PHYSICS COLLEGE

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.250 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.960 m above the floor. (a) Calculate his velocity (in m/s) when he leaves the floor. 4.33 Correct: Your answer is correct. m/s (b) Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.250 m.

Answers

Answer 1

Answer:

Answer:

a. 3.73 m/s b. 27.8 m/s²

Explanation:

(a) Calculate his velocity (in m/s) when he leaves the floor.

Using the conservation of energy principles,

Potential energy gained by basketball player = kinetic energy loss of basket ball player

So, ΔU + ΔK = 0

ΔU = -ΔK

mg(h' - h) = -1/2m(v'² - v²)

g(h' - h) = -1/2(v'² - v²) where g = acceleration due to gravity = 9.8 m/s², h' = 0.960 m, h = 0.250 m, v' =0 m/s (since the basketball player momentarily stops at h' = 0.960 m) and v = velocity with which the basketball player leaves the floor

Substituting the values of the variables into the equation, we have

9.8 m/s²(0.960 m - 0.250 m) = -1/2((0 m/s)² - v²)

9.8 m/s²(0.710 m) = -1/2(-v²)

6.958 m²/s² = v²/2

v² = 2 × 6.958 m²/s²

v² = 13.916 m²/s²

v = √(13.916 m²/s²)

v = 3.73 m/s

(b) Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.250 m.

Using v² = u² + 2as where u = initial speed of basketball player before lengthening = 0 m/s, v = final speed of basketball player after lengthening = 3.73 m/s, a = acceleration during lengthening and s = distance moved during lengthening = 0.250 m

So, making, a subject of the formula, we have

a = (v² - u²)/2s

Substituting the values of the variables into the equation, we have

a = ((3.73 m/s)² - (0 m/s)²)/(2 × 0.250 m)

a = (13.913 m²/s² - 0 m²/s²)/(0.50 m)

a = 13.913 m²/s²/(0.50 m)

a = 27.83 m/s²

a ≅ 27.8 m/s²

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A 1200-kg cannon suddenly fires a 100-kg cannonball at 35 m/s. what is the recoil speed of the cannon? assume that frictional forces are negligible and the cannon is fired horizontally.

Answers

Answer:

Recoil velocity of cannon = 2.92 m/s

Explanation:

By law of conservation of momentum, we have momentum of cannon = momentum of cannonball.

Mass of cannon = 1200 kg

Mass of cannon ball = 100 kg

Velocity of cannon ball = 35 m/s

We have, Momentum of cannon = momentum of cannon ball

1200 x v = 100 x 35

v =3500/1200 = 2.92 m/s

Recoil velocity of cannon = 2.92 m/s

Final answer:

The recoil speed of the cannon is 2.92 m/s.

Explanation:

To find the recoil speed of the cannon, we can use the conservation of momentum. The initial momentum of the cannon and cannonball system is zero since the cannon is at rest before firing. The final momentum is the sum of the momenta of the cannon and cannonball after firing. Using the equation:

Initial momentum = Final momentum

(mass of cannon) x (recoil speed of cannon) = (mass of cannonball) x (velocity of cannonball)

Plugging in the given values:

(1200 kg) x (recoil speed of cannon) = (100 kg) x (35 m/s)

Solving for the recoil speed of the cannon:

recoil speed of cannon = (100 kg x 35 m/s) / 1200 kg = 2.92 m/s

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If a single circular loop of wire carries a current of 61 A and produces a magnetic field at its center with a magnitude of 1.70 10-4 T, determine the radius of the loop.

Answers

Answer:

The radius is $R = 0.22 5 \ m$

Explanation:

From the question we are told that

The current is $I = 61 \ A$

The magnetic field is $B = 1.70 *10^(-4) \ T$

Generally the magnetic field produced by a current carrying conductor is mathematically represented as

$B = (\mu_o * I)/(2 * R )$

=> $R = (\mu_o * I )/( 2 * B )$

Here $\mu_o$ is the permeability of free space with value $\mu_o = 4\pi * 10^(-7) N/A^2$

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=> $R = 0.22 5 \ m$

In part (a), suppose that the box weighs 128 pounds, that the angle of inclination of the plane is θ = 30°, that the coefficient of sliding friction is μ = 3 /4, and that the acceleration due to air resistance is numerically equal to k m = 1 3 . Solve the differential equation in each of the three cases, assuming that the box starts from rest from the highest point 50 ft above ground. (Assume g = 32 ft/s2 and that the downward velocity is positive.)

Answers

Answer:

v(t) = 21.3t

v(t) = 5.3t

$v(t) = 48 -48 e ^{ (t)/(9)}$

Explanation:

When no sliding friction and no air resistance occurs:

$m(dv)/(dt) = mgsin \theta$

where;

$(dv)/(dt) = gsin \theta , 0 < \theta < ( \pi)/(2)$

Taking m = 3 ; the differential equation is:

$3 (dv)/(dt)= 128*(1)/(2)$

$3 (dv)/(dt)= 64$

$(dv)/(dt)= 21.3$

By Integration;

$v(t) = 21.3 t + C$

since v(0) = 0 ; Then C = 0

v(t) = 21.3t

ii)

When there is sliding friction but no air resistance ;

$m (dv)/(dt)= mg sin \theta - \mu mg cos \theta$

Taking m =3 ; the differential equation is;

$3 (dv)/(dt)=128*(1)/(2) -(√(3) )/(4)*128*(√(3) )/(4)$

$(dv)/(dt)= 5.3$

By integration; we have ;

v(t) = 5.3t

iii)

To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :

$m (dv)/(dt)= mg sin \theta - \mu mg cos \theta - kv$

The differential equation is :

= $3 (dv)/(dt)=128*(1)/(2) - ( √( 3))/(4)*128 *( √( 3))/(2)-(1)/(3)v$

= $3 (dv)/(dt)=16 -(1)/(3)v$

By integration

$v(t) = 48 + Ce ^{(t)/(9)$

Since; V(0) = 0 ; Then C = -48

$v(t) = 48 -48 e ^{ (t)/(9)}$

why did thomson's from experermenting with cathode rays cause a big change in scientific thought about atoms

Answers

Answer:

His results gave the first evidence that atoms were made up of smaller particles.

How far above the bottom of the tank could a second hole be cut so that the stream emerging from it could have the same range as for the first hole

Answers

Answer: hello your question is incomplete below is the complete question

Water stands at a depth H in a large open tank whose side walls are vertical . A hole is made in one of the walls at a depth h below the water surface. Part B How far above the bottom of the tank could a second hole be cut so that the stream emerging from it could have the same range as for the first hole

answer :

At Height ( h ) from the bottom of Tank

Explanation:

Determine how far above the bottom of the tank a second hole be cut

For the second hole to have the same range as the first hole

Range of first hole = Velocity of efflux of water * time of fall of water

= √ (2gh) * √( 2g (H - h) / g)

= √ ( 4(H-h) h)

Hence the Height at which the second hole should be placed to exercise same range of stream emerging = h from the bottom of the Tank

Final answer:

The second hole should be cut at the same height as the first hole to have the same range for the stream.

Explanation:

In order for the stream emerging from the second hole to have the same range as the first hole, the second hole should be cut at the same height as the first hole. This is because the range of the stream depends on the initial velocity and the vertical distance traveled. If the second hole is higher or lower than the first hole, the vertical distance traveled will be different and the range of the stream will be affected.

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As you are leaving a building, the door opens outward. If the hinges on the door are on your right, what is the direction of the angular velocity of the door as you open it?a. up to the ceiling/sky.
b. down to the floor/ground.
c. to your left.
d. to your right.

Answers

Answer:

b. down to the floor/ground

Explanation:

The direction of the angular velocity can be easily found by the use of right hand rule. This rule states that when we curl the fingers of right hand in the direction of the rotation of the object, the direction of the thumb gives the direction of angular velocity. We apply this rule to our situation. Since, the door opens outward and the hinges are on right, it means that the rotation of door is clockwise. So, when we curl the fingers of right hand in clock wise direction, the thumb will point in downward direction. So, the direction of angular velocity will be down to the floor/ground.

The correct option is:

b. down to the floor/ground

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