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pogonyaev
3 years ago
10

Explain why the light bulb isn't lighting up in the circuit pictured on the right.

Physics
1 answer:
FromTheMoon [43]3 years ago
8 0
The wires should be attached to each screw on the light not only one
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. If force (F), work (W) and velocity (v) are taken as fundamental quantities.
alex41 [277]

Answer:

∴ [T]=[WF−1V−1]

Hope this answer is right!!

7 0
2 years ago
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In nuclear physics wht units are used to measure the radius of an atom ?
Alecsey [184]
Angstrom = 10^-10 m
for nucleus size are used fermi (femtometer  10^-15 m )
6 0
3 years ago
A firecracker in a coconut blows the coconut into three pieces. Twopieces of equal mass fly off south and west, perpendicular to
loris [4]

Answer:

V = 13m/s

North-East (i.e. 45 degree from North)

Explanation:

This question deals with the idea of momentum.

Since the directions of a compass are fixed, we can take south as horizontal and west as vertical. Since both pieces of coconut are of equal mass, then the resultant of the two pieces would be in between South and West and the third piece would be opposite this direction and be in the North-East (i.e. 45 degree from North) direction

To find the speed of the third piece, we first find the speed on the two pieces of the coconut in terms of V_{x} (horizontal component of velocity) and V_{y} (vertical component of velocity)

We have

V_{y} = Velocity in South direction = 18m/s

m_{y} = Mass of piece in South direction = m

V_{x} = Velocity in West direction = 18m/s

m_{x} = Mass of piece in West direction = m

V_{ry} = Velocity of Third Piece in North direction = unknown

V_{rx} = Velocity of Third Piece in East direction = unknown

m = Mass of third piece = 2m

mV_{ry}  = m_{y} V_{y} \\ 2mV_{ry}  = m18\\ V_{ry} = 9

mV_{rx}  = m_{x} V_{x} \\ 2mV_{rx}  = m18\\ V_{rx} = 9

Since we now know the horizontal and vertical component of the velocity of the third piece of coconut we find the resultant velocity. Since we know the direction is North-East, we can imagine a right angled triangle with base of 9 and height 9 and the hypotenuse equal to the resultant velocity .

We can simply apply the Pythagoras theorem and find the hypotenuse

Hypotenuse^{2}  = Height^{2}  + Base^{2} \\ V_{3}^2  = 9^{2} + 9^{2} \\ V = 12.728 m/s

V = 13m/s

6 0
3 years ago
A large solar panel on a spacecraft in Earth orbit produces 1.0 kW of power when the panel is turned toward the sun. What power
Mandarinka [93]

Answer:

e*P_s = 11 W

Explanation:

Given:

- e*P = 1.0 KW

- r_s = 9.5*r_e

- e is the efficiency of the panels

Find:

What power would the solar cell produce if the spacecraft were in orbit around Saturn

Solution:

- We use the relation between the intensity I and distance of light:

                                  I_1 / I_2 = ( r_2 / r_1 ) ^2

- The intensity of sun light at Saturn's orbit can be expressed as:

                                  I_s = I_e * ( r_e / r_s ) ^2

                                  I_s = ( 1.0 KW / e*a) * ( 1 / 9.5 )^2

                                  I_s = 11 W / e*a

- We know that P = I*a, hence we have:

                                  P_s = I_s*a

                                  P_s = 11 W / e

Hence,                       e*P_s = 11 W

3 0
3 years ago
A horizontal spring is attached to a wall at one end and a mass at the other. The mass rests on a frictionless surface. You pull
LenKa [72]

Answer:

54%

Explanation:

So, we have that the "magnitude of its displacement from equilibrium is greater than (0.66)A—''. Thus, the first step to take in answering this question is to write out the equation showing the displacement in simple harmonic motion which is = A cos w×t.

Therefore, we will have two instances t the displacement that is to say at a point 2π/w - a2 and the second point at a = a2.

Let us say that 2π/w = A, then, we have that a = A cos ^-1 (0.66)/2π. Also, we have that a2 = A/2 - A cos^- (0.66) / 2π.

The next thing to do is to calculate or determine the total length of of the required time. Thus, the total length is given as:

2a1 + ( A - 2a2) = 2A{ cos^-1 (0.66)}/ π.

Therefore, the total percentage of the period does the mass lie in these regions = 100 × {2a1 + ( A - 2a2) }/A = 2 { cos^-1 (0.66)}/ π × 100 = 54%.

Thus, the total percentage of the period does the mass lie in these regions = 54%.

6 0
3 years ago
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