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3 years ago

What is the input of energy for a device that provides 50 units of useful energy and has an efficiency of 0.55?

1 answer:

3 0

If that's the case, then

      50 units = 0.55 x the input energy

Divide each side by 0.55 :

   50 units/0.55 = the input energy =

   <span> 90 and 10/11 units</span>

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Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at

sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

E = k*Q / r^2

Where, k: Coulomb's Constant

Q: The charge of particle

r: The distance from source

- The Electric Field due to charge 1:

E_1 = k*Q_1 / r^2

E_1 = k*Q / (2*a)^2

E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

E_2 = k*Q_2 / r^2

E_2 = k*Q_2 / (13*a)^2

E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

E_net = E_1 + E_2

2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

1: 2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

2Q = Q / 4 + Q_2 / 169

Q_2 = 295.75*Q

2: 2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

2Q = Q / 4 - Q_2 / 169

Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0

3 years ago

You have a 3.00-liter container filled with N₂ at 25°C and 4.45 atm pressure connected to a 2.00-liter container filled with Ar

LuckyWell [14K]

Answer : The final pressure in the two containers is, 2.62 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

Thus, the expression for final pressure in the two containers will be:

$PV=P_1V_1+P_2V_2$

$P=\frac{P_1V_1+P_2V_2}{V}$

where,

$P_1$ = pressure of N₂ gas = 4.45 atm

$P_2$ = pressure of Ar gas = 2.75 atm

$V_1$ = volume of N₂ gas = 3.00 L

$V_2$ = volume of Ar gas = 2.00 L

P = final pressure of gas = ?

V = final volume of gas = (4.45 + 2.75) L = 7.2 L

Now put all the given values in the above equation, we get:

$P=\frac{(4.45atm)\times (3.00L)+(2.75atm)\times (2.00L)}{7.2L}$

$P=2.62atm$

Thus, the final pressure in the two containers is, 2.62 atm

8 0

3 years ago

What is the range of a ball thrown horizontally at 12 m/s if its time of flight is 3.0 s?

vladimir1956 [14]

Answer:

Explanation:

because I did this assignment, :) your welcome

Next time do it by yourself, but here's the answer kid

7 0

3 years ago

Read 2 more answers

What is formed when 2 or more elements combine chemically

ira [324]

Answer:

compound

Explanation:

a substance made from two or more different elements that have been chemically joined. Examples of compounds include water (H2O), which is made from the elements hydrogen and oxygen, and table salt (NaCl), which is made from the elements sodium and chloride.

6 0

2 years ago

What are the coordinates of the point that is 1/5 of the way from A(-7,-4) to B(3, 6)?

Olegator [25]

Answer:

coordinate will be

r = (1.33,4.33)

Explanation:

As we know that if a point will divide two given coordinates in m : n ratio

then in that case the point is given as

$x = \frac{mx_1 + nx_2}{m+n}$