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3 years ago

If we increase the driving frequency in a circuit with purely resistive load, how do the amplitudes VR and IR change?

Physics

1 answer:

Lady_Fox [76]3 years ago

3 0

Answer:

V = I R

expression does not depend on the frequency, so when changing this quantity does not change the voltage or the current in the circuit.

Explanation:

In a purely resistive circuit, the Ohm relation, m for which

V = I R

We see that the previous expression does not depend on the frequency, so when changing this quantity does not change the voltage or the current in the circuit.

This does not happen in circuits with capacitors or inductors where there is an explicit dependence on frequency.

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A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

8 0

1 year ago

During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 6.22 s, how high does it rise? The accel

BigorU [14]

Answer:

47.4 m

Explanation:

When an object is thrown upward, it rises up, it reaches its maximum height, and then it goes down. The time at which it reaches its maximum height is half the total time of flight.

In this case, the time of flight is 6.22 s, so the time the ball takes to reach the maximum height is

$t=\frac{6.22}{2}=3.11 s$

Now we consider only the downward motion of the ball: it is a free fall motion, so we can find the vertical displacement by using the suvat equation

$s=ut+\frac{1}{2}gt^2$

where

s is the vertical displacement

u = 0 is the initial velocity

t = 3.11 s is the time

$g=9.8 m/s^2$ is the acceleration of gravity (taking downward as positive direction)

Solving the formula, we find

$s=\frac{1}{2}(9.8)(3.11)^2=47.4 m$

7 0

3 years ago

If you are watching TV programs from 3:50AM-5:30AM for how much hour you are watching

andreev551 [17]

Answer:

You would be watching tv for 1 hour and 40 mins

Explanation:

How much time take it take for 3:50 am to 5:30

6 0

3 years ago

Read 2 more answers

An object moving at a constant speed of 25 m/s is making a turn with a radius of curvature of 7 m (this is the radius of the "ki

prisoha [69]

Answer:

- 278.34 kg m/s^2

Explanation:

The rate of the change of momentum is the same as the force.

The force that an object feels when moviming in a circular motion is given by:

F = -mrω^2

Where ω is the angular speed and r is the radius of the circumference

Aditionally, the tangential velocity of the body is given as:

v = rω

The question tells us that

v = 25 m/s

r = 7m

mv = 78 kg m/s

Therefore:

m = (78 kg m/s) / (25 m/s) = 3.12 kg

ω = (25 m/s) / (7 m) = 3.57 (1/s)

Now, we can calculate the force or rate of change of momentum:

F = - (3.12 kg) (7 m)(3.57 (1/s))^2

F = - 278.34 kg m/s^2

4 0

3 years ago

A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of thebuilding. Ignore air resis

Bumek [7]

Answer:

a)3.5s

b)28.57m/S

c)34.33m/S

d)44.66m/S

Explanation:

Hello!

we will solve this exercise numeral by numeral

a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration

$Y= VoT+0.5gt^{2}$

where

Vo = Initial speed =0

T = time

g=gravity=9.81m/s^2

y = height=60m

solving for time

$Y=0.5gt^2\\t=\sqrt{\frac{Y}{0.5g} } \\t=\frac{60}{0.5(9.81)} \\$

T=3.5s

b)The horizontal speed remains constant since there is no horizontal acceleration. with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed

$V=\frac{x}{t} =\frac{100}{3.5}=28.57m/s$

to find the final vertical velocity we use the equations for motion with constant velocity as follows

Vf=Vo+g.t

Vf=0+(9.81 )(3.5)=34.335m/S

d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares

$V=\sqrt{Vx^2+Vy^2} =\sqrt{34.33^2+28.57^2} =44.67m/S$

7 0

2 years ago