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storchak [24]
3 years ago
7

Plzzzz help 50 points thanks Super easy I will mark brainlest

Physics
1 answer:
Vladimir79 [104]3 years ago
8 0
Yeah, which one of the three
You might be interested in
Please help me! Thank you!
amid [387]
Question 1. To solve this we are going to take advantage of the fact that 1 horsepower = 745.7 Watts, so to convert watts to horsepower, we just need to multiply by the conversion factor \frac{1horsepower}{747.5Watts}.
Lets convert the power of our refrigerator from Watts to horsepower:
500Watts* \frac{1hosepower}{747.5Watts} =0.67horsepower

We can conclude that the power rating of a 500 W refrigerator in horsepower is 0.67 horsepower.

Question 2. An electric field tends to be strongest at the ends of pointed objects; the strongest the electric field the more charge it builds up, so at the ends of pointed objects there is enough charge to strip the atoms in the air form its electrons producing plasma. That glowing of that plasma is what we call <span>St. Elmo's fire.

Question 3. 
</span>- Even though both phenomena are <span>static discharges, they nature is completely different. </span><span>Lightning is an electric discharge: a discharge that occurs when opposite charges accumulate until the electric field becomes strong enough to allow a current to flow. St. Elmo's fire, on the other hand, is a coronal discharge: a luminous phenomena, similar to a neon tube, that occurs when a pointed object in a strong electric field creates plasma.
- Even tough their origin an behavior are different, both of them create plasma in the air.
</span><span>- Lightening is extremely hot, about the same temperature as the surface of the sun, whereas St. Elmo's fire is relatively cold.

Question 4. Since the glow in the neon tube is actually a coronal discharge that creates red plasma -due to the nature of the gas, it is basically a red St. Elmo's fire inside a tube. if we replace the neon gas with plasma, we will basically create a St. Elmo's fire inside the tube; the air inside the tube will be now a mixture of oxygen and nitrogen, so it will emit a bluish glow like St. Elmo's fire.

Question 5. St. Elmo's fire only appear during thunderstorms because it is the only time when </span><span>electrons accumulate on the bottoms of clouds and induce a positive charge in the ground creating a strong electric field. The coronal discharge that creates the St. Elmo fire can only happen whiting a strong local electric field, so it can only happen during thunderstorms.

</span>Question 6. No, air is not a good conductor of electric charge; otherwise phenomena like Sr Elmo's fire or lightening will occur outside thunder storms, which is not the case. Those phenomena only occur whiting strong electric fields, and those electric fields are only present during thunder storms. If air was a good conductor of electric charge we couldn't use electricity without killing ourselves in the process.
4 0
3 years ago
A 1200 kg truck is moving to the right at the speed of 30 m/s. It hit another identical truck that was at rest. The two trucks s
Marrrta [24]

Answer:

ok wht is the question

Explanation:

4 0
3 years ago
A hydrogen atom has a radius of 2.5 x 10-11 m<br> Determine the radius of a magnesium atom.
baherus [9]

Answer:

R = 1.5* 10^{-10}m

Explanation:

Given

r = 2.5 * 10^{-11}m -- radius of hydrogen atom

<em>See attachment</em>

Required

Determine the radius of magnesium atom (R)

From the attachment, the ratio of a hydrogen atom to a magnesium atom is:

Ratio = 6mm : 36mm

Simplify

Ratio =1 : 6

Represent the radius as ratio:

Ratio = r : R

Substitute r = 2.5 * 10^{-11}m

Ratio = 2.5 * 10^{-11}m : R

Equate both ratios

2.5 * 10^{-11}m : R = 1 : 6

Express as fraction

\frac{2.5 * 10^{-11}m}{R} = \frac{1}{6}

Cross Multiply

R * 1 = 2.5 * 10^{-11}m * 6

R * 1 = 2.5 * 6* 10^{-11}m

R * 1 = 15* 10^{-11}m

R = 15* 10^{-11}m

R = 1.5*10* 10^{-11}m

R = 1.5* 10^{1-11}m

R = 1.5* 10^{-10}m

Hence, the radius of the magnesium atom is: 1.5* 10^{-10}m

6 0
3 years ago
I need a bit of help on this one, can anyone answer?
emmainna [20.7K]

Potential energy = m · g · h

-- When you held the ball at 2.0 meters above the floor, it had

(0.5 kg) · (9.8 m/s²) · (2.0 m) = 9.8 Joules of potential energy.

-- After it bounced and went back up as high as it could, it was only 1.8 meters above the floor.  Its potential energy was

(0.5 kg) · (9.8 m/s²) · (1.8 m) = 8.82 Joules

-- Between the drop and the top of the bounce, it lost

(9.8 - 8.82) = <em>0.98 Joule</em> .

-- The energy was lost when the ball hit the floor.  During the hit, 0.98 joule of kinetic energy turned to <em>thermal energy</em>, which slightly heated the ball and the floor.

5 0
3 years ago
"If E = 7.50V and r=0.45Ω, find the minimum value of the voltmeter resistance RV for which the voltmeter reading is within 1.0%
Virty [35]

Answer:

The  minimum value is R_V =44.552\  \Omega

Explanation:

From the question we are given that

                  The voltage is E = 7.5V

                  The internal  resistance is r = 0.45

The objective of this solution is to obtain the minimum value of the voltmeter resistance for which the voltmeter reading is within 1.0% of the emf of the battery

  What is means is that we need to obtain voltmeter resistance such that

                                V = (100% -1%) of E

Where E  is the  e.m.f of the battery and V is the voltmeter reading

                          i.e    V = 99% of E = 0.99 E = 7.425  

Generally

                E = V + ir

     where ir is the internal potential difference of the voltmeter and

                V is the voltmeter reading

 Making i the subject of the formula above

            i = \frac{(E-V)}{r}

               =\frac{7.50-7.425}{0.45}

              = 0.1667 A

Now the current is constant through out the circuit so,

                  V = iR_V

Where  R_V is the value of voltmeter resistance

                Hence R_V = \frac{V}{i}  = \frac{7.425}{0.1667}

                                  =44.552\  \Omega

                       

8 0
4 years ago
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