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umka2103 [35]
3 years ago
5

A force of 1000N is used to kick a football of mass 0.8kg find the velocity with which the ball moves if it takes 0.8 sec to be

kicked.​
Physics
2 answers:
Ostrovityanka [42]3 years ago
5 0

The force used to kick the ball is 1000N

The mass of the ball is 0.8 kg

Time is 0.8 seconds

Therefore the velocity can be calculated as follows

F= Mv-mu/t

1000= 0.8(v) - 0.8(0)/0.8

1000= 0.8v- 0.8/0.8

Cross multiply both sides

1000(0.8) = 0.8v

800= 0.8v

divide both sides by the coefficient of v which is 8

800/0.8= 0.8v/0.8

v= 1000m/s

Hence the velocity is 1000m/s

Eduardwww [97]3 years ago
5 0

Answer:

1000 m/s

Explanation:

From the question given above, the following data were obtained:

Force (F) = 1000 N

Mass (m) = 0.8 Kg

Time (t) = 0.8 s

Velocity (v) =?

Next, we shall determine the the acceleration of the ball. This can be obtained as follow:

Force (F) = 1000 N

Mass (m) = 0.8 Kg

Acceleration (a) =?

F = ma

1000 = 0.8 × a

Divide both by 0.8

a = 1000 / 0.8

a = 1250 m/s²

Finally, we shall determine the velocity. This can be obtained as follow:

Time (t) = 0.8 s

Acceleration (a) = 1250 m/s²

Velocity (v) =?

a = v/t

1250 = v / 0.8

Cross multiply

v = 1250 × 0.8

v = 1000 m/s

Therefore, the ball will move with a velocity of 1000 m/s.

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a backpack has a mass of 8 kg. it is lifted and given 54.9 J of gravitational potential energy. how high is is lifted? accelerat
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P=mgh
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mash [69]

Answer:

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Explanation:

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5 0
3 years ago
ou are out stargazing with your 13.4-cm telescope. You point your telescope at an interesting formation in the sky, which you th
Alinara [238K]

Answer:

θ = 4.716 10⁻⁶ rad

Explanation:

In order for the releases to be considered separate, they must meet the Rayleigh criterion that establishes that the maximum diffraction of one star must coincide with the first minimum of the diffraction pattern of the second star.

We use the diffraction equation for a slit

            a sin θ = m λ

The minimum occurs at m = 1

             sin θ = λ / a

Since the angles in these systems are very small, we can approximate the sine to its angle in radians

             θ = λ / a

The telescope has a circular aperture whereby polar cords should be used, which introduces a constant number

           θ = 1.22 λ / a

Let's calculate

          θ = 1.22 518 10⁻⁹ / 13.4 10⁻²

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8 0
3 years ago
An airplane is flying at a speed of 45 m/s when it drops a 40 kg food package to the Polar exploration team. If the plane drops
Alina [70]

To solve this problem we will apply the concepts related to energy conservation, so the potential energy in the package must be equivalent to its kinetic energy. From there we will find the speed of the package in the vertical component. The horizontal component is given, as it is the same as the one the plane is traveling to. Vectorially we will end up finding its magnitude. So,

PE = KE

mgh = \frac{1}{2}mv^2

Here,

m = Mass

g = Gravity

h = Height

v = Velocity

Rearranging to find the velocity

v = \sqrt{2gh}

Replacing,

v = \sqrt{2(9.8)(120)}

v = 48.49m/s

Using the vector properties the magnitude of the velocity vector would be given by,

|V| = \sqrt{v_x^2+v_y^2}

|V| = \sqrt{45^2+48.42^2}

|V| = 66.2m/s

Therefore the package is moving to 66.2m/s

3 0
3 years ago
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