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3 years ago

Lightning is an example of what kind of electricity

Physics

2 answers:

n200080 [17]3 years ago

8 0

Answer:

The answer is A. Static discharge

Explanation:

ki77a [65]3 years ago

7 0

the answer would be DISCHARGE

I ʝυʂƚ ƚσσƙ ƚԋҽ TEST

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Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, t

NemiM [27]

Answer:

their final velocity is 0.091 m/s

Explanation:

Given;

mass of the first train, m₁ = 138,000 kg

mass of the second train, m₂ = 123,000 kg

initial velocity of the first train, u₁ = 0.288 m/s

initial velocity of the seocnd train, u₂ = -0.131 m/s (opposite direction to the first)

Let their common final velocity after been coupled = v

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(138,000 x 0.288) + (-0.131 x 123,000) = v(138,000 + 123,000)

39,744 - 16,113 = v(261,000)

23,631 = v(261,000)

v = 23,631 / 261,000

v = 0.091 m/s

Therefore, their final velocity is 0.091 m/s

5 0

2 years ago

A figure skater skates across a rink of length 50 m in 12.1 seconds. a. What is the average speed of the skater? (2 points) b. I

melamori03 [73]

(a) The skater covers a distance of S=50 m in a time of t=12.1 s, so its average speed is the ratio between the distance covered and the time taken:
$v= \frac{S}{t}= \frac{50 m}{12.1 s}=4.13 m/s$

(b) The initial speed of the skater is
$v_i = 4 m/s$
while the final speed is
$v_f = 5.3 m/s$
and the time taken to accelerate to this velocity is t=2 s, so the acceleration of the skater is given by
$a= \frac{v_f - v_i}{t}= \frac{5.3 m/s-4.0 m/s}{2.0 s}=0.65 m/s^2$

(c) The initial speed of the skater is
$v_i = 13.0 m/s$
while the final speed is
$v_f=0$
since she comes to a stop. The distance covered is S=8 m, so we can use the following relationship to find the acceleration of the skater:
$2aS=v_f^2 -v_i^2$
from which we find
$a= \frac{-v_i^2}{2S}= \frac{-(13.0m/s)^2}{2 \cdot 8.0 m}=-10.6 m/s^2$
where the negative sign means it is a deceleration.

4 0

3 years ago

If earth did not rotate how would air at the equator move?

Dmitry_Shevchenko [17]

Heat rises, and it is warmer at the equator, so I think warm air would rise at the equator and move towards the cooler poles.

3 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP BEST ANSWER WILL BE MARKED BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

pantera1 [17]

Answer:

It is the 3 one

Explanation:

3 0

3 years ago

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.