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ollegr [7]
3 years ago
9

A moving electron accelerates at 5200 m/s^2 in a 55 degree direction. After 0.530 seconds, it has a velocity of 6598 m/s in a -2

0.5 degree direction. What is the y-component of the initial velocity?
Physics
1 answer:
Natasha2012 [34]3 years ago
5 0

Answer:

-4568.25 m /s

Explanation:

Consider all velocity in y direction .

Acceleration in y direction = 5200 sin55 = 4259.59 m /s²

y component of final velocity = - 6598 sin 20.5 = - 2310.67 m /s

Let y component of initial velocity be u

v = u + at

- 2310.67 = u + 4259.59  x .53

- 2310.67 = u + 2257.58

u = - 4568.25 m /s

So y component of initial velocity

= -4568.25 m /s

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Answer:

0.13 seconds

Explanation:

Since 1 Km = 0.621 miles

3.84 x 105 km = 3.84 x 105 × 0.621 = 23846.4 miles

Speed = distance/time

time= distance/speed

Time= 23846.4/186,000

Time= 0.13 seconds

3 0
3 years ago
If the pressure in a gas is doubled while its volume is held constant, by what factor do vrms change
Nat2105 [25]

Answer is given below

Explanation:

given data

pressure = double

volume = constant

solution

As we know that an Average velocity and rms velocity is directly proportional to square root of PV ..................1

so if we take P is doubled while keeping V constant

than Velocity increases by a factor \sqrt{2}  

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8 0
3 years ago
Can you harvest energy by impact force in a battery? Explain your answer. ​
Pavlova-9 [17]

Answer:

1. Resistance

2. Store chmical energy and transfer it to electrical energy when a circuit is connected

3.Electric current

4. Series and parallel

5. can't see diagram

6. series

7. Parallel

8. can't not see diagram

Explanation:

6 0
3 years ago
A ball is thrown upward in the air by a passenger on a train that is moving with constant velocity. Describe the path of the bal
cestrela7 [59]

Answer:

a) For the passenger the ball is seen to go up in a straight line path, and fall back to the hands of the passenger in a straight line path.

b) For a stationary observer on the ground, the ball is seen to take a parabolic path from when it is thrown up to when it fall down back on the palms of the passenger.

Explanation:

If a ball is thrown up in a vehicle moving with a constant velocity, the ball will be seen as the passenger, who is on the same frame of reference as the ball, to go up and down in a vertical straight line path. For an observer on the ground, this is different, as the ball is seen to to have both a relative vertical and horizontal component of motion, making the ball take a parabolic path from the time it was thrown, to when it falls back to the hands of the passenger.

8 0
4 years ago
During a baseball game, a batter hits a pop-up to a fielder 93 m away.The acceleration of gravity is 9.8 m/s2.If the ball remain
velikii [3]

Explanation:

It is given that, a batter hits a pop-up to a fielder 93 m away, range of the projectile, R = 93 m

The ball remains in the air for 5.4 s, the time of flight is 5.4 s

Time of flight : T=\dfrac{2v\sin\theta}{g}

5.4=\dfrac{2v\sin\theta}{g}\\\\v\sin\theta=\dfrac{5.4\times 9.8}{2}\\\\v\sin\theta=26.46

Maximum height of the projectile : H=\dfrac{v^2\sin^2\theta}{2g}

We need to find H.

So,

H=\dfrac{(v\sin\theta)^2}{2g}\\\\H=\dfrac{(26.46)^2}{2\times 9.8}\\\\H=35.72\ m

So, it will rise to a height of 35.72 m.

6 0
3 years ago
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