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2 years ago

Can a golf ball bounce like a rubber ball? Due in 1 hour. 15 points.

Physics

1 answer:

klasskru [66]2 years ago

5 0

Answer:

So, I tested it out and came to a conclusion.

The rubber ball is quite bouncy due to the fact that it is made of rubber which is what makes it bounce back and high, but although the golfball does in fact bounce back it does not bounce as high as the rubber ball due to the fact that it is made of plastic.

Explanation:

Hope this helps you out!

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An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

5 0

3 years ago

An exoplanet with one half of Earth's mass and 50% of Earth's radius is discovered.

Georgia [21]

Answer:

The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet

Explanation:

The given parameters are;

The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M

The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R

The weight of the cadet on Earth = 800 N

$The \ weight, W =G\dfrac{M \times m}{R^{2}} = 800 \ N$

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

$W_1 =G\dfrac{\dfrac{M}{2} \times m}{ \left ( \dfrac{R}{2} \right ) ^{2}} = G\dfrac{\dfrac{M}{2} \times m \times 4}{ R ^{2}} = 2 \times G \times \dfrac{M \times m}{R^{2}} = 2 \times 800 \, N = 1,600 \, N$

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.

7 0

2 years ago

Would it be true that if you double the distance of an astronaut from a planet, the gravitational pull between them would be hal

velikii [3]

Answer:

Yes

Explanation:

Newton's law of universal gravitation is usually stated that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses(m1 and m2) and inversely proportional to the square of the distance between their centers(r).

F = Gm1m2/r²

This is a general physical law derived from

empirical observations by what Isaac Newton called inductive reasoning.

when distance is doubled the gravitational force will be reduced by quarter not half.

5 0

3 years ago

Read 2 more answers

The area of a rectangular park is 4 mi^2. The park has a width that is equal to "w", and a length that is 3 mi longer than the w

goldfiish [28.3K]

Answer:

l= 4 mi : width of the park

w= 1 mi : length of the park

Explanation:

Formula to find the area of the rectangle:

A= w*l Formula(1)

Where,

A is the area of the rectangle in mi²

w is the width of the rectangle in mi

l is the width of the rectangle in mi

Known data

A = 4 mi²

l = (w+3)mi Equation (1)

Problem development

We replace the data in the formula (1)

A= w*l

4 = w* (w+3)

4= w²+3w

w²+3w-4= 0

We factor the equation:

We look for two numbers whose sum is 3 and whose multiplication is -4

(w-1)(w+4) = 0 Equation (2)

The values of w for which the equation (2) is zero are:

w = 1 and w = -4

We take the positive value w = 1 because w is a dimension and cannot be negative.

w = 1 mi :width of the park

We replace w = 1 mi in the equation (1) to calculate the length of the park:

l= (w+3) mi

l= ( 1+3) mi

l= 4 mi

8 0

3 years ago

Which one?? please someone quick!

LiRa [457]

Answer:

i think its second law of motion.

Explanation:

4 0

2 years ago

Read 2 more answers