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Firdavs [7]
4 years ago
8

Can someone solve this problem and explain to me how you got it​

Physics
1 answer:
evablogger [386]4 years ago
4 0

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒F\alpha\frac{q1*q2}{r^{2}}

∴F=k\frac{q1*q2}{r^{2}}

where F is the force of attraction or repulsion

k is Coulumb's constant=9*10^{9}Nm^{2}C^{-2}

q1 and q2 are the magnitude of the charges

r is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

F=9*10^{9}*\frac{0.041*0.029}{12^{2}}

F=74312.5N

Question6:

Given: q1=3*10^{-18}C, r=5 m, F=4*10^{-11}N

therefore by Coulumb's law,

4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}

⇒q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C

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Answer: option 1 : the electric potential will decrease with an increase in y

Explanation: The electric potential (V) is related to distance (in this case y) by the formulae below

V = kq/y

Where k = 1/4πε0

Where V = electric potential,

k = electric constant = 9×10^9,

y = distance of potential relative to a reference point, ε0 = permittivity of free space

q = magnitude of electronic charge = 1.609×10^-19 c

From the formulae, we can see that q and k are constants, only potential (V) and distance (y) are variables.

We have that

V = k/y

We see the potential(V) is inversely proportional to distance (y).

This implies that an increase in distance results to a decreasing potential and a decrease in distance results to an increase in potential.

This fact makes option 1 the correct answer

3 0
3 years ago
A teacher wants to demonstrate that the radioactive source emits alpha beta and gamma radiation. Describe a method the teacher c
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By using an electric field, it is feasible to differentiate between these different forms of radiation.

<h3>What is a radioactive source?</h3>

A source that emits radiation like gamma, beta, and alpha rays is said to be radioactive. Using an electric field, we can discriminate between these different forms of radiation.

The field does not deflate the gamma rays, but it does deflate the alpha and beta rays, with the alpha being deflated to the field's negative portion and the beta to its positive part.

Hence, by using an electric field, it is feasible to differentiate between these different forms of radiation.

To learn more about the radioactive source refer;

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8 0
2 years ago
a wire with mass per unit length 75 g/m runs horizontally at right angles to a uniform horizontal 0.12 T magnetic field. what am
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Answer:

The amount of current that must flow through the wire for it to be suspended against gravity by magnetic force = 6.125 A

Explanation:

Force on a wire carrying current in an electric field is given by

F = (B)(I)(L) sin θ

For this question,

The magnetic force must match the weight of the wire.

F = mg

mg = (B)(I)(L) sin θ

(m/L)g = (B)(I) sin θ

Mass per unit length = 75 g/m = 0.075 kg/m

B = magnetic field = 0.12 T

I = ?

g = acceleration due to gravity = 9.8 m/s

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0.075 × 9.8 = 0.12 × I sin 90°

I = 0.075 × 9.8/0.12 = 6.125 A

3 0
3 years ago
Compute your average velocity in the following two cases: (a) You walk 50.2 m at a speed of 2.21 m/s and then run 50.2 m at a sp
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Answer:

a) 2.87 m/s

b) 3.23 m/s

Explanation:

The avergare velocity can be found dividing the length traveled d by the total time t.

a)

For the first part we easily know the total traveled length which is:

d = 50.2 m + 50.2 m = 100.4 m

The time can be found dividing the distance by the velocity:

t1 = 50.2 m / 2.21 m/s = 22.7149 s

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Therefore, the average velocity is:

v = d/t =2.87 m/s

b)

Here we can easily know the total time:

t = 1 min + 1.16 min = 129.6 s

Now the distance wil be found multiplying each velocity by the time it has travelled:

d1 = 2.21 m/s * 60 s = 132.6 m

d2 = 4.11 m/s *(1.16 * 60 s) = 286.056 m

d = 418.656 m

Therefore, the average velocity is:

v = d/t =3.23 m/s

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