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3 years ago

A body oscillates with 25hz what is its time period

Physics

1 answer:

Semmy [17]3 years ago

4 0

Answer:

4 seconds

Explanation:

The frequency of a body is the number of oscillations in one second. It is the number of cycles per unit time. The S.I unit of frequency is the Hertz (Hz).

The period of a body is the time taken to complete one oscillation. The period is inversely proportional to the frequency of the body. It is the reciprocal of frequency and the S.I unit is second (s).

A body oscillates with 25hz. Therefore the frequency (f) = 25 Hz.

The period (T) is given as:

$Period (T)=\frac{1}{frequency(f)} \\T=\frac{1}{f} =\frac{1}{0.25}=4\\T=4\ seconds$

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A 6- F capacitor is charged to 90 V and is then connected across a 700- resistor. What is the initial charge on the capacitor

Lorico [155]

Answer:

540C.

Explanation:

A capacitor of capacitance C when charged to a voltage of V will have a charge Q given as follows;

Q = CV ----------(i)

From the question, the initial charge on the capacitor is the charge on it before it was connected to the resistor. In other words, the initial charge on the capacitor will have a maximum value which can be calculated using equation (i) above.

Where;

C = 6F

V = 90V

Substitute these values into equation (i) as follows;

Q = 6 x 90

Q = 540 C

Therefore, the initial charge on the capacitor is 540C.

7 0

3 years ago

The force P is applied to the 45-kg block when it is at rest. Determine the magnitude and direction of the friction force exerte

cluponka [151]

Answer:

Check attachment for free body diagram of the question.

I used the free body diagram and the angles given in the diagram missing in the question above, but I used the data given in the above question.

Explanation:

Let frictional force be Fr acting down the plane

Let analyze the structure before inserting values

Using Newton's second law along the y-axis

ΣFy = Fnet = m•ay

Since the body is not moving in the y-direction, then ay = 0

N+PSinβ — WCosθ = 0

N+PSin20—441.45Cos15 = 0

N+PSin20—426.41 = 0

N = 426.41 — PSin20 , equation 1

The maximum Frictional force to be overcome is given as

Fr(max) = μsN

Fr(max) = 0.25(426.41 — PSin20)

Fr(max)= 106.6 —0.25•PSin20

Fr(max) = 106.6 — 0.08551P, equation 2

This is the maximum force that must be overcome before the body starts to move

Using Newton's law of motion in the x direction

Note, we took the upward direction up the plane as the direction of motion since the force want to move the block upward

Fnetx = ΣFx

Fnetx = P•Cosβ —W•Sinθ — Fr

Fnetx = P•Cos20—441.45•Sin15—Fr

Fnetx = 0.9397P — 114.256 — Fr

Equation 3

When Fnetx is positive, then, the body is moving up the plane, if Fnetx is negative, then, the maximum frictional force has not yet being overcome and the object is still i.e. not moving

a. When P = 0

From equation 2

Fr(max) = 106.6 — 0.08551P

Fr(max) = 106.6 — 0.08551(0)

Fr(max)= 106.6 N

So, 106.6N is the maximum force to be overcome

So, here the only force acting on the body is the weight and it acting down the plane, trying to move the body downward.

Wx = WSinθ

Wx = 441.45× Sin15

Wx = 114.256 N.

Since the force trying to move the body downward is greater than the maximum static frictional force, then the body is not in equilibrium, it is moving downward.

So, finding the magnitude of frictional force

From equation 1

N = 426.41 — PSin20 , equation 1

N = 426.41 N, since P=0

Then, using law of kinetic friction

Fr = μk • N

Fr = 0.22 × 426.41

Fr = 93.81 N.

b. Now, when P = 190N

From equation 2

Fr(max) = 106.6 — 0.08551(190)

Fr(max) = 106.6 —16.2469

Fr(max)= 90.353 N

So, 90.353 N is the maximum force to be overcome

Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight

Fnetx = P•Cosβ —W•Sinθ

Fnetx = 190Cos20 — 441.45Sin15

Fnetx = 64.29N

So the force moving the body up the incline plane is 64.29N

Fnetx < Fr(max)

Then, the frictional force has not being overcome yet.

Then, the body is in equilibrium.

Then, applying equation 3.

Fnetx = 0.9397P — 114.256 — Fr

Fnetx = 0, since the body is not moving

0 = 0.9397(190) —114.246 — Fr

Fr = 64.297 N

Fr ≈ 64.3N

c. When, P = 268N

From equation 2

Fr(max) = 106.6 — 0.08551(268)

Fr(max) = 106.6 —16.2469

Fr(max)= 83.68 N

So, 83.68 N is the maximum force to be overcome

Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight

Fnetx = P•Cosβ —W•Sinθ

Fnetx = 268Cos20 — 441.45Sin15

Fnetx = 137.58 N

So the force moving the body up the incline plane is 137.58 N

Fnetx > Fr(max)

Then, the frictional force has being overcome.

Then, the body is not equilibrium.

So, finding the magnitude of frictional force

From equation 1

N = 426.41 — 268Sin20 , equation 1

N = 334.75 N, since P=268N

Then, using law of kinetic friction

Fr = μk • N

Fr = 0.22 × 334.75

Fr = 73.64 N

d. The required force to initiate motion is the force when the block want to overcome maximum frictional force.

So, Fnetx = Fr(max)

Px — Wx = Fr(max)

From equation 1

Fr(max) = 106.6 — 0.08551P,

P•Cosβ-W•Sinθ = 106.6 — 0.08551P

P•Cos20 — 441.45•Sin15 = 106.6 — 0.08551P

P•Cos20—114.256=106.6 - 0.08551P

PCos20+0.08551P =106.6 + 114.256

1.025P=220.856

P = 220.856/1.025

P = 215.43 N

3 0

3 years ago

A delta connection has a voltage of 560 V connected to it. How much voltage is dropped across each phase

SashulF [63]

Answer:

E_Phase = 560V

Explanation:

The computation of the voltage i.e. dropped across each phase is shown below:

Given that

The delta connection line voltage is

E_line = 560 V

And, in the case of delta connection, the line voltage would be equivalent to the phase voltage

That means

E_Phase = E_Line

= 560 V

Hence, the voltage i.e. dropped across each phase is

E_Phase = 560V

5 0

3 years ago

This chart shows characteristics of three different waves, all with the same wavelength of 10 m but moving at different Which st

JulijaS [17]

Answer:

Wave X is moving the fastest.

Explanation:

As we know that the speed of the wave is given as the product of frequency and wavelength

so here we will have

$\lambda = 10 m$

now for wave X

$f = 6 Hz$

now the speed of this wave is

$v_x = 10 \times 6 = 60 m/s$

now for wave Y

$f = 0.5 Hz$

now the speed of this wave is

$v_y = 10 \times 0.5 = 5 m/s$

now for wave Z

$f = 2 Hz$

now the speed of this wave is

$v_z = 10 \times 2 = 20 m/s$

So correct answer will be

Wave X is moving the fastest.

3 0

4 years ago

Read 2 more answers

You are exploring a planet and drop a small rock from the edge of a cliff. In coordinates where the +y direction is downward and

Lelu [443]

Answer:

value of the acceleration of gravity on the planet is 5.00 m/s²

Explanation:

The problem is similar to a free fall exercise, with another gravity value, the expression they give us is the following:

y-yo = ½ gₐ t² (1)

They tell us that they make a squared time graph with the variation of the distance, it is appropriate to clarify this in a method to linearize a curve, which is plotted the nonlinear axis to the power that is raised, specifically, the linearization of a curve The square is plotted against the other variable.

Let's continue our analysis, as we have a linear equation, write the equation of the line.

y1 = m x1 + b (2)

where “y1” the dependent variable, “x1” the independent variable, “m” the slope and “b” the short point

In this case as the stone is released its initial velocity is zero which implies that b = 0,

We plot on the “y” axis the time squared “t²” and on the horizontal axis we place “y-yo”. To better see the relationship we rewrite equation 1 with this form

t² = 2 /gₐ (y-yo)

With the two expressions written in the same way, let's relate the terms one by one

y1 = t²

x1 = (y-yo)

m = 2/gap

b= 0

We substitute and calculate

m = 2/gp

gₐ = 2/m

gₐ = 2/ 0.400

gₐ = 5.00 m / s²

This is the value of the acceleration of gravity on the planet, note that the decimals are to keep the figures significant

6 0

3 years ago