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Jet001 [13]
3 years ago
5

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
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