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max2010maxim [7]

3 years ago

7

If you put something like a piece of cardboard between a magnet and an iron nail, the magnet still holds the nail in place, even

though the magnet is not touching the nail.
Explain how that happens.

1 answer:

anastassius [24]3 years ago

6 0

Answer:

Hiiii

Explanation:

Helllooooooo

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A 85 kg skier races down a hill from a height of 100 m. Ignore friction.

Ede4ka [16]

Answer:

(a)158.9 N up the ski slope

(b) 0.19

Explanation:

PLEASE MARK ME AS BRAINLIEST

F^k=MkN

158.9=817.7 M*s

Mk=0.19

4 0

3 years ago

a wrench weighs 5.24 newtons on earth. when it is taken to the Moon, where g =1.16 m/s2 how much does it weigh?

Andrew [12]

“Weight of the wrench” on “the moon” is “6.07 kg”.

<u>Explanation</u>:

Weight of the wrench is 5.24 N

Weight of the wrench in kilograms = W × g

Taken “g” on the moon is $1.16 \mathrm{m} / \mathrm{s}^{2}$

$=5.24 \mathrm{N} \times 9.81 \mathrm{m} / \mathrm{s}^{2}=51.352 \mathrm{kg}$

Weight of the wrench in kilograms is 51.352 kg.

Formula to calculate weight of the object on the moon is

$\frac{\text {weight of the object on earth}}{9.81 \mathrm{m} / \mathrm{s}^{2}} \times 1.16 \mathrm{m} / \mathrm{s}^{2}$

Substitute the values given,

$=\frac{51.352 \mathrm{kg}}{9.81 \mathrm{m} / \mathrm{s}^{2}} \times 1.16 \mathrm{m} / \mathrm{s}^{2}$

$=5.234 \times 1.16$

= 6.07 kg

Therefore, weight of the wrench on the moon is 6.07 kg.

6 0

3 years ago

Use the drop-down menus to complete the sentences.

Musya8 [376]

Answer:

yea i need this 2

Explanation:

3 0

3 years ago

Two hockey players are about to collide on the ice, One player has a mass of

Furkat [3]

Answer:

19 kg x m/s East

Just did it!

3 0

3 years ago

Unpolarized light with intensity 370 W/m2 passes first through a polarizing filter with its axis vertical, then through a second

vampirchik [111]

To solve this problem it is necessary to apply the concepts given by Malus regarding the Intensity of light.

From the law of Malus intensity can be defined as

$I' = \frac{I_0}{2} cos^2 \theta$

Where

$\theta =$ Angle From vertical of the axis of the polarizing filter

$I_0 =$ Intensity of the unpolarized light

The expression for the intensity of the light after passing through the first filter is given by

$I = \frac{I_0}{2}$

Replacing we have that

$I = \frac{370}{2}$

$I = 185W/m^2$

Re-arrange the equation,

$I'= \frac{I_0}{2}cos^2\theta$

Re-arrange to find \theta

$cos^2\theta = \frac{2I'}{I_0}$

$cos^2\theta = \frac{2*138}{370}$

$\theta = cos^{-1}(\sqrt{\frac{2*138}{370}})$

$\theta = 0.5282rad$

$\theta = 30.27\°$

The value of the angle from vertical of the axis of the second polarizing filter is equal to 30.2°

4 0

3 years ago