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lana [24]
3 years ago
8

At high speeds, a particular automobile is capable of an acceleration of about 0.540 m/s^2. At this rate, how long (in seconds)

does it take to accelerate from 91.0 km/hr to 104 km/hr?
Physics
1 answer:
Mama L [17]3 years ago
5 0

Answer:

t = 6.68 seconds

Explanation:

The acceleration of the automobile, a=0.54\ m/s^2

Initial speed of the automobile, u = 91 km/hr = 25.27 m/s

Final speed of the automobile, v = 104 km/hr = 28.88 m/s

Let t is the time taken to accelerate from u to v. It can be calculated as the following formula as :

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

t=\dfrac{28.88-25.27}{0.54}

t = 6.68 seconds

So, the time taken by the automobile to accelerate from u to v is 6.68 seconds. Hence, this is the required solution.

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Which of the following is not a part of the modern theory of evolution?
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Answer:

D

Explanation:

D) Evolution proceeds in the direction desired by members of a generation is not the modern theory of evolution.

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A parked car begins to roll down a hill, what can you conclude from that observation?
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its The rolling friction is greater than the force of the car’s weight against the hill.

and A force was required to start the car rolling.

Explanation:

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Determine the potential difference between two charged parallel plates that are 0.50 cm apart and have an electric field strengt
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What happens to the radiation coming from the Sun and heading towards Earth?
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7 0
3 years ago
A sled slides along a horizontal surface on which the coefficient of kinetic friction is 0.25. Its velocity at point A is 8.6 m/
jenyasd209 [6]

Answer:

\Delta t =1.31\ s

Explanation:

given,

coefficient of kinetic friction, μ = 0.25

Speed of sled at point A = 8.6 m/s

Speed of sled at point B = 5.4 m/s

time taken to travel from point A to B.

we know,

J = F Δ t

J is the impulse

where  F is the frictional force.

t is the time.

we also know that impulse is equal to change in momentum.

J = m(v_f - v_i)

frictional force

F = μ N

where as N is the normal force

now,

F\Delta t = m(v_f -v_i)

\mu m g \times \Delta t = m(v_f-v_i)

\mu g \times \Delta t = v_f-v_i

\Delta t =\dfrac{v_f-v_i}{\mu g}

\Delta t =\dfrac{8.6-5.4}{0.25\times 9.8}

\Delta t =1.31\ s

time taken to move from A to B is equal to 1.31 s

3 0
3 years ago
Read 2 more answers
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