What is one use for infrared waves? A radar B Medical imaging C Thermal imaging cameras

The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the c

maria [59]

To solve this problem we will start using the concepts related to the electric field, from there we will find the load exerted on the body. Through this load it will be possible to make a sum of forces in balance to find the load that a human supports. Finally with these values it will be possible to find the repulsive force. We will proceed as follows,

The electric field is

$E= \frac{kQ}{R^2}$

Here,

k = Coulomb's Constant

Q = Charge

R = Distance (At this case from the center of mass of the earth to the surface)

Rearranging to find the charge,

$Q = \frac{ER^2}{k}$

Replacing,

$Q = frac{(150)(6.38*10^6)}{8.99*10^9}$

$Q = 6.79*10^5 C$

Since the electric field is directed towards the center of earth, the charge is negative.

PART A) Once the load is found we can proceed to apply the balance of Forces, for which the electrostatic force must be equivalent to the weight, this in order to satisfy the balance, therefore

$F_w = F_e$

$mg = \frac{kQq}{R^2}$

Replacing,

$(62)(9.8) = \frac{(8.99*10^9)(q)(-6.79*10^5)}{(6.38*10^6)^2}$

Solving for q,

$q = -4.056C$

PART B) Finally using the given distance and the values of the found load we can find the repulsive Force, which is

$F =\frac{kq^2}{d^2}$

$F = \frac{(8.99*10^9)(-4.056)^2}{110^2}$

$F = 1.22*10^7N$

PART C) The answer is no. According to the information found, we can conclude that traveling through an electric field is not viable because there is a repulsive force of great magnitude acting on the body.

3 0

3 years ago

What is the magnitude (in N/C) and direction of an electric field that exerts a 3.50 ✕ 10−5 N upward force on a −1.55 µC charge?

IRISSAK [1]

Answer:

The magnitude of electric field is 22.58 N/C

Solution:

Given:

Force exerted in upward direction, $\vec{F_{up}} = 3.50\times 10^{- 5} N$

Charge, Q = $- 1.55\micro C = - 1.55\times 10^{- 6} C$

Now, we know by Coulomb's law,

$F_{e} = \frac{1}{4\pi\epsilon_{o}\frac{Qq}{R^{2}}$

Also,

Electric field, $E = \frac{1}{4\pi\epsilon_{o}\frac{q}{R^{2}}$

Thus from these two relations, we can deduce:

F = QE

Therefore, in the question:

$\vec{E} = \frac{\vec F_{up}}{Q}$

$\vec{E} = \frac{3.50\times 10^{- 5}}{- 1.55\times 10^{- 6}}$

$\vec{E} = - 22.58 N/C$

Here, the negative side is indicative of the Electric field acting in the opposite direction, i.e., downward direction.

The magnitude of the electric field is:

$|\vec{E}| = 22.58\ N/C$

6 0

3 years ago

Andrew pokes a marble, and the marble rolls down a ramp. The marble moves with speed. Which forces are acting on the marble in t

stich3 [128]

The forces are Andrew poking the marble and then gravity pulling the marble downward

4 0

3 years ago

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A student rides a bicycle for 15 miles in 3 hours. What is the student's speed? What else would you need to know for the velocit

kaheart [24]

Answer:

5 miles per hour

Explanation:

if you divide 15 by 3 you get 5, therefore the student is going 5 miles per hour.

3 0

2 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.16

ycow [4]

We assign the variables: T as tension and x the angle of the string
The centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²)/0.9 = T+0.163gcosx, giving T=(0.163x4.87²)/0.9 – 0.163x9.8cosx.

(1)At the bottom of the circle x=π and T=(0.163x4.87²)/0.9 – .163*9.8cosπ=5.893N. 
(2)Here x=π/2 and T=(0.163x4.87²)/0.9 – 0.163x9.8cosπ/2=4.295N. 
(3)Here x=0 and T=(0.163x4.87²)/0.9 – 0.163x9.8cos0=2.698N. 
(4)We have T=(0.163v²)/0.9 – 0.163x9.8cosx.

This minimum v is obtained when T=0 and v verifies (0.163xv²)/0.9 – 0.163x9.8=0, resulting to v=2.970 m/s.

4 0

3 years ago

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