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Viefleur [7K]
3 years ago
8

The particle accelerator at CERN can accelerate an electron through a potential

Physics
1 answer:
asambeis [7]3 years ago
8 0

Answer:

K.E = 1.28 × 10^-17 KeV

Explanation:

Given that a particle accelerator at CERN can accelerate an electron through a potentialdifference of 80 kilovolts.

To Calculate the kinetic energy (in keV) of the electron​, let us first find the electron charge which is 1.60 × 10^-19C

The kinetic energy = work done

K.E = e × kV

Substitute e and the voltage into the formula

K.E = 1.60 × 10^-19 × 80

K.E = 1.28 × 10^-17 KeV

Therefore, the kinetic energy is approximately equal to 1.28 × 10^-17 KeV

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Kinetic and potential; what type of energy is each of the points
miskamm [114]

Answer:

1 potential

2 kenitic

3 kenitic

4 potential

5 kenitic

6 potential

7 0
4 years ago
How long does a car (1000 kg) have a speed of 30 m/s from a rest if the engine power is 10kw
Lemur [1.5K]

Answer: 90.1 s

Explanation:

Use equation for power:

P=F*V

Use eqation for force:

F=ma

F---force

V---velocity

Vr=om/s

V=30m/s

m=1000kg

P=10000W

---------------------------

P=FV

F=P/V

F=10000W/30m/s

F=333.33N

Use equation for force to find accelartaion.

F=ma

a=F/m

a=333.33N/1000kg

a=0.333 m/s²

Use equation for accelaration to find out time:

a=(V-Vs)/t

t=(V-Vs)/a

t=(30m/s)/(0.333m/s²)

t=90.09 s≈90.1 s

------------------------

5 0
3 years ago
Which statements best describe displacement? Check all that apply.
xxMikexx [17]

Answer:

the last one, the third one, and the first one.

8 0
3 years ago
Read 2 more answers
An object is pulled with two forces, 10 N northward and 15 N southward. The direction of the net force is to the An object is pu
ValentinkaMS [17]

Answer:

check image

Explanation:

For any question related to newons law of motion first draw the free body diagram(FBD),

7 0
4 years ago
A missile is moving 1350 m/s at a 25.0° angle. It needs to hit a target 23,500 m away in a 55.0° direction in 10.20 s. What is d
QveST [7]

Answer:

  The target's velocity is about 1320 m/s in the direction 265.7°.

Explanation:

In order for there to be a collision between missile and target, we must have ...

  (target starting position) + (target movement) = (missile movement)

assuming the missile starts from the origin of all measurements. The missile moves 10.2 seconds before impact, so moves a distance of ...

  (10.2 s)(1350 m/s) = 13,770 m

__

We are interested in the target movement, so we can solve for that:

  (target movement) = (missile movement) - (target starting position)

In terms of meters, this is ...

  (target movement) = 13770∠25° - 23500∠55° ≈ 13467.74∠-94.3°

The target covers this distance in the same 10.2 seconds before collision, so its speed is (13467.74 m)/(10.2 s) ≈ 1320.4 m/s.

As a positive angle, the target's direction is ...

  -94.3° +360° = 265.7°

The direction of the target's velocity is 265.7°.

_____

If you're calculating this by hand, there are a couple of ways you can do it. You can convert to rectangular coordinates and back (perhaps least confusing), or you can use the law of cosines to solve the triangle, then translate angles back to the x-y coordinate plane.

Using rectangular coordinates, we have ...

  13770∠25° = 13770(cos(25°), sin(25°)) ≈ (12479.9, 5819.45)

  23500∠55° = 23500(cos(55°), sin(55°)) ≈ (13479.0, 19250.1)

Then the difference is ...

  (12479.9, 5819.45) -(13479.0, 19250.1) ≈ (-999.188, -13430.6)

and the (3rd-quadrant) angle is ...

  target direction = arctan(-13430.6/-999.188) ≈ -94.3° = 265.7°

__

The target's speed is found by dividing the distance it covers by the time it takes.

  √(13430.6² +999.188²)/10.2 ≈ 1320.36 . . . m/s

3 0
3 years ago
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