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BlackZzzverrR [31]
3 years ago
15

a force is applied to a box of 10.0 kg for 4.0 s. the box goes from rest to 25 m/s in that time. What is the magnitude of that f

orce?
Physics
1 answer:
Paul [167]3 years ago
8 0

Given:

m(mass of the box)=10 Kg

t(time of impact)=4 sec

u(initial velocity)=0.(as the body is initially at rest).

v(final velocity)=25m/s

Now we know that

v=u+at

Where v is the final velocity

u is the initial velocity

a is the acceleration acting on the body

t is the time of impact

Substituting these values we get

25=0+a x 4

4a=25

a=6.25m/s^2

Now we also know that

F=mxa

F=10 x6.25

F=62.5N

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Answer:Visible lighlight

Explanation:

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2 years ago
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ohaa [14]
For this, we need the formula:
V = k q / r
where k is the Coulombs law constant = 9 x 10^9 N
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Simply plug in the values and solve for V
5 0
2 years ago
Read 2 more answers
Two light bulbs are wired in series and one bulb burns out (opens.) Technician A says that the other bulb will still work. Techn
Radda [10]

Answer:

Neither technician

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Neither technician is correct.

two bulbs are connected in series one bulb burn out  

If one bulb in the series burns out then the circuit will break.

In a series circuit same current passes each resistor.

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6 0
3 years ago
Two plane mirrors are separated by 120°, as the drawing illustrates. If a ray strikes mirror M1 at a θ1 = 64° angle of incidence
tino4ka555 [31]

Angle, θ2 at which the light leaves mirror 2 is 56°

<u>Explanation:</u>

Given-

θ1 = 64°

So, α will also be 64°

According to the figure:

α + β = 90°

So,

β = 90° - α

  = 90° - 64°

  = 26°

β + γ + 120° = 180°

γ = 180° - 120° - β

γ = 180° - 120° - 26°

γ = 34°

γ + δ = 90°

δ = 90° - γ

δ = 90° - 34°

δ = 56°

According to the law of reflection,

angle of incidence = angle of reflection

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Therefore, angle θ2 at which the light leaves mirror 2 is 56°

8 0
2 years ago
Starting at 1.0 m/s, a cheetah runs with a constant acceleration for 4.8 s reaching a speed of 28 m/s. What is the acceleration
LenaWriter [7]

Answer:

c.5.6m/s^2

Explanation:

Initial velocity of cheetah,u=1 m/s

Time taken by cheetah =4.8 s

Final velocity of cheetah,v=28 m/s

We have to find the acceleration of this cheetah.

We know that

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

Where v=Final velocity of object

u=Initial velocity of object

t=Time taken by object

Using the formula

Then, we get

Acceleration, a=\frac{28-1}{4.8}=\frac{27}{4.8} m/s^2

Acceleration=a=5.6 m/s^2

Hence, the acceleration of cheetah=5.6m/s^2

5 0
3 years ago
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