1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ale4655 [162]
2 years ago
13

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!

You might be interested in
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
Other questions:
  • Five hundred joules of heat are added to a closed system. The initial internal energy of the system is 87 J, and the final inter
    5·2 answers
  • What type of wave passes through the spring in the frog toy? Why?
    7·1 answer
  • Can someone please help with this?
    13·1 answer
  • a 100gm copper block is heated in boiling water for 10min and then it is dropped into 150gm of water at 30 C in a 200gm calorime
    10·1 answer
  • A 140 g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.537 m. The ball makes 2.27 revolut
    5·1 answer
  • what is the result of 6.2×10 to the fourth power times 3.3×10 to the second power express in scientific notation
    14·1 answer
  • A yellow ball with a mass of 2 kg is rolling across the floor at 3 m/s. A red ball with a mass of 3 kg is rolling across the sam
    15·2 answers
  • Ohm's law is not applicable to​
    13·1 answer
  • Why is the sky blue?
    15·2 answers
  • A lightbulb has a power rating of 6.75 W and draws a current of 0.75 A when connected to a battery.
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!