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3 years ago

He output voltage and current of a transformer are determined by the

2 answers:

4 0

The output voltage of a transformer is

   (input voltage) x (number of turns of wire in secondary)
   divided by
      (number of turns of wire in the primary winding) .

The output current is

   (output voltage)
   divided by
   (total resistance in the circuit connected to the output).

Ede4ka [16]3 years ago

3 0

Answer: the number of turns in the primary and secondary coils

Explanation:

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A railroad car moving at a speed of 3.46 m/s overtakes, collides and couples with two coupled railroad cars moving in the same d

I am Lyosha [343]

Answer:

$2.09\ \text{m/s}$

$22298.4\ \text{J}$

Explanation:

m = Mass of each the cars = $1.6\times 10^4\ \text{kg}$

$u_1$ = Initial velocity of first car = 3.46 m/s

$u_2$ = Initial velocity of the other two cars = 1.4 m/s

v = Velocity of combined mass

As the momentum is conserved in the system we have

$mu_1+2mu_2=3mv\\\Rightarrow v=\dfrac{u_1+2u_2}{3}\\\Rightarrow v=\dfrac{3.46+2\times 1.4}{3}\\\Rightarrow v=2.09\ \text{m/s}$

Speed of the three coupled cars after the collision is $2.09\ \text{m/s}$ .

As energy in the system is conserved we have

$K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}$

The kinetic energy lost during the collision is $22298.4\ \text{J}$ .

6 0

3 years ago

An airplane flies in a loop (a circular path in a vertical plane) of radius 160 m . The pilot's head always points toward the ce

notka56 [123]

Answer:

a) 39.6 m/s b) 4123 N

Explanation:

a) At the top of the loop, all of the forces point downwards (force of gravity and normal force).

Fnet=ma

ma=m(v^2/R) (centripetal acceleration)

mg=m(v^2/R)

m cancels out (this is why pilot feels weightless) so,

g=(v^2/R)

9.8 m/s^2 = v^2/160 m

v^2=1568 m^2/s^2

v=39.6 m/s

b) At the bottom of the loop, the normal force and the force of gravity point in opposite directions. The normal force is the weight felt.

Convert 300 km/hr to m/s

300 km/hr=83.3 m/s

Convert pilot's weight into mass:

760 N = 77.55 kg

Fnet=ma

n-mg=m(v^2/R)

n=(77.55 kg)(((83.3 m/s)^2)/160 m)+(77.55 kg)(9.8 m/s^2)

n=3363.2 N+760 N=4123 N

5 0

2 years ago

A charge +1.9 μC is placed at the center of the hollow spherical conductor with the inner radius 3.8 cm and outer radius 5.6 cm.

Archy [21]

To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.

PART A)

The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:

$Q_{int}=-Q1=-1.9*10^{-6} C$ . This is the total charge on the inner surface of the conducting shell.

PART B)

The positive charge (of the same value) on the external surface of the conducting shell is:

$Q_{ext}=+Q_1=1.9*10^{-6} C$

The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,

$Q_{ext, Total}=Q_2+Q_{ext}$

$Q_{ext, Total}=1.9+3.8$

$Q_{ext, Total}=5.7 \mu C$

5 0

3 years ago

A marble resting near the edge of .90 m high table is given an initial horizontal speed of 1.24 m/s. What will be it’s horizonta

Maru [420]

Answer:

0.53 m

Explanation:

First of all, we have to consider the vertical motion of the ball, in order to find the time it takes for the marble to reach the ground. The initial height is $h=0.90 m$ , the initial vertical velocity is zero, while the acceleration is $g=-9.81 m/s^2$ , so the vertical position at time t is given by

$y(t)=h-\frac{1}{2}gt^2$

By demanding y(t)=0, we find the time t at which the ball reaches the ground:

$0=h-\frac{1}{2}gt^2$

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(0.9 m)}{9.81 m/s^2}}=0.43 s$

Now we can find the horizontal range of the marble: we know the initial horizontal speed (v=1.24 m/s), we know the total time of the motion (t=0.43 s), and since the horizontal speed is constant, the total distance traveled on the horizontal direction is

$x=vt=(1.24 m/s)(0.43 s)=0.53 m$

8 0

3 years ago

A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a

Rasek [7]

Answer:

$W=173.48J$

Explanation:

information we know:

Total force: $F=45N$

Weight: $w=100N$

distance: $4m$

vertical component of the force: $F_{y}=12N$

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: $F_{x}=Fcos\theta$