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serg [7]
2 years ago
5

A football player kicks a 0.41-kg football initially at rest; and the ball flies through the air. If the kicker's foot was in co

ntact with the ball for 0.051 s and the ball's initial speed after the collision is 21 m/s, what was the magnitude of the average force on the football?
Physics
1 answer:
LUCKY_DIMON [66]2 years ago
4 0

Answer:

Average force on the football = 168.82 N

Explanation:

Force = Mass x Acceleration

F = ma

Mass, m = 0.41 kg

We have equation of motion, v = u + at

Initial velocity, u = 0 m/s

Final velocity, v = 21 m/s

Time, t = 0.051 s

Substituting

                  21 = 0 + a x 0.051

                     a = 411.76 m/s²

Substituting in force equation,

                   F = ma = 0.41 x 411.76 = 168.82 N

Average force on the football = 168.82 N

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(a) Calculate the absolute pressure at the bottom of a fresh- water lake at a depth of 27.5 m. Assume the density of the water i
ddd [48]

Answer:

a) P = 370.993\,kPa, b) F = 25.948\,kN

Explanation:

a) The absolute pressure at a depth of 27.5 meters is:

P = P_{atm} + P_{man}

P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (27.5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)

P = 370.993\,kPa

b) The force exerted by the water is:

F = (P - P_{atm})\cdot A

F = (370.993\,kPa-101.3\,kPa)\cdot \left(\frac{\pi}{4} \right)\cdot (0.35\,m)^{2}

F = 25.948\,kN

5 0
2 years ago
Read 2 more answers
A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta
BartSMP [9]

Answer:

35%

Explanation:

Assuming no external torques present during the collision between the record and the turntable, total angular momentum must be conserved.

For a rotating body with some angular velocity and moment of inertia, the angular momentum can be expressed as follows;

L = I* ω

So, as initial angular momentum and final angular momentum must be the same, we have:

Li = Lf

⇒ I₁ * ω₁ = I₂ * ω₂ (1)

where I₁ is the rotational inertia of the turntable, and I₂, is the combined rotational inertia of the turntable and the record:

I₂ = I₁ + 0.54 I₁ = 1.54 I₁

We can solve (1) for the new common angular speed, as follows:

ω₂ = ω₁ / 1.54 (2)

The initial rotational kinetic energy is given by definition for the following equation:

Kroti = 1/2 * I₁ * ω₁² (3)

The final rotational kinetic  energy takes into account the new rotational inertia and the common final angular speed:

Krotf = 1/2* I₂ * ω₂² = 1/2* 1.54 I₁* (ω₁/1.54)² (4)

Dividing both sides in (3) and (4), we get:

Krotf/Kroti = 1/1.54 = 0.65

This means that the final rotational kinetic energy, has reduced to 0.65 of the initial value, or that has lost 35% of the initial kinetic energy.

8 0
2 years ago
|| Climbing ropes stretch when they catch a falling climber, thus increasing the time it takes the climber to come to rest and r
Otrada [13]

To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.

Newton's second law is defined as

F = ma

Where,

m = mass

a = acceleration

From this equation we can figure the acceleration out, then

a = \frac{F}{m}

a = \frac{11*10^3}{80}

a = 137.5m/s

From the cinematic equations of motion we know that

v_f^2-v_i^2 = 2ax

Where,

v_f =Final velocity

v_i =Initial velocity

a = acceleration

x = displacement

There is not Final velocity and the acceleration is equal to the gravity, then

v_f^2-v_i^2 = 2ax

0-v_i^2 = 2(-g)x

v_i =\sqrt{2gx}

v_i = \sqrt{2*9.8*4.8}

v_i = 9.69m/s

From the equation of motion where acceleration is equal to the velocity in function of time we have

a = \frac{v_i}{t}

t = \frac{v_i}{a}

t =\frac{9.69}{137.5}

t = 0.0705s

Therefore the time required is 0.0705s

4 0
2 years ago
Read 2 more answers
Do as physics instructor fred cauthen does and place a tennis ball close to and above the top of a basketball. drop the balls to
Ksivusya [100]

Answer:

after shock

creating a system for the conservation of the energy of the basketball ball and creating a system for the tennis ball only, the conservation of energy should be applied to each system independently

Explanation:

When the two balls fall they acquire the same speed since they are accelerated by the same force, their weight and the acceleration of the acceleration of gravity. When reaching the floor the mechanical energy of the system is conserved.

Upon reaching the floor, the first ball (basketball) collides with the floor, this process is very fast, at the end of the process the basketball comes out with a velicad up and collides with the much lighter tennis ball that is still descending .

we assume that the shocks are elastic, when solving the momentary and kinetic energy findings, we find the velocities after each shock

     

In this clash the tennis ball acquires a high kinetic speed with an upward direction that makes a very high height high. Again this shock is very fast and the tennis ball almost does not move.

Here we must separate the system, creating a system for the conservation of the energy of the basketball ball and another system for the tennis ball only, the conservation of energy should be applied to each system independently

Em₀ =K = 1/2 m v²

                    Em_{f} = U = m g h

As in the elastic shock the final speed of the tennis ball is approximately 2 vo, we can calculate the maximum height

                 m g h = 1/2 m (2v₀)²

                 h = 2 v₀²/g

To reconcile this with the conservation of energy we must calculate the energy for the tennis ball at two points, the first when the crash with the tennis ball ends and at the end point at its maximum height.

6 0
2 years ago
An electron moves with a constant horizontal velocity of 3.0 × 106 m/s and no initial vertical velocity as it enters a deflector
Ghella [55]

Answer:

a = 5.05 x 10¹⁴ m/s²

Explanation:

Consider the motion along the horizontal direction

v_{x} = velocity along the horizontal direction = 3.0 x 10⁶ m/s

t = time of travel

X = horizontal distance traveled = 11 cm = 0.11 m

Time of travel can be given as

t = \frac{X}{v_{x}}

inserting the values

t = 0.11/(3.0 x 10⁶)

t = 3.67 x 10⁻⁸ sec

Consider the motion along the vertical direction

Y = vertical distance traveled = 34 cm = 0.34 m

a = acceleration = ?

t = time of travel  = 3.67 x 10⁻⁸ sec

v_{y} = initial velocity along the vertical direction = 0 m/s

Using the kinematics equation

Y = v_{y} t + (0.5) a t²

0.34 = (0) (3.67 x 10⁻⁸) + (0.5) a (3.67 x 10⁻⁸)²

a = 5.05 x 10¹⁴ m/s²

7 0
2 years ago
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