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Gemiola [76]
2 years ago
13

10 m/s in 10 3. A car accelerates from 0 m/s to 50 m/s in 10 seconds. What is the acceleration?

Physics
1 answer:
son4ous [18]2 years ago
3 0

Answer:

5ms⁻²

Explanation:

Given that,

Initial velocity of car (at t=0) u=0 m/s

Final velocity of car (after 10 sec) v=50 m/s

Time taken (t)= 10 sec.

acceleration(a)=?

now,

v = u + at

at = v-u

a=  (v-u) / t

a= (50-0) / 10

a = (50/10)

a=5 ms⁻²

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The impulse J is equal to the magnitude of the force applied to the cannonball times the time it is applied:
J=F \Delta t
But the impulse is also equal to the change in momentum of the cannonball:
J=\Delta p
If we put the two equations together, we find
F \Delta t= \Delta p
And since we know the magnitude of the average force and the time, we can calculate the change in momentum:
\Delta p= F \Delta t=(8.0 \cdot 10^3 N)(1.0 \cdot 10^{-1} s)=800 kg m/s
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3 years ago
Will upvote!!!!
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D. potential energy, because there is a bunch of water pent up, essentially stationary, waiting to roll down the steep mountain from the peak, so to say. if the dam were to be removed it would become kinetic.
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3 years ago
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n outer space, a constant net force with a magnitude of 140 N is exerted on a 32.5 kg probe initially at rest. A) What accelerat
musickatia [10]

Answer:

a) 4.31 m/s²

b) 215.5 m

Explanation:

a) According to Newton's first law of motion

The net force applied to particular mass produced acceleration, a, according to

F = ma

F = 140 N

m = 32.5 kg

a = ?

140 = 32.5 × a

a = 140/32.5 = 4.31 m/s²

b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s

u = initial velocity of the probe = 0 m/s (since it was initially at rest)

a = 4.31 m/s²

t = 10 s

s = distance travelled = ?

s = ut + at²/2

s = 0 + (4.31×10²)/2 = 215.5 m

7 0
3 years ago
Why does it take much longer time and distance to stop a moving ship in water than a moving car on the road
Naddik [55]
The force on the ship is more than a car
6 0
3 years ago
What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL
lozanna [386]

Answer:

\displaystyle \Delta x=1.74\ m

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

\displaystyle PE = \frac{1}{2}k(\Delta x)^2

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

\displaystyle \Delta x=\swrt{\frac{2PE}{k}}

Substituting:

\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}

Calculating:

\displaystyle \Delta x=\sqrt{3.0175}

\boxed{\displaystyle \Delta x=1.74\ m}

6 0
3 years ago
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