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soldier1979 [14.2K]
3 years ago
6

Exposure to what kind of energy is sensed by human skin as warmth?

Physics
1 answer:
olganol [36]3 years ago
5 0
Ultraviolet rays.

The sun emits UV rays which reach the earth by radiation, heats the earth. The harmful rays are reflected back.
Hope it helped!
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Help asap please I will give you 5stars
boyakko [2]

Explanation:

In the parallel combination, the equivalent resistance is given by :

\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+....

4. When three 150 ohms resistors are connected in parallel, the equivalent is given by :

\dfrac{1}{R}=\dfrac{1}{150}+\dfrac{1}{150}+\dfrac{1}{150}\\\\R=50\ \Omega

5. Three resistors of 20 ohms, 40 ohms and 100 ohms are connected in parallel, So,

\dfrac{1}{R}=\dfrac{1}{20}+\dfrac{1}{40}+\dfrac{1}{100}\\\\=11.76\ \Omega

Hence, this is the required solution.

6 0
3 years ago
The wind pushes a paper cub along the sand at a beach. The cup has a mass of 25 grams (= ? Kg's) and accelerates at a rate of 5
bixtya [17]

Force = (mass) x (acceleration)

= (0.025 kg) x (5 m/s²)

=   0.125 Newton
7 0
3 years ago
An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p
igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

Explanation:

Given:

Distance from the wire to the field point r = 2.83 \times 10^{-2} m

Speed of electron v = 35.5 \%c

Current I = 17.7 A

For finding the acceleration,

First find the magnetic field due to wire,

  B = \frac{\mu _{o}I }{2\pi r }

Where \mu_{o} = 4\pi   \times 10^{-7}

  B = \frac{4\pi \times 10^{-7}  \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }

  B = 12.50 \times 10^{-5} T

The magnetic force exerted on the electron passing through straight wire,

  F = qvB  

  F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}

  F = 21.3 \times 10^{-16} N

From the newton's second law

  F = ma

Where m = mass of electron = 9.1 \times 10^{-31} kg

So acceleration is given by,

   a = \frac{F}{m}

   a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }

   a = 2.34 \times 10^{15} \frac{m}{s^{2} }

Therefore, the magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

7 0
3 years ago
To accelerate, an object must *
Sonja [21]
Answer: Not 100% sure but I think it’s C.

Hope this helps! ^^
4 0
3 years ago
How long would it take a leopard, running at an average speed of 20 m/s to travel 500 m?
alexandr1967 [171]

Answer:

25 seconds

Explanation:

500/20

4 0
3 years ago
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