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kotegsom [21]
3 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]3 years ago
6 0
The answer is 0.68 meters per second, swagteam
vova2212 [387]3 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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You have a small piece of iron at 25 °C and place it into a large container of water at 75 °C. Which of these could be the tempe
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What is the difference between a circuit and a current?
dimulka [17.4K]

Answer:

a circuit is a complete loop that carries a current in the form of electrons from negative to positive

Explanation:

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3 0
3 years ago
A horizontal force of 40N is needed to pull a 60kg box across the horizontal floor at which coefficient of friction between floo
Mazyrski [523]

Answer:

Coefficient of friction is 0.068.

Work done is 320~J.

Explanation:

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Mass of the box (m): 60 kg

Force needed (F): 40 N

The formula to calculate the coefficient of friction between the floor and the box is given by

F=\mu mg...................(1)

Here, \mu is the coefficient of friction and g is the acceleration due to gravity.

Substitute 40 N for F, 60 kg for m and 9.80 m/s² for g into equation (1) and solve to calculate the value of the coefficient of friction.

40 N=\mu\times60 kg\times9.80 m/s^{2} \\~~~~~\mu=\frac{40 N}{60 kg\times9.80 m/s^{2}}\\~~~~~~~=0.068

The formula to calculate the work done in overcoming the friction is given by

W=Fd..........................(2)

Here, W is the work done and d is the distance travelled.

Substitute  40 N for F and 8 m for d into equation (2) to calculate the work done.

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8 0
3 years ago
An elementary particle of mass m completely absorbs a photon, after which its mass is 1.01m. (a) what was the energy of the inco
sdas [7]
A.) We use the famous equation proposed by Albert Einstein written below:

E = Δmc²
where
E is the energy of the photon
Δm is the mass defect, or the difference of the mass before and after the reaction
c is the speed of light equal to 3×10⁸ m/s

Substituting the value:

E = (1.01m - m)*(3×10⁸ m/s) = 0.01mc² = 3×10⁶ Joules

b) The actual energy may be even greater than 3×10⁶ Joules because some of the energy may have been dissipated. Not all of the energy will be absorbed by the photon. Some energy would be dissipated to the surroundings.
8 0
3 years ago
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