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Anna [14]
3 years ago
15

Can you help me doing an essay about actual self​

Physics
1 answer:
Rudik [331]3 years ago
4 0

Answer:

yes absolutely

Explanation:

why not

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Please Help On These 2 Questions!!!!! I severely need help!!!!!!
lys-0071 [83]
3. In a uniform electric field, the equation for the magnitude of the magnetic field is E=(V/d). V= voltage d= distance. If the magnetic field magnitude is constant , as stated in your problem, then the voltage must stay the same otherwise the value of "E" would change". And the problem already told us the "E" is uniform and so, not changing. Does that make sense?

4a. If the magnetic field lines are equally spaced apart, in other words share the same density. Then we know that the magnitude of the magnetic field is unchanging. This is because the density of of the magnetic field lines(how many are in a certain area) is related to the magnitude being expressed by the electric field. Greater magnitude is expressed by the presence of more lines (higher line density) 

4b. The electric potential is measured in Volts(V) and is uniform along  the same equipotential line. What is an equipotential line(gray)? It is a line drawn perpendicular(forms a right angle with) to the magnetic field lines(black) to show the changes in electric potential. One space where electric potential will always be the same because it will always be equal to 0 Volts is exactly in between a positive and negative charges of equal charge value I have pointed to this line with a purple arrow in my picture.

I really hope this makes sense to you and that my pictures help! :)

3 0
3 years ago
What number on this diagram refers to the tick marks for the dependent variable?
givi [52]

Answer:

I'm pretty sure it's 3.

Explanation:

Because if you look at your options the only that would be relevant to tick marks would be either 4 or 3. And it said in the question that we're looking for the one for the dependent variable. And the dependent variable is on the Y- Axis and the 3 is the tick marks for the y-axis. So your answer is 3.

4 0
3 years ago
What amount of heat is required to raise the temperature of 25 grams of copper to cause a 15ºC change? The specific heat of copp
Lina20 [59]

The amount of heat required is B) 150 J

Explanation:

The amount of heat energy required to increase the temperature of a substance is given by the equation:

Q=mC\Delta T

where:

m is the mass of the substance

C is the specific heat capacity of the substance

\Delta T is the change in temperature of the substance

For the sample of copper in this problem, we have:

m = 25 g (mass)

C = 0.39 J/gºC (specific heat capacity of copper)

\Delta T = 15^{\circ}C (change in temperature)

Substituting, we find:

Q=(25)(0.39)(15)=146 J

So, the closest answer is B) 150 J.

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

3 0
3 years ago
A car travelling 10m/s increases its speed to 20m/s over a time of 4s. What is the acceleration of the car? A car is traveling a
Alexxx [7]

Answer:

Explanation:

Acceleration is the change in velocity with respect to time.

Acceleration = change in velocity/Time

Acceleration = final velocity - initial velocity/Time

Given initial velocity = 10m/s

Final velocity = 20m/s

Time taken = 4s

Acceleration = 20-10/4

Acceleration = 10/4

Acceleration =2.5m/s²

For the second part of the question:

Given parameters

initial velocity = 15m/s

acceleration = -3m/s²

time = 4seconds

a = v-u/t

-3 = v-15/4

cross multiply

-12 = v-15

add 15 from both sides

-12+15 = v-15+15

3 = v

<em>Hence the final velocity of the car is 3m/s</em>

6 0
3 years ago
A thin lens with a focal length of 6.0 cm is used as a simple magnifier by (a) what angular magnification is obtainable with the
Serhud [2]

Answer:

4.167

4.83871 cm

Explanation:

u = Object distance

v = Image distance = 25 cm

f = Focal length = 6 cm

Angular magnification is given by

m=\frac{25}{f}\\\Rightarrow m=\frac{25}{6}\\\Rightarrow m=4.167

The angular magnification of the lens is 4.167

Lens equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{6}=\frac{1}{u}+\frac{1}{-25}\\\Rightarrow \frac{1}{6}+\frac{1}{25}=\frac{1}{u}\\\Rightarrow \frac{1}{u}=\frac{31}{150}\\\Rightarrow u=4.83871\ cm

The closest distance by which the object can be examined is 4.83871 cm

5 0
3 years ago
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