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3 years ago

14

What quantity measures the number of complete cycles an oscillation makes per second? A. period B. amplitude C. frequency D. for

ce

1 answer:

Free_Kalibri [48]3 years ago

8 0

Hello, there Jcparris

Your answer is going to be C. Frequency

If my answer helped please leave a thank rate 5 stars and the most important rank me braiinliest thank you and have a great day!

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A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the mer

son4ous [18]

Solution :

Given data :

Mass of the merry-go-round, m= 1640 kg

Radius of the merry-go-round, r = 7.50 m

Angular speed, $\omega = \frac{1}{8}$ rev/sec

$=\frac{2 \pi \times 7.5}{8}$ rad/sec

= 5.89 rad/sec

Therefore, force required,

$F=m.\omega^2.r$

$$$=1640 \times (5.89)^2 \times 7.5$

= 427126.9 N

Thus, the net work done for the acceleration is given by :

W = F x r

= 427126.9 x 7.5

= 3,203,451.75 J

6 0

2 years ago

Car A hits car B (initially at rest and of equal mass) from behind while going 15 m/s Immediately after the collision, car B mov

Mamont248 [21]

Given :

Initial speed of car A is 15 m/s and initial speed of car B is zero.

Final speed of car A is zero and final speed of car B is 10 m/s.

To Find :

What fraction of the initial kinetic energy is lost in the collision.

Solution :

Initial kinetic energy is :

$K.E_i = \dfrac{15^2m}{2} + 0\\\\K.E_i = \dfrac{225 m}{2}$

Final kinetic energy is :

$K.E_f = \dfrac{10^2m}{2} + 0\\\\K.E_f = \dfrac{100m}{2}$

Now, fraction of initial kinetic energy loss is :

$Loss = \dfrac{\dfrac{225m}{2}-\dfrac{100m}{2}}{\dfrac{100m}{2}}\\\\Loss = \dfrac{125}{100}\\\\Loss = 1.25$

Therefore, fraction of initial kinetic energy loss in the collision is 1.25 .

6 0

3 years ago

The spacing between the plates of a 1.0 μF capacitor is 0.050 mm. a) What is the surface area of the plates? b) How much charge

nataly862011 [7]

Answer:

(a) surface area of the plate will be equal to $1.129m^2$

(b) Charge on the capacitor is equal to $1.5\times 10^{-6}C$

Explanation:

We have given spacing between the plates d = 0.05 mm = $5\times 10^{-5}m$

Value of capacitance $C=1\mu F=10^{-6}F$

(A) Capacitance of a parallel plate capacitor is equal to $C=\frac{\epsilon _0A}{d}$

So $10^{-6}=\frac{8.85\times 10^{-12}\times A}{10^{-5}}$

$A=1.129m^2$

So surface area of the plate will be equal to $1.129m^2$

(B) It is given that capacitor is charged by 1.5 volt

So voltage V = 1.5 volt

Charge on the capacitor is equal to $Q=CV$

So $Q=1.5\times 10^{-6}C$

5 0

3 years ago

Addition of a lubricant such as oil to a surface will __ friction ?

Mashcka [7]

Answer:

decrease

Explanation:

Addition of a lubricant such as oil to a surface will decrease friction. This makes the surface more greasy or slippery. This is very much efficient in machines, for example.

5 0

3 years ago

A small block of mass m on a horizontal frictionless surface is attached to a horizontal spring that has force constant k. The b

Rashid [163]

Answer:

$v_{max} = |d|\cdot \sqrt{\frac{k}{m} }$

Explanation:

The maximum speed of the block occurs when spring has no deformation, that is, there is no elastic potential energy, which can be remarked from appropriate application of the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot d^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$k \cdot d^{2} = m\cdot v^{2}$

$v_{max} = |d|\cdot \sqrt{\frac{k}{m} }$

5 0

3 years ago