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DaniilM [7]
2 years ago
13

) What is the electric potential due to the nucleus of hydrogen at a distance of 7.50× 10-11 m? Assume the potential is equal to

zero as r→∞. (Express your answer to three significant figures.)
Physics
2 answers:
ohaa [14]2 years ago
5 0
For this, we need the formula:
V = k q / r
where k is the Coulombs law constant = 9 x 10^9 N
q is the charge of the hydrogen nucleus (proton) = <span>1.6 x 10^-19 C</span> 
r is the distance
Simply plug in the values and solve for V
Kipish [7]2 years ago
5 0

Answer:

Electric potential, V = 19.2 volts

Explanation:

In this case we need to find the electric potential due to the nucleus of hydrogen at a distance of 7.5\times 10^{-11}\ m. For this we must know the concept of electric potential. It is the work done in moving a charged particle in electric field from zero to a particular point. Its formula is given by :

V=\dfrac{kq}{r}

q is the charge of nucleus of hydrogen i.e. proton

V=\dfrac{9\times 10^9\times 1.6\times 10^{-19}}{7.5\times 10^{-11}}

V=19.2\ \text{volt}

So, the electric potential due to the nucleus of hydrogen at a given distance is 19.2 volts. Hence, this is the required solution.

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The MAC is 58 inches, The CG limits are from 26% to 43% MAC. If the CG is found to be
eimsori [14]

By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.

<h3>What are the limits?</h3>

First, we need to find the limits.

We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.

So if 58 in is the 100%, the 26% and 43% of that are:

  • 26% → (26%/100%)*58in = 0.26*58 in = 15.08 in
  • 43% → (43%/100%)*58in = 0.43*58 in = 24.94 in.

But we know that the CG is found to be 45.5% MAC, then it measures:

(45.5%/100%)*58in = 0.455*58in = 26.39 in

We need to compare it with the largest limit, so we get:

26.39 in - 24.94 in = 1.45 in

This means that the CG is 1.45 inches out of limits.

If you want to learn more about percentages, you can read:

brainly.com/question/14345924

6 0
2 years ago
a rocket, initially at rest, is fired vertically with a net upward acceleration of 12 m/s2 . at an altitude of 0.50 km, the engi
kobusy [5.1K]

The rocket travelled a maximum height at 1.0102 km.

Given,

The acceleration of a rocket (a) = 12 m/s²

The altitude of the rocket (s) =  0.50 km = 0.5×10³m

The maximum height of the rocket (h) = ?

Solution,

A rocket is a spacecraft, aircraft, vehicle or projectile that obtains thrust from a rocket engine.

The rate of change of the velocity of an object with respect to time is known as acceleration. It is denoted by (a).i.e. unit is m/s²

(a) = Δv/Δt

Where , Δv is change in velocity and Δt is change in time.

The rate of change in position with respect to time is known as velocity. i.e. Its unit is m/s.

(v)= Δx/Δt

Where,Δx is the change in position and Δt is change in time & v is velocity.

Therefore we know the equation of motion is written as,

v² = u² +2as

Where, v  is final velocity , u is initial velocity , a is acceleration and s is altitude of the rocket.

Then putting the value ,

v² = 0 + ( 2× 10 × 0.5×10³)m/s

v² = \sqrt{10000} m/s

v = 100 m/s

Therefore, at altitude of 0.50 km the initial velocity of rocket (u) will be 100 m/s, final velocity v become zero and under free falling the acceleration will be taken (-g) then equation of motion can be given as ,

v² = u² - 2(g)h

h = (v²- u² ) / 2g

h = 10,000/2×9.8

h = 510.2 m

So that the rocket travelled the maximum height ,

(h)= (0.5 km + 510.2m)

(h) = 1.0102 km

Hence, the rocket travelled at the maximum height h is 1.0102 km

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Answer:

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Explanation:

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