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ehidna [41]
2 years ago
15

A compact disc has a radius of 6 cm. If the disc rotates about its axis at a constant angular speed of 5 rev/s, what is the tota

l distance traveled by a point on the rim in 40 minutes
Physics
1 answer:
mojhsa [17]2 years ago
3 0

The total distance traveled  is mathematically given as

d=4.5km

<h3>What is the total distance traveled?</h3>

Question Parameters:

A compact disc has a radius of 6 cm.

the constant angular speed of 5 rev/s

time 40 min

Generally, the equation for the linear speed   is mathematically given as

v=w*r

Therefore

v=10 pi * 6

v=60pi cm/s

In conclusion, the equation of the distance

distance = v*t

d=60pi * 2400

d=452389.342 cm

d=4.5km

Read more about Distance

brainly.com/question/4931057

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State Newtons first law of motion.
Fofino [41]

Answer:

An object at rest will stay at rest and an object in motion will stay in motion unless it is acted upon by an unbalanced/external force.

8 0
3 years ago
Transverse waves are sent along a 4.50 m long string with a speed of 85.00 m/s. The string is under a tension of 20.00 N. What i
frutty [35]

Answer:

m = 0.0125 kg

Explanation:

Let us apply the formula for the speed of a wave on a string that is under tension:

v = \sqrt{\frac{F}{\mu} }

where F = tension force

μ = mass per unit length

Mass per unit length is given as:

μ  = m / l

where m = mass of the string

l = length of the string

This implies that:

v = \sqrt{\frac{F}{m/l} }\\ \\v = \sqrt{\frac{F * l}{m} }

Let us make mass, m, the subject of the formula:

v^2 = \frac{F * l}{m}\\\\m = \frac{F * l}{v^2}

From the question:

F = 20 N

l = 4.50 m

v = 85 m/s

Therefore:

m = \frac{20 * 4.5}{85^2}\\\\m = \frac{90}{7225}\\ \\m = 0.0125 kg

5 0
2 years ago
Which type of force can act through empty space?
katrin2010 [14]

D. Magnetism this is shown by the planet's magnetic pull on moons and space debris.

8 0
3 years ago
Olivia put a glass of water in the freezer. She left it there for three hours. When she returned, the water had turned to ice. W
Darya [45]
A. freezing, when water turns to ice the water is turning from a liquid to a solid.
7 0
3 years ago
A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia
aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the <u>initial point</u> is 19820 J

2) The potential energy at <u>point A</u> is 19620 J

3) The kinetic energy at <u>point B</u> is 10010 J

4) The potential energy at <u>point C</u> is zero

5) The kinetic energy at <u>point C</u> is 19820 J

6) The velocity of the roller coaster at <u>point C</u> is 19.91 m/s

1) The total energy for the roller coaster at the <u>initial point</u> can be found as follows:

E_{t} = KE_{i} + PE_{i}

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The<em> total energy</em> is:

E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J

Hence, the total energy for the roller coaster at the <u>initial point</u> is 19820 J.

   

2) The <em>potential energy</em> at point A is:

PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J

Then, the potential energy at <u>point A</u> is 19620 J.

3) The <em>kinetic energy</em> at point B is the following:

KE_{A} + PE_{A} = KE_{B} + PE_{B}

KE_{B} = KE_{A} + PE_{A} - PE_{B}

Since

KE_{A} + PE_{A} = KE_{i} + PE_{i}

we have:

KE_{B} = KE_{i} + PE_{i} - PE_{B} =  19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J

Hence, the kinetic energy at <u>point B</u> is 10010 J.

4) The <em>potential energy</em> at <u>point C</u> is zero because h = 0 meters.

PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J

5) The <em>kinetic energy</em> of the roller coaster at point C is:

KE_{i} + PE_{i} = KE_{C} + PE_{C}            

KE_{C} = KE_{i} + PE_{i} = 19820 J      

Therefore, the kinetic energy at <u>point C</u> is 19820 J.

6) The <em>velocity</em> of the roller coaster at point C is given by:

KE_{C} = \frac{1}{2}mv_{C}^{2}

v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s

Hence, the velocity of the roller coaster at <u>point C</u> is 19.91 m/s.

Read more here:

brainly.com/question/21288807?referrer=searchResults

I hope it helps you!

3 0
3 years ago
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