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makvit [3.9K]
3 years ago
7

In a given lightning flash, the potential difference between a cloud and the ground is 2.86 times 109 V and the quantity of char

ge transferred is 23.1 C. What is the decrease in energy of the transferred charge? (Please give the magnitude.) If all that energy could be used to accelerate a 1519 kg automobile from rest, what would be the final speed of the automobile?
Physics
1 answer:
Nadya [2.5K]3 years ago
3 0

Answer:

Explanation:

decrease in energy of the transferred charge

= Voltage x charge

= 2.86 x 10⁹ x 23.1

= 66.067 x 10⁹ J

the final speed of the automobile be V

1/2 m v² = 66.067 x 10⁹

v² = 66.067 x 10⁹ x 2 / 1519

= .08698 x10⁹

= 87 x 10⁶

v = 9.32 x 10³ m / s

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Two positive point charges, each of which has a charge of 2.5 × 10−9 C, are located at y = + 0.50m and y = − 0.50m. Find the mag
Salsk061 [2.6K]

Answer:

The resultant electric force is 14.8N to the right.

Explanation:

Since the three charges aren't in the same line, we have to break down the force in components. First, we need to know the distance from the third charge to the other ones. That is made using the Pythagorean Theorem. As the figure is symmetric with respect to the x-axis, the two distances are the same:

r=\sqrt{(0.50m)^{2}+(0.70m)^{2}}=0.86m

Now, we use the Coulomb's Law to obtain the magnitude of the individual forces caused by each charge on the third charge:

|F_{13}|=k\frac{q_1q_3}{r^{2}} \\\\|F_{13}|=(9*10^{9}Nm^{2}/C^{2})\frac{(2.5*10^{-9}C)(3.0*10^{-9}C)}{(0.86m)^{2}}\\\\|F_{13}|=9.1N

For the same reason the distances are the same, the magnitude of the forces are the same:

|F_{23}|=|F_{13}|=9.1N

So, to get the resultant force, we have to break down this forces in components. To do this, we need their angles with respect to the x-axis. Let θ₁ and θ₂ be these angles, respectively. Then, we calculate them using trigonometry:

\theta_1=\arctan(\frac{-0.50m}{0.70m})=-35.5\°\\\\\theta_2=\arctan(\frac{0.50m}{0.70m})=35.5\°

Now, we calculate the components of the forces:

F_{13}_x=F_{13}\cos\theta_1=9.1N\cos(-35.5\°)=7.4N\\\\F_{13}_y=F_{13}\sin\theta_1=9.1N\sin(-35.5\°)=-5.3N\\\\F_{23}_x=F_{23}\cos\theta_2=9.1N\cos(35.5\°)=7.4N\\\\F_{23}_y=F_{23}\sin\theta_2=9.1N\sin(35.5\°)=5.3N

Evidently, the y-components cancel out, and the resultant electric force on the third charge is 7.4N+7.4N=14.8N along the x-axis (to the right, because it's positive).

8 0
3 years ago
What is epsilon zero and why does it come into use, particularly in the case of Gauss's Law?
Paha777 [63]
<span>Epsilon zero is permittivity of free space means how much air or vacuum permits electric field to travel from one charge to other.It is constant in the coulomb law. This allow Gauss's a lot easier to solve rather than using K</span>
5 0
4 years ago
Two 2 kg masses is placed at either end of a rod that has a mass of .5 kg and a length of 3 m. What is the moment of inertia if
sergij07 [2.7K]

Explanation:

a) I=\displaystyle \sum_{i}m_ir_i^2

where r_i is the distance of the mass m_i from the axis of rotation. When the axis of rotation is placed at the end of the rod, the moment of inertia is due only to one mass. Therefore,

I= mr^2 = (2\:kg)(3\:m)^2 = 18\:kg-m^2

b) When the axis of rotation is placed on the center of the rod, the moment is due to both masses and the radius r is 1.5 m. Therefore,

\displaystyle I= \sum_{i}m_ir_i^2 = 2(2\:kg)(1.5\:m)^2 = 9\:kg-m^2

7 0
3 years ago
A 500 kg rollercoaster car starts from rest at the top of a 10.0 m tall hill. it then travels down the track and up a loop. the
malfutka [58]

Speed of the roller coaster at the top of the loop= 7.67 m/s

Explanation:

using the law of conservation of energy

KEi + PEi= KEf + PEf

KEi= kinetic energy at the top of the hill=0 because the car is at rest there.

PEi= potential energy at the top of the hill

PEf= potential energy at the top of the loop

KEf= kinetic energy at the top of the loop

Also kinetic energy= 1/ 2m v² and potential energy= mgh

m= mass

h= height

v= velocity

so 0+ mghi = 1/2mv² + mg h

500 (9.8)(10)+ 1/2 (500) v²= 500 ( 9.8) (7)

49000+250 v²= 34300

250v²= 14700

v²=58.8

v=7.67 m/s

4 0
3 years ago
The combined focal length of two thin lens is 24 cm and the focal length of one converging lens is 8
qwelly [4]

Answer: f = -12 cm

Explanation: <u>Combined</u> <u>lenses</u> is an array of  simple lenses with a common axis. The combination is useful for correction of optical aberrations which cannot be corrected by simple lenses.

When two lenses are in contact and are thin, focal lengths are related as:

\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}}

If there is a distance between the lenses, the focal length will be:

\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}} -\frac{d}{f_{1}f_{2}}

Since the lenses in the question above are thin and in contact, the focal length of one of them will be:

\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}}

\frac{1}{f_{2}} =\frac{1}{f_{1}} -\frac{1}{F}

\frac{1}{f_{2}} =\frac{1}{8} -\frac{1}{24}

\frac{1}{f_{2}} =\frac{-2}{24}

f_{2}= -12

The focal length of the other lens is -12 cm, with the negative sign meaning it's a converging lens.

7 0
3 years ago
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