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3 years ago

A metal bar magnet produces a magnetic field in space surrounding it. The point

Physics

1 answer:

MatroZZZ [7]3 years ago

3 0

I think the answer is D but i could be wrong

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A student whirls a rubber stopper attached to a string in a horizontal circle above her head. If the radius of the stopper in ci

alexandr402 [8]

Answer:

$v = 12.57 m/s$

Explanation:

As we know that the radius of the circular motion is given as

$R = 50.00 cm$

time period of the motion is given as

$T = 0.2500 s$

now we know that it is moving with uniform speed

so it is given as

$v = \frac{2\pi R}{T}$

now plug in all data

$v = \frac{2\pi(0.50)}{0.25}$

$v = 12.57 m/s$

3 0

3 years ago

Identify the Following as physical properties or chemical properties.

sveticcg [70]

A.) is chemical, B.) is physical, C.) is physical, D.) is chemical, E.) is physical, F.) is physical, G.) is physical, and H.) is chemical.

5 0

3 years ago

Read 2 more answers

Convert the mass to kgs thanks

Natali [406]

Memorize this and you'll be able to do ALL of these: <em>1 kg = 1,000 g</em>

So if you have some grams, divide the number by 1,000 to get kilograms.

1,000 g = 1.000 kg

500 g = 0.500 kg

100 g = 0.100 kg

50 g = 0.050 kg

20 g = 0.020 kg

10 g = 0.010 kg

4 0

3 years ago

Which statement correctly describes the relationship between thermal energy and particle movement?

zhuklara [117]

As thermal energy increases, there is more particle movement. As thermal energy increases, there is more particle movement. As thermal energy increases, there is less particle movement.

Sure hope this helps you

5 0

2 years ago

A small logo is embedded in a thick block of crown glass (n = 1.52), 4.70 cm beneath the top surface of the glass. The block is

harkovskaia [24]

The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

$d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})$

Where,

$d_g =$ Depth of glass

$n_w =$ Refraction index of water

$n_g =$ Refraction index of glass

$n_{air} =$ Refraction index of air

$d_w =$ Depth of water

I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to

$d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})$

$d'w = (1.7cm) (\frac{1}{1.33})+(4.2cm)(\frac{1}{1.52})$

$d'w = 4.041cm$

Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm

3 0

3 years ago