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Lynna [10]
3 years ago
5

A circuit is made up of a power source, a coil of copper wire, a moving bar

Physics
2 answers:
ale4655 [162]3 years ago
7 0

Answer:the power source

Explanation:

olganol [36]3 years ago
4 0

Answer:

A. The power source

Explanation:

Just took the quiz

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In this section of a circuit, a current of 2.0 A flows across R1. What is the current across R2? Let R1 = 4.0 ohm, R2 = 8.0 ohm,
Liono4ka [1.6K]

Answer:

I=1A

Explanation:

In parallel combination voltage is same

In R1

V=IR

V=2×4

V=8V

In R2

V=IR

I=V/R

I=8/8

I=1A

3 0
4 years ago
A total electric charge of 6.75 nC is distributed uniformly over the surface of a metal sphere with radius 20.0 cm. If the poten
djverab [1.8K]

Answer:

a) 60 V

b) 125 V

c) 125 V

Explanation:

<u>Given</u>

We are given the total electric charge q = 6.75 nC = 6.75x 10^-9 C distributed uniformly over the surface of a metal sphere with a radius of R = 20.0 cm = 0.020 m.  

<u>Required </u>

We are asked to calculate the potential at the distances

(a) r = 10.0 cm

(b) r = 20.0 cm

(c) r = 40.0 cm  

<u>Solution</u>

(a) Here, the distance r > R so, we can get the potential outside the sphere (r > R) where the potential is given by

V = q/4\pi∈_o                       (1)

r is the distance where the potential is measured and the term 1/4\pi∈_o equals  9.0 x 10^9 Nm^2/C^2. Now we can plug our values for q and r into equation (1) to get the potential V where r = 0.10 m  

V= 1*q/4\pi∈_o*r

 =60 V

(b) Here the distance r is the same for the radius R, so we can get the potential inside the sphere (r = R) where the potential is given by  

V = 1*q/4\pi∈_o*R                (2)    

Now we can plug our values for q and R into equation (2) to get the potential V where R = 0.20 m  

V = 1*q/4\pi∈_o*R

    = 125 V

(c) Inside the sphere the electric field is zero therefore, no work is done on a test charge that moves from any point to any other point inside the sphere. Thus the potential is the same at every point inside the sphere and is equal to the potential on the surface. and it will be the same as in part (b)  

V= 125 V

7 0
3 years ago
An object is 50 cm from a converging lens with a focal length of 40 cm . A real image is formed on the other side of the lens, 2
Leno4ka [110]

Answer:

d) -4.0

Explanation:

The magnification of a lens is given by

M=-\frac{q}{p}

where

M is the magnification

q is the distance of the image from the lens

p is the distance of the object from the lens

In this problem, we have

p = 50 cm is the distance of the object from the lens

q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct

Also, q is positive since the image is real

So, the magnification is

M=-\frac{200 cm}{50 cm}=-4.0

7 0
3 years ago
A tank is 6 m long, 4 m wide, 5 m high, and contains kerosene with density 820 kg/m3 to a depth of 4.5 m. (Use 9.8 m/s2 for the
Mazyrski [523]

The pressure at the bottom of the tank is given by:

P = ρgh

P = pressure, ρ = fluid density, g = gravitational acceleration, h = depth

Given values:

ρ = 820kg/m³, g = 9.8m/s², h = 4.5m

Plug in and solve for P:

P = 820(9.8)(4.5)

P = 36000Pa

5 0
3 years ago
A mass m = 0.6 kg is released from rest at the top edge of a hemispherical bowl with radius = 1.1 meters. The mass then slides w
Aneli [31]

Answer:

The  angle is  \theta   = 36.24 ^o

Explanation:

From the question we are told that

    The  mass is  m  =  0.6 \ kg

     The radius is  r =  1.1 \ m

     The speed is  v =  3.57 \ m /s

According to  the law of energy conservation

  The  potential energy of the mass at the top is equal to the kinetic energy at the bottom i.e

      m * g  *  h  =  \frac{1}{2}  *  m * v^2

 =>    h  =  \frac{1}{2 g } *  v^2

Here h is the vertical distance traveled by the mass  which is also mathematically represented as

      h  =  r * sin (\theta )

So

     \theta   = sin ^{-1} [ \frac{1}{2* g* r } *  v^2]

substituting values

     \theta   = sin ^{-1} [ \frac{1}{2* 9.8* 1.1 } *  (3.57)^2]

     \theta   = 36.24 ^o

3 0
4 years ago
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