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vesna_86 [32]
4 years ago
10

1) An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and

inductor, or a capacitor. At time t = 0 the voltage is zero and increasing toward a maximum. At time t = T/4 the current in the unknown element is equal to zero and at time t = T/2 the current is I = -Imax (where Imax is the maximum current amplitude). What is the unknown element?
Physics
1 answer:
Lady_Fox [76]4 years ago
6 0

Answer: Capacitor

Explanation:

Given the following ;

At time, t = T/4 ; I = 0

At time, t = T/2 ; I = - Imax

Where T = period

We can confirm what the unknown element is by using the relation, such that the parameters satisfy the stated condition ;

Recall;

I = Iosin(wt + π/2) [for CAPACITOR]

w = 2π/T

At t = T/4

I = Iosin(2π/T(T/4) + π/2)

I = Iosin(π/2 + π/2)

I = Iosin(0)

I = 0

At t = T/2

I = Iosin(2π/T(T/2) + π/2)

I = Iosin(π + π/2)

I = Iosin(0)

I = Iosin(3π/2)

I = Iosin(540/2)

I = Iosin(270)

I = -Io

Note : Io = Imax

Both conditions are met, Hence, the unknown element is a CAPACITOR.

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6.
Vikentia [17]

Answer:

Explanation:

Givens

vi = 0

a = 9.81

d = 4.50 m

vf = ?

Formula

vf^2 = vi^2 + 2 * a * d

Solution

Substitute the knowns into the formula

vf^2 =0 +  2 * 9.81 * 4.50

vf^2 = 88.29                          Take the square root of both sides.

sqrt(vf^2) = sqrt(88.29)    

vf = 9.40 m/s

3 0
3 years ago
Two people, one with mass m1 and the other with mass m2, stand on a stationary sled with mass M on a frozen lake. Assume that th
lozanna [386]

Answer:

Part a)

Velocity of sled

v = \frac{m_1 s}{m_1 + m_2 + M}

velocity of first man who jump off

v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}

Part b)

Velocity of sled

v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s

Also the speed of second person is given as

v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}

Part c)

change in kinetic energy of sled + two people is given as

KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2

Explanation:

As we know that here we we consider both people + sled as a system then there is no external force on it

So here we can use momentum conservation

since both people + sled is at rest initially so initial total momentum is zero

now when first people will jump with relative velocity "s" then let say the sled + other people will move off with speed v

so by momentum conservation we have

0 = m_1(v - s) + (m_2 + M)v

v = \frac{m_1 s}{m_1 + m_2 + M}

so velocity of the sled + other person is

v = \frac{m_1 s}{m_1 + m_2 + M}

velocity of first man who jump off

v_1 = \frac{m_1 s}{m_1 + m_2 + M} - s

v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}

Part b)

now when other man also jump off with same relative velocity

so let say the sled is now moving with speed vf

so by momentum conservation we have

(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) = m_2(v_f - s) + Mv_f

(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) + m_2s = (m_2 + M)v_f

Now we have

v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s

Also the speed of second person is given as

v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s - s

v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}

Part c)

change in kinetic energy of sled + two people is given as

KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2

here we know all values of speed as we found it in part a) and part b)

4 0
3 years ago
A Texas rancher wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in
xz_007 [3.2K]
The answer:
the full question is as follow:
 <span>A Texas rancher wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in Figure below , where A = 4.90 km and θC = 15°. He then correctly calculates the length and orientation of the fourth side D. What is the magnitude and direction of vector D? 

As shown in the figure, 
A + B + C + D = 0, so to find the </span>magnitude and direction of vector D, we should follow the following method:
 D = 0 - (A + B + C) , 
let  W = - (A + B + C), so the magnitude and direction of vector D is the same of the vector W characteristics

Magnitude
 A + B + C = <span> (4.90cos7.5 - 2.48sin16 - 3.02cos15)I</span>
<span>+ (-4.9sin7.5 + 2.48cos16 + 3.02sin15)J
</span>= 1.25I +2.53J
the magnitude of W= abs value of (A + B + C) = sqrt (1.25² + 2.53²)
= 2.82
 
the direction of D can be found by using Dx and Dy value
we know that     tan<span>θo = Dx / Dy = 1.25 / 2.53 =0.49
</span>tanθo =0.49 it implies θo = arctan 0.49 = 26.02°

direction is 26.02°

4 0
3 years ago
a 3.5 kg mass has an initial velocity of 4.0 m/s. How much force is required to increase the velocity of the mass to 8.0 m/s in
Nadya [2.5K]

Answer:

<em>4.67 N of force are required</em>

Explanation:

The second Newton's law states the net force exerted on a body of mass m that has an acceleration a, is given by:

F=m.a

On the other hand, the kinematics equations relate the acceleration with the change of speed over time, expressed as:

\displaystyle a=\frac{v_f-v_o}{t}

We are given the initial speed of vo=4 m/s on a mass of m=3.5 Kg, the final speed of vf=8 m/s which took t=3 seconds.

The acceleration is:

\displaystyle a=\frac{8-4}{3}=1.33

a=1.33\ m/s^2

Thus, the force is:

F=3.5\ Kg\cdot 1.33\ m/s^2

F=4.67\ N

4.67 N of force are required

6 0
3 years ago
A stone is thrown horizontally at a speed of 10.0 m/s from the top of a cliff 139.4 m high.
Lesechka [4]

Answer:

a) 14.2sec

b) 1394m away if horizontal speed never changes

c) 9.8m/s

Explanation:

5 0
3 years ago
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