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Fed [463]
2 years ago
5

……………………………………………………………..?

Physics
1 answer:
kotegsom [21]2 years ago
3 0

Answer:

C

Explanation:

First find the electrical wattage

W = I^2 * R

R = 12 ohms

I = 2 amps

Wattage = 2^2 * 12

Wattage = 4* 12

Wattage = 48 watts.

Now you need to use the power formula

Work = Power * Time

Work = ?

Power = 48 watts

Time = 3 minutes = 3 * 60 = 180 seconds.

Work = 48 * 180

Work = 8640 J

That's C

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The velocity versus time graph of particle A is tangent to the velocity versus time graph for particle B at point O. What is the
Fudgin [204]

Answer: C. -1.16 meters/second2

Explanation:

A= v/t (velocity/time)

in this case: v=7 and t=6

So, A= 7/6

A=1.16

The graph is decreasing so accelleration would be negative

A= <u>-1.16 meters/second2</u>

<u>Option C!</u> ; )

<u></u>

3 0
3 years ago
A rubber bullet of mass m=0.025\,\mathrm{kg}m=0.025kg traveling at velocity v_0 = 50\,\mathrm{m/s}v 0 ​ =50m/s hits an iron bloc
Pavel [41]

Answer: 0.001 m

Explanation:

In order to get the máximum height reached by the iron block, we can use the energy conservation principle, as all the kinetic energy impressed upon the iron blck by the rubber bullet during the collision, becomes gravitational potential energy, as follows:

½ m v2 = m. g. h (1)

We don´t know the value of v, but if we look to the collision, and we asume no external forces act during it, the total momentum must be conserved.

The initial momentum, as the block is at rest, is just the one due to the rubber bullet:

P1 = mb . vb = 0.025 kg. 50 m/s = 1.25 kg. m/s

The final momemtum, is just the sum of the one due to the bullet (after being bounced back) and the one for the iron block:

P2 = mb . vfb  + mib . vib= 0.025 Kg. (-35 m/s) + 15 Kg.vib

As we have already said, P1 = P2, so we can write the following equation:

0.025 Kg. (-35 m/s) + 15 Kg. vib = 1.25 Kg. m/s.

Solving for vib, we have:

vib = 0.14 m/s

Now, we can replace this value in the equation (1) above:

½ . 15 Kg. (0.14)2  (m/s)2 = 15 Kg. 9.8 m/s2. H

Solving for H, we have:

H = 0.001 m

3 0
3 years ago
A uniform electric field exists in the region between two oppositely charged plane parallel plates. a proton is released from re
dedylja [7]

According to Newton's second law

E.e = a * mp  ..... (1)

where

E is the magnitude of the electric field; e = 1.6 * 10^-19 is the elementary charge; mp = 1.67*10^-27 kg is the proton mass; a is the acceleration.

So, the distance

l = at^2/2 .......(2)

The proton accelerated

a = 2l / t^2 ...........(3)

From equations (1) and (3)

E= 32.51 V/m

Electric field

The physical field that surrounds electrically charged particles and exerts a force on all other charged particles in the field, either attracting or repelling them, is known as an electric field (also known as an E-field).  It can also refer to a system of charged particles' physical field. Electric charges and time-varying electric currents are the building blocks of electric fields. The electromagnetic field, one of the four fundamental interactions (also known as forces) of nature, manifests itself in both electric and magnetic fields.

To learn more about an electric field refer here:

brainly.com/question/15800304

#SPJ4

5 0
2 years ago
On both the spring and fall equinoxes, the sun's rays are vertically overhead at the _______
Anni [7]

It is overhead at the equator, it is because the sun ray’s will be moving vertically as this will be directed at the equator. It is because if it moves vertically, it will hit or overhead the equator and this usually happens in spring and fall.

5 0
3 years ago
Light travels at 3 × 108 m/s, and it takes about 8 min for light from the sun to travel to Earth. Based on this, the order of ma
N76 [4]

Answer:

The order of magnitude of the distance from the sun to Earth is 10⁸ km.

Explanation:

The order of magnitude of the distance from the sun to Earth can be calculated as follows:

c = \frac{x}{t}

Where:

c: is the speed of light = 3x10⁸ m/s

t: is the time = 8 min

Hence, the distance is:

x = c*t = 3 \cdot 10^{8} m/s*8 min*\frac{60 s}{1 min} = 1.44 \cdot 10^{11} m = 1.44 \cdot 10^{8} km

Therefore, the order of magnitude of the distance from the sun to Earth is 10⁸ km.

I hope it helps you!

5 0
3 years ago
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