s$ and $g = -9.807\,\frac{m}{s^{2}}$ , the initial velocity of the softball is:

$v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0$

$3\cdot v_{o} -44.132\,m= 0$

$v_{o} = 14.711\,\frac{m}{s}$

The initial velocity of the softball is 14.711 meters per second.

8 0

3 years ago

A man does 4,780 J of work in the process of pushing his 2.70 103 kg truck from rest to a speed of v, over a distance of 25.5 m.

wolverine [178]

Answer:

(A) Velocity will be 1.88 m/sec

(b) Force will be 187.45 N

Explanation:

We have given work done = 4780 j

Distance d = 25.5 m

(A) Mass of the truck m = $m=2.70\times 10^3kg$

We know that kinetic energy is given by

$KE=\frac{1}{2}mv^2$

So $v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2\times 4780}{2.7\times 10^3}}=1.88m/sec$

(B) We know that work done is given by

W = Fd

So $F=\frac{W}{d}=\frac{4780}{25.5}=187.45N$

4 0

3 years ago

_________ are exchanged or shared during the formation of a chemical bond.

Thepotemich [5.8K]

<span>A chemical bond is a lasting attraction between atoms, ions or molecules that enables the formation of chemical compounds. The bond may result from the electrostatic force of attraction between oppositely charged ions as in ionic bonds; or through the sharing of electrons as in covalent bonds.

</span>

6 0

3 years ago

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