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3 years ago

12

A car traveling at 34 m/s starts to decelerate steadily. It comes to a complete stop in 6 seconds. What is its acceleration?

1 answer:

swat323 years ago

3 0

The acceleration of the object described above is calculated by dividing the given initial velocity by the time it took for it to completely stop. That is,

a = Vi / t

Substituting the known values,

a = (34 m/s) / 6 s = 5.67 m/s²

The answer to this item is -5.67 m/s² because the motion of the object is that it is coming to a full stop from high velocity motion.

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Explanation:

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3 years ago

A particle, whose acceleration is constant, is moving in the negative x direction at a speed of 4.91 m/s, and 12.9 s later the p

Zinaida [17]

Answer:

The particle’s velocity is -16.9 m/s.

Explanation:

Given that,

Initial velocity of particle in negative x direction= 4.91 m/s

Time = 12.9 s

Final velocity of particle in positive x direction= 7.12 m/s

Before 12.4 sec,

Velocity of particle in negative x direction= 5.32 m/s

We need to calculate the acceleration

Using equation of motion

$v = u+at$

$a=\dfrac{v-u}{t}$

Where, v = final velocity

u = initial velocity

t = time

Put the value into the equation

$a=\dfrac{7.12-(-4.91)}{12.9}$

$a=0.933\ m/s^2$

We need to calculate the initial speed of the particle

Using equation of motion again

$v=u+at$

$u=v-at$

Put the value into the formula

$u=-5.321-0.933\times12.4$

$u=-16.9\ m/s$

Hence, The particle’s velocity is -16.9 m/s.

4 0

3 years ago

Which units are used to express kinetic energy?

labwork [276]

Answer:

The SI units for energy is Joules.

4 0

4 years ago

Read 2 more answers

A rock is dropped from the top of a tall tower. Half a second later another rock, twice as massive as the first, is dropped. Ign

Gemiola [76]

B- the acceleration is greater for the more massive rock

5 0

2 years ago

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How far should the second antenna be moved in order to receive a minimum signal from a station that broadcasts at 99.4 MHz?

raketka [301]

Answer:

So the distance of the antenna from the station will be 3.018 m

Explanation:

We have given the frequency of the broadcast $f=99.4MHz=99.4\times 10^4Hz$

The speed of light $c=3\times 10^8m/sec$

The distance of the antenna to receive a minimum signal from the station is given by $d=\frac{v}{f}=\frac{3\times 10^{8}}{99.4\times 10^6}=3.018m$

So the distance of the antenna from the station will be 3.018 m

5 0

3 years ago