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ladessa [460]
3 years ago
13

Calculate the speed (in m/s) of an electron and a proton with a kinetic energy of 1.85 electron volt (eV). (The electron and pro

ton masses are me = 9.11 ✕ 10−31 kg and mp = 1.67 ✕ 10−27 kg. Boltzmann's constant is kB = 1.38 ✕ 10−23 J/K.)
Physics
1 answer:
Natali5045456 [20]3 years ago
6 0

Answer:

ve = 8.06 x 10^5 m/s

vp = 1.882 x 10^4 m/s

Explanation:

K = 1.85 eV = 1.85 x 1.6 x 10^-19 J = 2.96 x 10^-19 J

me = 9.11 ✕ 10^-31 kg,

mp = 1.67 ✕ 10^-27 kg

Let the speed of electron is ve and the proton is vp.

For electron :

K = 1/2 me x ve^2

2.96 x 10^-19 = 0.5 x 9.11 x 10^-31 x ve^2

ve^2 = 6.498 x 10^11

ve = 8.06 x 10^5 m/s

For proton:

K = 1/2 mp x vp^2

2.96 x 10^-19 = 0.5 x 1.67 x 10^-27 x vp^2

vp^2 = 3.5449 x 10^8

vp = 1.882 x 10^4 m/s

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A 500kg elevator is raised to a height of 10 m in 10 seconds. Calculate the power of the motor.
Hunter-Best [27]

Answer:

Correct answer: Third statement  P = 4900 W

Explanation:

Given:

m = 500 kg  the mass of the elevator

h = 10 m  reached height after t = 10 seconds

P = ? power of the motor

The formula for the calculating power of the motor is:

P = W / t

since work is a measure of change in this case of potential energy then it is:

W = ΔEp = Ep - 0 = Ep

In this case we must take g = 9.81 m/s²

Ep = m g h = 500 · 9.81 · 10 = 49,050 W ≈ 49,000 W

Ep ≈ 49,000 W

P = Ep / t = 49,000 / 10 = 4,900 W

P =4,900 W

God is with you!!!

3 0
3 years ago
A 2.00 kg block hangs from a spring balance calibrated in Newtons that is attached to the ceiling of an elevator.(a) What does t
tresset_1 [31]

Answer:

Part a)

Reading = 2.00 kg

Part b)

Reading = 2.00 kg

Part c)

Reading = 4.04 kg

Part d)

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

Explanation:

Part a)

When elevator is ascending with constant speed then we will have

F_{net} = 0

T - mg = 0

T = mg

So it will read same as that of the mass

Reading = 2.00 kg

Part b)

When elevator is decending with constant speed then we will have

F_{net} = 0

T - mg = 0

T = mg

So it will read same as that of the mass

Reading = 2.00 kg

Part c)

When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have

F_{net} = ma

T - mg = ma

T = mg + ma

Reading is given as

Reading = \frac{mg + ma}{g}

Reading = 2.00\frac{9.81 + 10}{9.81}

Reading = 4.04 kg

Part d)

Here the speed of the elevator is constant initially

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

3 0
3 years ago
1. A student practicing for a track meet ran 263 m in 30 sec. What was her average speed?
faltersainse [42]

Answer: By 47

Explanation: subtract 310 and 263

6 0
3 years ago
Why is the sky blue?
ladessa [460]

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5 0
3 years ago
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A basket has a mass of 5.5 kg. Find the magnitude of the normal force if the basket is at rest on a ramp inclined above the hori
storchak [24]
Here is the correct answer of the given problem above.
Given that the basket has a mass of 5.5kg, the magnitude of the normal force if the basket is at rest on a ramp inclined above the horizontal is at 12 degrees. The solution is simple: 
<span>Fn at rest = lmgl </span>
<span>= 5.5kg (9.80N/kg) 
=</span><span> mgCos12degrees 
Hope this answer helps. </span>
8 0
4 years ago
Read 2 more answers
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