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faltersainse [42]
3 years ago
14

A circular loop of wire is positioned so the normal to its area points vertically upward. A bar magnet is dropped from above the

loop so that its north pole falls through the loop first. When looking at the loop from above, in which direction is the induced current in the loop just as the north pole of the magnet enters the loop?
a. Clockwise
b. There is no induced current Incorrect.
Physics
1 answer:
Kruka [31]3 years ago
8 0
B cause there is no induced current incorrect
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The temperature of a smelting furnace is found to be 2000 degree Celsius.find the temperature on Fahrenheit scale​
Dima020 [189]

Answer:

3632.

Explanation:

Based on the formula:

2000×(9/5)+32=3632

6 0
2 years ago
Match the following vocabulary words with their definitions:
Ronch [10]

Correct matching:


1 acceleration --> rate of change in velocity, which is the change in velocity divided by the change in time

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4. gravity --> force of attraction between all masses in the universe

5. Inertia --> an object´s resistance to a change in motion

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6 0
3 years ago
Nerve impulses in the human body travel at a speed of about 100 m/s. A 1.6 m tall man accidentally drops a hammer on his toe. Ho
EastWind [94]

time=distance/speed

1.6/100 secs = 0.016secs=16millisecs

4 0
3 years ago
PLZ HELP I WILL GIVE BRAINLIEST
Vaselesa [24]

Answer:

12.7m/s

Explanation:

Given parameters:

Mass of diver  = 77kg

Height of jump  = 8.18m

Unknown:

Final velocity  = ?

Solution:

To solve this problem, we apply the motion equation below:

             v²   = u²  + 2gH

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

H is the height

 Now insert the parameters and solve;

       v² = 0²  +  2 x 9.8 x 8.18

     v  = 12.7m/s

8 0
2 years ago
The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 440g . The vibra
ANEK [815]

Answer:

Tension, T = 2038.09 N

Explanation:

Given that,

Frequency of the lowest note on a grand piano, f = 27.5 Hz

Length of the string, l = 2 m

Mass of the string, m = 440 g = 0.44 kg

Length of the vibrating section of the string is, L = 1.75 m

The frequency of the vibrating string in terms of tension is given by :

f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}

\mu=\dfrac{m}{l}

\mu=\dfrac{0.44}{2}=0.22\ kg/m

T=4L^2f\mu

T=4\times (1.75)^2\times (27.5)^2 \times 0.22

T = 2038.09 N

So, the tension in the string is 2038.09 N. Hence, this is the required solution.

6 0
3 years ago
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