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2 years ago

A current of 0.2A flows through a component with a resistance of 40ohms calculate the potential difference

Physics

1 answer:

erastova [34]2 years ago

4 0

Answer:

Explanation:

Given parameters:

Current = 0.2A

Resistance = 40ohms

Unknown:

Potential difference = ?

Solution:

Potential difference is force that drives current in an electric circuit.

According to Ohms law, it is given as;

V = IR

V is the potential difference

I is the current

R is the resistance

Now, insert the parameters and solve;

V = 0.2 x 40 = 8V

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A spacecraft at rest has moment of inertia of 100 kg-m^2 about an axis of interest. If a 1 newton thruster with a 1 meter moment

snow_lady [41]

Answer:

18 radians

Explanation:

The computation is shown below:

As we know that

Torque = Force × Moment arm

= 1N × 1M

= 1N-M

Torque = $I\alpha$

$\alpha = \frac{torque}{I}\\\\= \frac{1}{100}\\\\= 0.01 rad/s^2$

Now

$\theta = w_ot + \frac{1}{2} \alpha t^2\\\\w_o = 0\\\\\theta = 0 \times 60 + \frac{1}{2} \times 0.01 \times 60^2\\\\= 18\ radians$

Here t = 1 minutes = 60 seconds

3 0

3 years ago

If 14:15=42:x find the value of x

Yuliya22 [10]

Answer:

x=45

Explanation:

by taking 42 and dividing it by 14 you get 1/3, because 14 is 1/3 of 42 you can then see that the ratio is multiplied by 3. si then you can just multiply 15 by 3 to get 45

3 0

2 years ago

Read 2 more answers

(a) What is the ionization energy of a hydrogen atom that is in the n = 6 excited state? (b) For a hydrogen atom, determine the

crimeas [40]

Answer:

(a) 0.3778 eV

(b) Ratio = 0.0278

Explanation:

The Bohr's formula for the calculation of the energy of the electron in nth orbit is:

$E=\frac {-13.6}{n^2}\ eV$

(a) The energy of the electron in n= 6 excited state is:

$E=\frac {-13.6}{6^2}\ eV$

$E=-0.3778\ eV$

Ionisation energy is the amount of this energy required to remove the electron. Thus, |E| = 0.3778 eV

(b) For first orbit energy is:

$E=\frac {-13.6}{1^2}\ eV$

$E=-13.6\ eV$

$Ratio=\frac {E_6}{E_1}$

$Ratio=\frac {-0.3778}{-13.6}$

Ratio = 0.0278

7 0

3 years ago

Which is normally greater, the energy in ordinary sound or the energy in ordinary light? how does the speed of sound compare to

Thepotemich [5.8K]

The energy in ordinary light is greater than the energy in ordinary sound

Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc. Additionally, there is heat and work, which is energy being transferred from one body to another.

Energy is a physical system's ability to perform labor. The capital letter E is a typical sign for energy. The joule, denoted by the letter J, is the common unit. The energy produced by one newton's (1 N) worth of force acting over one meter's (1 m) worth of displacement is measured in joules (1 J). Because it is a fundamental human requirement, energy plays a significant role in our daily lives.

To learn more about Energy please visit-
brainly.com/question/1932868
#SPJ4

8 0

2 years ago

A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co

Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0

2 years ago