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olchik [2.2K]
3 years ago
14

Two locomotives approach each other on parallel tracks. Each has a speed of 95 km per h with respect to the ground. If they are

initially 8.5 km apart, how long will it be before they reach each other?
Physics
1 answer:
jekas [21]3 years ago
4 0

Answer:

t = 0.045 hr

Explanation:

given,

speed of the car A = 95 Km/h

speed of car B = 95 Km/h

initial distance = 8.5 Km

time = ?

now,

the relative speed between the two vehicle

= 95 - (-95) = 190 Km/h

we have use negative speed because one car is moving opposite to other.

we know,

distance = speed x time

t = \dfrac{distance}{speed}

t = \dfrac{8.5}{190}

t = 0.045 hr

time taken the car to reach each other is equal to 0.045 hours.

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3 years ago
Which of the following is not a possible environmental effect of acid rain?
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Answer:

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Mrrafil [7]

Answer:

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Explanation:

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4 0
3 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
4 years ago
The Sankey diagrams below show the energy transfers in two light bulbs. How much energy is wasted by light bulb 2?
Law Incorporation [45]

Answer:

The amount of energy wasted by light bulb 2 = 70 J

Explanation:

Sankey diagrams are used to graphically indicate the proportional flow rate of a quantity

The labelled Sankey diagrams, here, indicate the energy transfers in two light bulbs, light bulb 1 and light bulb 2

From the labelled Sankey diagrams, we have;

For light bulb 1, the input energy = 80 J

The useful energy = 40 J

The wasted energy = 80 - 40 = 40 J

For  light bulb 2, the input energy = 80 J

The useful energy = 10 J

Therefore, the wasted energy = 80 - 10 = 70 J

The amount of energy wasted by light bulb 2 = 70 J.

5 0
4 years ago
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