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Romashka-Z-Leto [24]
3 years ago
13

A research van de graaff generator has a 2.00-mdiameter metal sphere with a charge of 5.00 mc on it. (a) what is the potential n

ear its surface? (b) at what distance from its center is the potential 1.00 mv? (c) an oxygen atom with three missing electrons is released near the van de graaff generator. what is its energy in mev when the atom is at the distance found in part b?
Physics
1 answer:
Yuki888 [10]3 years ago
5 0
(a) The potential on the surface of a charged sphere of radius R is equal to
V(R) = k_e  \frac{Q}{R}
where k_e = 8.99 \cdot 10^9 N m^2 C^{-2} is the Coulomb's constant, Q is the charge on the sphere's surface.
For the generator mentioned in the problem, the charge is Q= 5 mC=5 \cdot 10^{-3} C, while the radius is R= \frac{d}{2}= \frac{2.0 m}{2} =1.0 m. Using these values in the formula, we can calculate the potential at the surface:
V(R)=8.99 \cdot 10^9 N m^2 C^{-2}  \frac{5 \cdot 10^{-3} C}{1.0 m}=4.5 \cdot 10^7 V

(b) The potential generated by the sphere at a certain distance r from the centre of the sphere is given by
V(r) = k_e  \frac{Q}{r}
the problem asks at which distance V(r) = 1 mV=1\cdot 10^{-3} V. Substituting in the previous formula we can find the value of r:
r=k_e  \frac{Q}{V(r)}= 8.99 \cdot 10^9 N m^2 C^{-2} \frac{5 \cdot 10^{-3}}{1\cdot 10^{-3} V}=4.5 \cdot 10^{10} m

(c) An oxygen atom with 3 missing electrons has a positive charge of +3e, with e being the elementary charge.
The electric potential energy of a charged particle located at some point with voltage V is
U=q V
where q is the charge of the particle, which is in our case q=+3e. So we can calculate the energy of the oxygen atom at the distance found in part b, which corresponds to r=4.5 \cdot 10^{10}m and a voltage of V=1 mV:
U=(3 e)(1 mV) = 3 meV
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notka56 [123]

The distance x will the ball land after flies off with a horizontal initial velocity  is 3.0635 m.

<h3>What is mechanical energy?</h3>

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.

M.E = KE +PE

A small metal ball with a mass of m = 62.0 g is attached to a string of length l = 1.85 m. It is held at an angle of θ = 48.5° with respect to the vertical.

The ball is then released. When the rope is vertical, the ball collides head-on and perfectly elastically with an identical ball originally at rest. This second ball flies off with a horizontal initial velocity from a height of h = 3.76 m, and then later it hits the ground.

The conservation of energy principle states that total mechanical energy remains conserved in all situations where there is no external force acting on the system.

Kinetic energy  = Potential energy

1/2 mv² =mgh₁

The velocity at the bottom, when the height h = 5m, is

v= √2gh₁...................(1)

The vertical height h₁ = l- lcosθ

h₁ = l- lcosθ

h₁ = 1.85 - 1.85cos48.5°

h₁ =0.6241 m

Putting the values in equation (1), we get

v = √2x 9.81 x0.6241

v = 3.499 m/s

The horizontal distance traveled is

x = vt

x = v x √2h/g

Plug the values, we get

x =  3.499 x √2x3.76 / 9.81

x = 3.0635 m

Thus, the horizontal distance ball travels is  3.0635 m.

Learn more about mechanical energy.

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6 0
1 year ago
When she rides her bike, she gets to her first classroom building 36 minutes faster than when she walks. Of her average walking
postnew [5]

Answer:

d=2.4\ miles

Explanation:

Given:

  • average walking speed, v_w=3\ mph
  • average biking speed, v_b=12\ mph

<u>According to given condition:</u>

t_w=t_b+\frac{36}{60}

where:

t_w= time taken to reach the building by walking

t_b= time taken to reach the building by biking

We know that,

\rm time=\frac{distance}{speed}

so,

\frac{d}{v_w} =\frac{d}{v_b} +\frac{36}{60}

\frac{d}{3}=\frac{d}{12} +\frac{3}{5}

d=2.4\ miles

7 0
3 years ago
Read 2 more answers
A copper telephone wire has essentially
Lunna [17]

Answer:

128.21 m

Explanation:

The following data were obtained from the question:

Initial temperature (θ₁) = 4 °C

Final temperature (θ₂) = 43 °C

Change in length (ΔL) = 8.5 cm

Coefficient of linear expansion (α) = 17×10¯⁶ K¯¹)

Original length (L₁) =.?

The original length can be obtained as follow:

α = ΔL / L₁(θ₂ – θ₁)

17×10¯⁶ = 8.5 / L₁(43 – 4)

17×10¯⁶ = 8.5 / L₁(39)

17×10¯⁶ = 8.5 / 39L₁

Cross multiply

17×10¯⁶ × 39L₁ = 8.5

6.63×10¯⁴ L₁ = 8.5

Divide both side by 6.63×10¯⁴

L₁ = 8.5 / 6.63×10¯⁴

L₁ = 12820.51 cm

Finally, we shall convert 12820.51 cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

12820.51 cm = 12820.51 cm × 1 m / 100 cm

12820.51 cm = 128.21 m

Thus, the original length of the wire is 128.21 m

5 0
3 years ago
A rocket ship starts from rest and turns on its forward booster rockets causing it to have a constant acceleration of 4 meters p
bagirrra123 [75]

Complete question is;

A rocket ship starts from rest and turns on its forward booster rockets, causing it to have a constant acceleration of 4 m/s² rightward. After 3s, what will be the velocity of the rocket ship?

Answer:

v = 12 m/s

Explanation:

We are given;

Initial velocity; u = 0 m/s (because ship starts from rest)

Acceleration; a = 4 m/s²

Time; t = 3 s

To find velocity after 3 s, we will use Newton's first equation of motion;

v = u + at

v = 0 + (4 × 3)

v = 12 m/s

6 0
3 years ago
Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P P and volume V V satisfy th
Verdich [7]

Answer:

the volume decreases at the rate of 500cm³ in 1 min

Explanation:

given

v = 1000cm³, p = 80kPa, Δp/t= 40kPa/min

PV=C

vΔp + pΔv = 0

differentiate with respect to time

v(Δp/t) + p(Δv/t) = 0

(1000cm³)(40kPa/min) + 80kPa(Δv/t) = 0

40000 + 80kPa(Δv/t) = 0

Δv/t = -40000/80

= -500cm³/min

the volume decreases at the rate of 500cm³ in 1 min

3 0
2 years ago
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