Which part of vernier caliper is used to measure internal diameter of cylinder

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

An electron enters a region of space containing a uniform 0.0000193-T magnetic field. Its speed is 121 m/s and it enters perpend

Mandarinka [93]

Answer

Given,

Magnetic field, B = 0.0000193 T

speed, v = 121 m/s

mass of electron, m = 9.11 x 10⁻³¹ Kg

charge of electron, q = 1.6 x 10⁻¹⁹ C

radius of the electron path, r = ?

$r = \dfrac{mv}{qB}$

$r = \dfrac{9.31\times 10^{-31}\times 121}{1.6\times 10^{-19}\times 0.0000193}$

r = 3.64 x 10⁻⁵ m

We know frequency is inverse of time period

d = v t

$2 \pi r = v \times t$

$t = \dfrac{2 \pi r}{v}$

$t = \dfrac{2 \pi \times 3.64\times 10^{-5}}{121}$

t = 1.889 x 10⁻⁶ s.

now, frequency

$f = \dfrac{1}{t}$

$f = \dfrac{1}{1.889\times 10^{-6}}$

$f = 529380\ Hz$

3 0

3 years ago

The most important barriers to diffusion today are physical features of Earth's surface.

nikdorinn [45]

Answer:

False

Explanation:

Physical features of earth's surface like mountains, deserts and oceans are not the barriers in today's modern world. Technological improvements in the transportation and communication has removed these barriers. So they are not barriers anymore.

7 0

3 years ago

Lighting from the sun travels through space to earths atmosphere.

Lady bird [3.3K]

Yes it does do all that

3 0

3 years ago

I will be so thankful if u answer correctly!!

olga_2 [115]

<h2>Answer:</h2>

(a) 10N

<h2>Explanation:</h2>

The sketch of the two cases has been attached to this response.

<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>

In this case, a frictional force $F_{r}$ is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;

∑F = ma -------------------(i)

Where;

∑F = effective force acting on the object (box)

m = mass of the object

a = acceleration of the object

∑F = F - $F_{r}$

m = 50kg

a = 0 [At constant velocity, acceleration is zero]

<em>Substitute these values into equation (i) as follows;</em>

F - $F_{r}$ = m x a

F - $F_{r}$ = 50 x 0

F - $F_{r}$ = 0

F = $F_{r}$ -------------------(ii)

<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>

In this case, the same frictional force $F_{r}$ is opposing the movement of the box.

∑F = 1.5F - $F_{r}$

m = 50kg

a = 0.1m/s²

<em>Substitute these values into equation (i) as follows;</em>

1.5F - $F_{r}$ = m x a

1.5F - $F_{r}$ = 50 x 0.1

1.5F - $F_{r}$ = 5 ---------------------(iii)

<em>Substitute </em> $F_{r}$ <em> = F from equation (ii) into equation (iii) as follows;</em>

1.5F - F = 5

0.5F = 5

F = 5 / 0.5

F = 10N

Therefore, the value of F is 10N

<em />

4 0

2 years ago

Which part of vernier caliper is used to measure internal diameter of cylinder ​

Which part of vernier caliper is used to measure internal diameter of cylinder