When landing from a jump, a basketball player of mass 82 kg has a velocity of 1.2 m/s right before they hit the ground. The player then lands on the floor with their feet and is quickly brought to a stop. If the the floor deforms by 0.025 m while the player lands on it, what was the average force under the player's feet during the landing

Answer:

What are the options?

Explanation:

Answer:

stopping potential is the negative potential applied to the circuit to stop the moving electrons so as to stop the flow of current

for high current high negative potential is applied

Answer:

yes

Explanation:

i hope this helps not sure im right

Answer:

c

Explanation:

Complete Question

the maximum force a pilot can stand is about seven times his weight. what is the minimum radius of curvature that a jet plane's pilot, pulling out of a vertical dive, can tolerate at a speed of 250m/s?

Answer:

The value is    [tex]r = \frac{250^2 }{6 * 9.8 }[/tex]

Explanation:

From the question we are told that

 The  weight of the pilot is   [tex]W = mg[/tex]

 The maximum force a pilot can withstand is  [tex]F_{max} = 7 W = 7 (mg)[/tex]

 The speed is  [tex]v = 250 \ m/s[/tex]  

Generally the centripetal force acting on the pilot is equal to the net force acting on the pilot i.e

      [tex]F_c = F_{max} - mg[/tex]

Here N  is the normal force acting on the pilot

Now

      [tex]F_c = \frac{m v^2 }{r}[/tex]

So

      [tex]\frac{m v^2 }{r} = 7(mg) - mg[/tex]  

=>  [tex]r = \frac{v^2 }{6g}[/tex]

=>  [tex]r = \frac{250^2 }{6 * 9.8 }[/tex]

=>  [tex]r = 1063 \ m[/tex]

Answer:

38.40

Explanation:

This is the answer for this question

Answer:

4.5 × 10^6 Nm

Explanation:

Given that a 52.0-kg woman wearing high-heeled shoes is invited into a home in which the kitchen has vinyl floor covering. The heel on each shoe is circular and has a radius of 0.600 cm. If the woman balances on one heel, what pressure does she exert on the floor?

Let's first calculate the area covered by the heel shoe by using the area of a circle.

Convert cm to m

0.6 / 100 = 0.006

A = πr^2

A = 22/7 × 0.006^2

A = 1.131 × 10^-4 m^2

The force exerted by the woman = mg

Force exerted = 52 × 9.8

Force exerted = 509.6 N

Pressure = force / Area

Pressure = 509.6 / 1.131 × 10^-4

Pressure = 4505853.3 N/m

Therefore, she exerts pressure of 4.5 × 10^6 Nm approximately on the floor.

Answer:

A jet pilot puts an aircraft with a constant speed into a vertical loop is explained below in complete details.

Explanation:

Well, the difficulty does not provide the pilot's mass (or weight in regular gravity), but the difficulty can be resolved and declared in courses of m (the pilot's mass).

When the jet is at the foundation of the circuit, a free-body chart displays the centripetal energy working upward approaching the middle of the loop, and the sound force of the chair and the pilot also upward. The pilot's weight (mg) is earthward. From Newton's second law:

?F(c) = ma(c) = n - mg

n = mg + ma(c)

= m[g + a(c)]

Since centripetal acceleration equals v² / r, the equalization enhances:

n = m[g + (v² / r)]

Answer:

0 because j×j =0

Explanation:

so a×b is 0

This question is incomplete, the complete question is;

A football quarterback throws a 0.408 kg football for a long pass. While in the motion of throwing, the quarterback moves the ball 1.909 m, starting from rest, and completes the motion in 0.439 s. Assuming the acceleration is constant, what force does the quarterback apply to the ball during the pass ;

a) F_throw = 8.083 N

b) F_throw = 9.181 N

c) F_throw = 2.284 N

d) F_throw = 16.014 N

e) None of these is correct

Answer:

the quarterback applied a force of 8.083 N to the ball during the pass

so Option a) F_throw = 8.083 N is the correct answer

Explanation:

Given that;

m = 0.408 kg

d = 1.909 m

u = 0 { from rest}

t = 0.439 s

Now using Kinetic equation

d = ut + 1/2 at²

we substitute

1.909 = (0 × 0.439) + 1/2 a(0.439)²

1.909 = 0 + 0.09636a

1.909 = 0.09636a

a = 1.909 / 0.09636

a = 19.8111 m/s²

Now force applied will be;

F = ma

we substitute

F = 0.408 ×  19.8111

F = 8.0828 ≈ 8.083 N

Therefore the quarterback applied a force of 8.083 N to the ball during the pass

so Option a) F_throw = 8.083 N is the correct answer

Answer:

The second distance of the sound from the source is 431.78 m..

Explanation:

Given;

first distance of the sound from the source, r₁ = 1.48 m

first sound intensity level, I₁ = 120 dB

second sound intensity level, I₂ = 70.7 dB

second distance of the sound from the source, r₂ = ?

The intensity of sound in W/m² is given as;

[tex]dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 = Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2[/tex]

[tex]For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 = Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2[/tex]

The second distance, r₂, can be determined from sound intensity formula given as;

[tex]I = \frac{P}{A}\\\\I = \frac{P}{\pi r^2}\\\\Ir^2 = \frac{P}{\pi }\\\\I_1r_1^2 = I_2r_2^2\\\\r_2^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2}} \\\\r_2 = \sqrt{\frac{(1)(1.48^2)}{(1 \times \ 10^{-4.93})}}\\\\r_2 = 431.78 \ m[/tex]

Therefore, the second distance of the sound from the source is 431.78 m.

Answer:

(A)

Explanation:

Answer:

275 x 10"N

Explanation:

Answer:

The final velocity will be half of the initial velocity of the spacecraft.

Explanation:

Angular momentum is conserved for the circular force motion and central force motion.

Considering

L = MVR = Constant

Where

M = Mass of the object

V = Velocity of the object

r = radius of circle

We know that

V = [tex]\frac{1}{R}[/tex]

So,

[tex]\frac{V_{2} }{V_{1} }[/tex] = [tex]\frac{R_{1} }{R_{2} }[/tex]

As per the given data

[tex]R_{1}[/tex] = Initial Radius = 300 km

[tex]R_{2}[/tex] = Final Radius = 600 km

[tex]V_{1}[/tex] = Initial Velocity =

[tex]V_{2}[/tex] = Final Velocity =

Placing values in the formula

[tex]\frac{V_{2} }{V_{1} }[/tex] = [tex]\frac{300 km}{600 km }[/tex]

[tex]\frac{V_{2} }{V_{1} }[/tex] = [tex]\frac{1}{2}[/tex]

[tex]{V_{2}[/tex] = [tex]\frac{1}{2} V_{1}[/tex]

Hence, The final velocity will half of the initial velocity of the spacecraft.

Answer:

Explanation: about 2 Mph when it falls and sorry if I’m wrong

Answer:

1.25s

1) 90.9m

Explanation:

From the question, we are given the following;

Speed v = 30m/s

Maximum height H = 45m

Required;

Time

Using the equation of motion S = ut + 1/2gt²

Substitute;

45= 30t + 1/2(9.8)t²

45 = 30t + 4.9t²

4.9t² + 30t - 45 = 0

Factorize;

t = -30 ±√30²-4(-45)(4.9)/2(4.9)

t = -30 ±√900+882/9.8

t = -30 ±42.21/9.8

t = -30 + 42.21/9.8

t = 12.21/9.8

t = 1.25secs

Hence the ball spent 1.25secs in air.

i) To get the horizontal distance, we will use the formula;

R = U√2H/g

R = 30√2(45)/9.8

R = 30√90/9.8

R = 30√9.18

R = 30(3.03)

R = 90.9m

Hence the horizontal distance is 90.9m

Answer:

The correct option is  c

Explanation:

From the question we are told that

    The ratio of the noise to human hearing threshold is  [tex]I = 150 000[/tex]

Generally from the equation given we have that

       dB =  10 log I

So

      dB =  10 log 150000  

The expression that shows a valid step in the process of finding the decibel level of the noise is dB = 10 log 150,000. Option C is correct

Given the equation for calculating the intensity of sound which is measured in decibel expressed as:

dB=10 log I

where;

I is the ratio of the sound to the human hearing threshold

Given that noise is 150,000 times greater than the human hearing threshold.

Substitute I = 150,000 into the expression above;

dB =  10 log 150,000

Hence the expression that shows a valid step in the process of finding the decibel level of the noise is dB = 10 log 150,000

Learn more here: https://brainly.com/question/14468501

Answer:

2.When they reach the bottom of the fall

Explanation:

The potential energy of the waterfall is maximum at the maximum height and decreases with decrease in height. Based on the law of conservation of mechanical energy, as the potential energy of the water fall is decreasing with  decrease in height of the fall, its kinetic energy will be increasing and the kinetic energy will be maximum at zero height (bottom of the fall).

Thus, the correct option is "2" When they reach the bottom of the fall

Answer:

yes it also does have but not in the exact form but it does 30 percent of respiration to produce rain

Explanation:

Force, F = 60 N

Angle, [tex]\theta=30^{\circ}[/tex]

We need to find horizontal and vertical component of the force.

Horizontal component,

[tex]F_x=F\cos\theta\\\\=60\times \cos(30)\\\\=51.96\ N[/tex]

Vertical component,

[tex]F_=F\sin\theta\\\\=60\times \sin(30)\\\\=30\ N[/tex]

So, the horizontal and vertical component are 51.96 N and 30 N respectively.

Answer:

The Starship Enterprise is powered by combining matter with antimatter. Suppose 1 kg of each are combined and ejected backward at the speed of light, what is the final speed of the Enterprise starting from rest? Assume that the mass of the Enterprise is 10,000 kg and the spaceship does not reach relativistic speed.

Answer:

Explanation:

The the mass of the car can be found by using the formula

[tex]m = \frac{f}{a} \\ [/tex]

f is the force

a is the acceleration

From the question we have

[tex]m = \frac{400}{0.85} \\ = 470.58823...[/tex]

We have the final answer as

Hope this helps you

Answer:

The value is  [tex]\frac{\Delta S }{ L} = - 0.0721 \ J / km[/tex]

Explanation:

From the question we are told that

   The  force is  [tex]F = 22.7 \ N[/tex]

    The value of room temperature is [tex]T = 298 \ K[/tex]

Generally the rate at which its entropy changes as it stretches is mathematically represented as

         [tex]\frac{\Delta S }{ L} = - \frac{F}{T}[/tex]

=>      [tex]\frac{\Delta S }{ L} = - \frac{21.5}{ 298 }[/tex]

=>      [tex]\frac{\Delta S }{ L} = - 0.0721 \ J / km[/tex]

Answer:

The mass of the other ball is 397.775 kg.

Explanation:

Gravitation is the force of mutual attraction that bodies experience due to the fact that they have a certain mass.

The universal law of gravitation is a classical physical law that describes the gravitational interaction between different bodies with mass.

The law was formulated by Newton, who deduced that the force with which two bodies of different masses are attracted only depends on the value of their masses and the square of the distance that separates them.

In other words, the Law of Universal Gravitation predicts that the force exerted between two bodies of masses M1 and M2 separated by a distance "d" is proportional to the product of their masses and inversely proportional to the square of the distance, that is:

[tex]F=G\frac{M1*M2}{d^{2} }[/tex]

where:

In this case:

Replacing:

[tex]1.65*10^{-7} N=6.67384*10^{-11} \frac{N*m^{2} }{kg^{2} }\frac{81 kg*M2}{(3.61 m)^{2} }[/tex]

Solving for M2:

[tex]M2=\frac{1.65*10^{-7} N*(3.61 m)^{2}}{6.67384*10^{-11} \frac{N*m^{2} }{kg^{2} }*81 kg}[/tex]

M2= 397.775 kg

The mass of the other ball is 397.775 kg.

Answer:

The answer is true, as gravity affects everything.

Answer:

Energy is applied on the charge to do work.

Explanation:

Work is only be done when one charge moves against the electric field of anther charge that require huge amount of energy because both have same charge and the force of repulsion occurs between them. Work is defined as the product of force and displacement so from this equation we can conclude that work is done when a force is applied on an object and it moves in the direction in which force is applied. If the force or energy is removed from the charge which is present in the electric field of another charge so it moves away from that charge and the work is also be done..

Answer:

5× 3 over 10 = 1.5

Explanation:

F = 1m/s × kg over n

The net work done on the block in 1 minutes is most nearly 1800 Joule.

A force must be applied in order for work to be completed, and there must also be motion or displacement in the force's direction. The amount of force multiplied by the distance moved in the force's direction is known as the work done by a force acting on an item.

Work has no direction and only magnitude. Work is a scalar quantity as a result.

Given that: A horizontal force F is used to pull a 5 kg block across a floor at constant speed of 3 m/s.

As the block is moving with uniform velocity; this force is equal to the frictional force in magnitude.

So, The net work done on the block in 1 minutes is =

force× displacement

= force × velocity × time

= 10 × 3 × 60 Joule

= 1800 Joule.

Learn more about work here:

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Answer:

120 km/hr

Explanation:

Let D be the distance between the rocket and the camera as the rocket is moving upwards. Let d be the distance the rocket moves and L be the distance between the camera and the base of the rocket = 4 km.

Now, at any instant, D² = d² + L²

= d² + 4²

= d² + 16 since the three distances form a right-angled triangle with the distance between the rocket and the camera as the rocket is moving upwards as the hypotenuse side.

differentiating the expression to find the rate of change of D with respect to time, dD/dt ,we have

d(D²)/dt = d(d² + 16)/dt

2DdD/dt = 2d[d(d)/dt]

dD/dt = 2d[d(d)/dt] ÷ 2D

Now d(d)/dt = vertical speed of rocket = 200 km/hr

dD/dt = 200d/D  [D = √(d² + 16)]

dD/dt = 200d/[√d² + 16]

Now substituting d = 3 km, the distance the rocket has risen into the equation, we have

dD/dt = 200(3)/[√(3² + 16)]

dD/dt = 600/[√(9 + 16)]

dD/dt = 600/√25

dD/dt = 600/5

dD/dt = 120 km/hr

So,  the speed at which the distance from the camera to the rocket changing when the rocket has risen 3 km is 120 km/hr.