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PolarNik [594]
3 years ago
10

An 877 kg drag race car accelerates from rest to 116 km/h in 0.951 s. What change in momentum does the force produce? Answer in

units of kg. m/s.​
Physics
1 answer:
Eduardwww [97]3 years ago
7 0

Answer:

The change in momentum is 28265.71  kg-m/s.

Explanation:

Given that,

Mass of a car, m = 877 kg

Initial velocity of the car, u = 0 (at rest)

Final velocity of the car, v = 116 km/h = 32.23 m/s

Time, t = 0.951 s

We need to find the change in momentum produced by the force. It can be calculated as the difference of final momentum and the initial momentum.

\Delta P=m(v-u)\\\\=877\times (32.23-0)\\\\=28265.71\ kg-m/s

So, the change in momentum is 28265.71  kg-m/s.

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How much Tim in minutes will it take a car driving at 90km/hr to travel 27 kilometers
astraxan [27]

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18min

Explanation:

v=d/t

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3 years ago
g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
Veseljchak [2.6K]

Answer:

The speed of q₂ is 4\sqrt{10}\ m/s

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

E_{i}=E_{f}

\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2

\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})

Put the value into the formula

\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})

0.00075(400-v_{f}^2)=0.18&#10;

400-v_{f}^2=\dfrac{0.18}{0.00075}

-v_{f}^2=240-400

v_{f}^2=160

v_{f}=4\sqrt{10}\ m/s

Hence, The speed of q₂ is 4\sqrt{10}\ m/s

7 0
3 years ago
Two forces (4N and 3N) pull your the left, while a 12N force pulls to the right. What is the net force
Lapatulllka [165]

Answer:

The answer is 11N to the right

Explanation:

Because 4N-3N= 1N

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8 0
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