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irina1246 [14]
3 years ago
9

A car is accelerated at a constant rate from 15 m/s to 25 m/s. It takes the car 6 s to reach its final speed. What is the car’s

acceleration?
Physics
1 answer:
sladkih [1.3K]3 years ago
5 0

Answer:

<h2>1.67 m/s²</h2>

Explanation:

The car’s acceleration can be found by using the formula

a =  \frac{v - u}{t}  \\

where

v is the final velocity

u is the initial velocity

t is the time taken

a is the acceleration

From the question we have

a =  \frac{25 - 15}{6}  =  \frac{10}{6}  =  \frac{5}{3}  \\  = 1.666666...

We have the final answer as

<h3>1.67 m/s²</h3>

Hope this helps you

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A man with a mass of 60 kg rides a bike with a mass of 13 kg. What is the force needed by the man to accelerate the bike at 0.90
Tanzania [10]
Bro it would be9999999 cuh cuh
8 0
3 years ago
If a net horizontal force of 0.8 N is applied to a toy whose mass is 1.2 kg what acceleration is produced?
julia-pushkina [17]

Answer:

<h2>0.67 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{0.8}{1.2}   \\  = 0.666666...

We have the final answer as

<h3>0.67 m/s²</h3>

Hope this helps you

6 0
3 years ago
A rail car moving at 10m/s collides with and connect to another stationary car. What is their final velocity?
ki77a [65]

Answer:

read this it might help some

When a moving object collides with a stationary object of identical mass, the stationary object encounters the greater collision force. When a moving object collides with a stationary object of identical mass, the stationary object encounters the greater momentum change.

Explanation:

6 0
3 years ago
(m = 4.0 kg) the cockroach rides on the rim (at radius R) of a disk (M = 6.0 kg) that turns about its center like a merry-go-rou
kvasek [131]

Answer:

ω' = 2.5 rad/s

Explanation:

mass of cockroach, m = 4 kg

mass of disk, M = 6 kg

Radius of disc= R

initial angular velocity, ω = 2 rad/s

Let the final angular velocity is ω'

As no external torque is applied, so the angular momentum is constant.

Angular momentum = Moment of inertia x angular velocity

I ω = I' ω'

\frac{1}{2}\left ( M+m \right )R^{2}\times {2} = \left (\frac{1}{2}MR^{2}+m(0.5R)^{2})  \right )\omega '

10R^{2} = 4R^{2}\omega '

ω' = 2.5 rad/s

5 0
4 years ago
Sam, whose mass is 75 kg, straps on his skis and starts down a 50-m-high, 20 frictionless slope. A strong headwind exerts a hori
LenaWriter [7]

Answer:

Explanation:

Sam mass=75kg

Height is 50m

20° frictionless slope

Horizontal force on Sam is 200N

According to the work energy theorem, the net work done on Sam will be equal to his change in kinetic energy.

Therefore

Wg - Ww =∆K.E

Note initial the body was at rest at top of the slope.

Then, ∆K.E is K.E(final) - K.E(initial)

K.E Is given as ½mv²

Since initial velocity is zero then, K.E(initial ) is zero

Therefore, ∆K.E=½mVf²

Wg is work done by gravity and it is given by using P.E formulas

Wg=mgh

Wg=75×9.8×50

Wg=36750J

Ww is work done by wind and it's is given by using formulae for work

Work=force × distance

Ww=horizontal force × horizontal distance

Using Trig.

TanX=opposite/adjacent

Tan20=h/x

x=h/tan20

x=50/tan20

x=137.37m

Then,

Ww=F×x

Ww=200×137.37

We=27474J

Now applying the formula

Wg - Ww =∆K.E

36750 - 27474 =½×75×Vf²

9276=37.5Vf²

Vf²=9275/37.5

Vf²= 247.36

Vf=√247.36

Vf=15.73m/s

3 0
4 years ago
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