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Sedbober [7]
3 years ago
12

If you drop a bouncing ball from a height of 40 centimeters, explain why it can only bounce back up to a height of less than 40

centimeters.
Physics
1 answer:
iren2701 [21]3 years ago
8 0

Answer:

Due to energy loss while collision ball will not reach to same height while if there is no energy loss then in that case ball will reach to same height

Explanation:

As we know that initially ball is held at height h = 40 cm

So here we can say that kinetic energy of the ball is zero and potential energy is given as

U = mgH

now when strike with the ground then its its fraction of kinetic energy is lost in form of other energies

So the ball will left rebound with smaller energy and hence it will reach to height less than the initial height

While if we assume that there is no energy loss during collision then in that case ball will reach to same height again

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You drive a car 1600 ft to the east, then 2500 ft to the north. The trip took 2.5 minutes. What was the magnitude of your averag
elena-14-01-66 [18.8K]

Answer:

Velocity=6.03m/s

Explanation:

Given data

Time t=2.5 minutes=150 seconds

Distance A=1600 ft=487.68 m........east

Distance B=2500 ft=762m ........north

To find

Average velocity

Solution

First we need to find the resultant distance magnitude.To find that we apply Pythagorean theorem to find hypotenuse

So

A^{2}+B^{2}=C^{2}\\  C=\sqrt{A^{2}+B^{2}}\\ C=\sqrt{(487.68m)^{2}+(762m)^{2}}\\ C=904.7m

Velocity=\frac{Distance}{Time}\\Velocity=\frac{904.7m}{150s}\\Velocity=6.03m/s

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3 years ago
a 1-kg discus is thrown with a velocity of 19 m/s at an angle of 35 degrees from the vertical direction. calculate the vertical
nlexa [21]

Answer:

Vx =  10.9 m/s ,  Vy = 15.6 m/s

Explanation:

Given velocity V= 19 m/s

the angle 35 ° is taken from Y-axis so the angle with x-axis will be 90°-35° = 55°

θ = 55°

to Find Vx = ? and Vy= ?

Vx = V cos θ

Vx = 19 m/s  × cos 55°

Vx =  10.9 m/s

Vx = V sin θ

Vy = 19 m/s  × sin 55°

Vy = 15.6 m/s

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3 years ago
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The magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is th
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Answer:

1.92 x 10⁻¹²J

Explanation:

The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e

F = ma

Where;

a = \frac{v^2}{r}             [v = linear velocity, r = radius of circular path]

=> F = m\frac{v^2}{r}           ------------(i)

We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;

F = q v B sin θ       ----------(ii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = the angle between the velocity and the magnetic field.

Combining equations (i) and (ii) gives

m\frac{v^2}{r} = q v B sin θ           [θ = 90° since the proton is orbiting at the maximum orbital radius]

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=> m\frac{v^2}{r} = q v B

Divide both side by v;

=> m\frac{v}{r} = qB

Make v subject of the formula

v = \frac{qBr}{m}

From the question;

B = 1.25T

m = mass of proton = 1.67 x 10⁻²⁷kg

r = 0.40m

q = charge of a proton = 1.6 x 10⁻¹⁹C

Substitute these values into equation(iii) as follows;

v = \frac{(1.6*10^{-19})(1.25)(0.4)}{(1.67*10^{-27})}

v = 4.79 x 10⁷m/s

Now, the kinetic energy, K, is given by;

K = \frac{1}{2}mv²

m = mass of proton

v = velocity of the proton as calculated above

K = \frac{1}{2}(1.67*10^{-27} * (4.79 * 10^7)^2 )

K = 1.92 x 10⁻¹²J

The kinetic energy is 1.92 x 10⁻¹²J

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