Ask question

Ask question

All categories

zhannawk [14.2K]

3 years ago

7

hich best explains why no current is induced? The wire needs to be coiled less tightly. The wire needs to be straight, not coile

d. The magnet needs to be moved through the coils of wire. The magnet needs to be held above the coils of wire.

1 answer:

Dahasolnce [82]3 years ago

7 0

The magnet needs to be held above the coils of wires

You might be interested in

A roundabout in a fairground requires an input power of 2.5 kW when operating at a constant angular velocity of 0.47 rad s–1 . (

natita [175]

Answer:

Explanation:

Since the roundabout is rotating with uniform velocity ,

input power = frictional power

frictional power = 2.5 kW

frictional torque x angular velocity = 2.5 kW

frictional torque x .47 = 2.5 kW

frictional torque = 2.5 / .47 kN .m

= 5.32 kN . m

= 5 kN.m

b )

When power is switched off , it will decelerate because of frictional torque .

5 0

3 years ago

WILL MARK BRAINLIEST AND THANK

timurjin [86]

Same temperature and difference in air pressure

6 0

3 years ago

Read 2 more answers

A man jogs at a speed of 1.6 m/s. His dog

FromTheMoon [43]

I believe it is
1.6x=2.7(x-1.8)
1.1x=2.7*1.8
x~4.4
4.4*1.6
~7.1m

5 0

3 years ago

Mrs. Perez added a room temperature copper cube and an aluminum cube she just removed from the freezer to a beaker of boiling wa

faust18 [17]

I think the answer is A

5 0

3 years ago

The stiffness of a particular spring is 42 N/m. One end of the spring is attached to a wall. A force of 2 N is required to hold

Liula [17]

Answer:

$L_{o}=0.1224m$

Explanation:

Given data

Force F=2 N

Length L=17 cm = 0.17 m

Spring Constant k=42 N/m

To find

Relaxed length of the spring

Solution

From Hooke's Law we know that

$F_{spring}=k_{s}|s|\\F_{spring}=k_{s}(L-L_{o})\\ 2N=(42N/m)(0.17m-L_{o})\\2=7.14-42L_{o}\\-42L_{o}=2-7.14\\42L_{o}=5.14\\L_{o}=(5.14/42)\\L_{o}=0.1224m$

6 0

4 years ago