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kotegsom [21]
4 years ago
15

A 0.023 kg beetle is sitting on a record player 0.15 m from the center of the record. if it takes 0.070 n of force to keep the b

eetle moving in a circle on the record, what is the tangential speed of the beetle? 0.68 m/s 0.46 m/s 0.33 m/s 0.11 m/s
Physics
2 answers:
Margaret [11]4 years ago
6 0
The answer is 0.68 meters per second, swagteam
vova2212 [387]4 years ago
6 0

Answer:

Tangential speed is 0.68 m/s

Explanation:

It is given that,

Mass of the beetle, m = 0.023 kg

It is placed at a distance of 0.15 m from the center of record i.e. r = 0.15 m.

If it takes 0.070 n of force to keep the beetle moving in a circle on the record i.e. centripetal force acting on it is, F = 0.070 N

We have to find the tangential speed of the beetle. The formula for centripetal force is given by :

F=\dfrac{mv^2}{r}

v is tangential speed

v=\sqrt{\dfrac{Fr}{m}}

v=\sqrt{\dfrac{0.070\times 0.15}{0.023}}

v = 0.675 m/s

or

v = 0.68 m/s

Hence, the correct option for tangential speed is (A).

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because the light bounces off of the water creating a bend. the answer is d.

6 0
3 years ago
→Fo
irakobra [83]

Answer:

The mass of the block, M =T/(3a +g)  Kg

Explanation:

Given,

The upward acceleration of the block a = 3a

The constant force acting on the block, F₀ = Ma = 3Ma

The mass of the block, M = ?

In an Atwood's machine, the upward force of the block is given by the relation

                                     Ma = T - Mg

                                      M x 3a = T - Ma    

                                    3Ma + Mg = T

                                       M = T/(3a +g)  Kg

Where 'T' is the tension of the string.

Hence, the mass of the block in Atwood's machine is, M = T/(3a +g)  Kg

3 0
3 years ago
Read 2 more answers
A rotating space station is said to create “artificial gravity”—a loosely-defined term used for an acceleration that would be cr
Grace [21]

Answer: 0.313 rad/s

Explanation:

The equation that relates the velocity V and the angular velocity \omega in the uniform circular motion is:

V=\omega.r   (1)

Where r=d/2=100m is the radius of the space station (with a diaeter of 200m) that describes the uniform circular motion.

Isolating \omega from (1):

\omega=\frac{V}{r}  (2)

On the other hand, we are told the “artificial gravity” produced by the cetripetal acceleration a_{c} is 9.8m/s^{2}, and is given by the following equation:

a_{c}=\frac{V^{2}}{r}   (3)

Isolating V:

V=\sqrt{a_{c}.r}   (4)

V=31.3049m/s   (5)

Substitutinng (5) in (2):

\omega=\frac{31.3049m/s}{100m}  (6)

\omega=0.313rad/s This is the angular velocity that would produce an “artificial gravity” of 9 9.8m/s^{2}.

6 0
3 years ago
A spring with spring constant k is suspended vertically from a support and a mass m is attached. The mass is held at the point w
garik1379 [7]

Answer:

The oscillation frequency of the spring is 1.66 Hz.

Explanation:

Let m is the mass of the object that is suspended vertically from a support. The potential energy stored in the spring is given by :

E_s=\dfrac{1}{2}kx^2

k is the spring constant

x is the distance to the lowest point form the initial position.

When the object reaches the highest point, the stored potential energy stored in the spring gets converted to the potential energy.

E_P=mgx

Equating these two energies,

\dfrac{1}{2}kx^2=mgx

\dfrac{k}{m}=\dfrac{2g}{x}.............(1)

The expression for the oscillation frequency is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

f=\dfrac{1}{2\pi}\sqrt{\dfrac{2g}{x}} (from equation (1))

f=\dfrac{1}{2\pi}\sqrt{\dfrac{2\times 9.8}{0.18}}

f = 1.66 Hz

So, the oscillation frequency of the spring is 1.66 Hz. Hence, this is the required solution.

8 0
4 years ago
What causes the earth to rotate
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The Earth rotates due to the manner in which it was formed. Earth began its rotation from the spinning movement of the accretion disk. In short, the earth rotates because of angular momentum caused by the accretion disk.
4 0
4 years ago
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