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aivan3 [116]
3 years ago
15

The magnetic field inside a 5.0-cm-diameter solenoid is 2.0 T and decreasing at 3.00 T/s . Part A What is the electric field str

ength inside the solenoid at a point on the axis?
Physics
1 answer:
Doss [256]3 years ago
7 0

Given Information:

Diameter of solenoid = 5.0 cm

Magnetic field = B = 2.0 T

Rate of change of magnetic field = 3.0 T/s

Required Information:

Electric field strength inside the solenoid at a point on the axis = ?

Answer:

The electric field strength inside the solenoid at a point on the axis is zero.

Explanation:

From the Faraday's Law

E = -r/2 dB/dt

Where dB/dt is the rate of change of magnetic field and r is zero because the distance from the axis is zero.

Therefore, electric field strength inside the solenoid at a point on the axis is 0.

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a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it wi
Lelu [443]

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s .The amplitude of the subsequent oscillations 48.13 cm/s

a 1.25 kilogram block is fastened to a spring with a 17.0 newtons per meter spring constant. Given that K is equal to 14 Newtons per meter and mass equals 10.5 kg. The block is then struck with a hammer by a student while it is at rest, giving it a speedo of 46.0 cm for a brief period of time. The required energy provided by the hammer, which is half mv squared, is transformed into potential energy as a result of the succeeding oscillations. This is because we know that energy is still available for consultation. So access the amplitude here from here. He will therefore be equal to and by. Consequently, the Newton's spring constant is 14 and the value is 10.5. The velocity multiplied by 0.49

Speed at X equals 0.35 into amplitude, or vice versa. At this point, the spirit will equal half of K X 1 squared plus half. Due to the fact that this is the overall energy, square is equivalent to half of a K square or an angry square. amplitude is 13 and half case 14 x one is 0.35. calculate that is equal to initial velocities of 49 squares and masses of 10.5. This will be divided in half and start at about 10.5 into the 49-square-minus-14. 13.42 into the entire square in 20.35. dividing by 10.5 and taking the square as a result. 231 6.9 Six centimeters per square second. 10.5 into 49 sq. 14. 2 into a 13.42 square entire. then subtract 10.5 from the result to get the square. So that is 48.13cm/s.

To learn more about oscillations Please click on the given link:

brainly.com/question/26146375

#SPJ4

This is incomplete question Complete Question is:

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s . what are The amplitude of the subsequent oscillations?

4 0
1 year ago
PLEASE HELP ASAP!!!
musickatia [10]

Answer:

22.2 W

Explanation:

First of all, we calculate the work done by moving the wagon, using the formula:

W=Fdcos \theta

where

F = 20 N is the magnitude of the force

d = 1000 m is the displacement of the wagon

\theta=0^{\circ} is the angle between the direction of the force and of the displacement (assuming the force is applied in the direction of motion)

Substituting, we find

W=(20)(1000)=20,000 J

Now we can find the power generated, which is equal to the ratio between the work done and the time taken:

P=\frac{W}{t}

where

W = 20,000 J

t = 15 min = 900 s

Substituting,

P=\frac{20000}{900}=22.2 W

And the same value in Joules/second (remember that 1 Watt = 1 Joule/second)

5 0
3 years ago
During the first 50 s a truck traveled at constant speed of 25 m/s. Find the distance that it is traveled.
Allushta [10]
Time=50s
speed=25m/s

Distance = speed×time
=25×50
=1250m

DISTANCE TRAVELLED IS =1250m

6 0
3 years ago
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