In which type of chemical reaction is heat absorbed from the surroundings?

The objects in the inner part of our solar system are composed mostly of

N76 [4]

“Inner Planets” – Mercury, Venus, Earth, and Mars – which are so named because they orbit closest to the Sun. ... For starters, the inner planets are rocky and terrestrial, composed mostly of silicates and metals, whereas the outer planets are gas giants.

4 0

3 years ago

Câu 1. Con lắc lò xo treo thẳng đứng, dao động điều hòa với biên độ 2cm và tần số góc 20 rad/s. Chiều dài tự nhiên của lò xo là

natka813 [3]

Answer:

28 cm and 32 cm

Explanation:

1. The spring pendulum hangs vertically, oscillates harmonic with amplitude 2cm and angular frequency 20 rad/s. The natural length of

a spring is 30cm. What is the minimum and maximum length of the spring during the oscillation? Take g = 10m/s2.

As the amplitude is 2 cm and the natural length is 30 cm. So, it oscillates between 30 -2 = 28 cm to 30 + 2 = 32 cm.

So, the minimum length is 28 cm and the maximum length is 32 cm.

8 0

3 years ago

How many water (in kg) can raise its temperature from 23°C to 100°C by adding 21500 Joules of energy?

igomit [66]

Answer: Approximately 8.0g of water

Explanation:

4 0

3 years ago

An engine causes a car to move 10 meters with a force of 100 N. The engine produces 10,000 J of energy. What is the efficiency o

kodGreya [7K]

Answer:

Part A

The efficiency of the engine is 10%

Part B

The change in internal energy is 300 J

Part C

The change in volume is 1 m³ which is one cubic meter

Explanation:

Part A

Efficiency is defined as the ratio of energy output to energy input;

The given parameters are;

The distance the car is moved, d = 10 meters

The force which moves the car, F = 100 N

The amount of energy produced by the engine = 1,000 J

Therefore, we have;

The energy output to the car = The work done on the car = Force applied to the car, F × The distance the car moves, d;

∴ The energy output to the car by the engine = F × d = 100 N × 10 m = 1,000 J

The energy input from the engine = The energy produced by the engine = 10,000 J

The efficiency of the engine = (The energy output)/(The energy input)= 1,000J/10,00J = 0.1

The efficiency in percentage = 0.1 × 100 = 10 %

The efficiency of the engine = 10%

Part B

The amount of heat added to the substance, ΔQ = 1,000J

The amount of work the substance does on the atmosphere, W = 700 J

The change in internal energy, ΔU is given as follows;

ΔQ = W + ΔU

∴ ΔU = ΔQ - W

For the substance, we have;

The change in internal energy, ΔU = 1,000 J - 700 J = 300 J

Part C

The work done by the piston, W = 1,000 J

The pressure, in the piston, P = 1,000 Pa = constant

The work done by the piston in a constant pressure process, W = P × ΔV

Where;

W = The work done

P = The constant pressure applied

ΔV = The change in volume = V₂ - V₁

V₂ = The final volume

V₁ = The initial volume

∴The change in volume ΔV = W/P = 1,000 J/(1,000Pa) = 1 m³

The change in volume ΔV = 1 m³

3 0

3 years ago

A closely wound search coil has an area of 3.21 cm2, 120 turns, and a resistance of 58.7 O. It is connected to a charge-measurin

Alexxx [7]

Answer:

The magnetic field in the System is 0.095T

Explanation:

To solve the exercise it is necessary to use the concepts related to Faraday's Law, magnetic flux and ohm's law.

By Faraday's law we know that

$\epsilon = \frac{NBA}{t}$

Where,

$\epsilon =$ electromotive force

N = Number of loops

B = Magnetic field

A = Area

t= Time

For Ohm's law we now that,

V = IR

Where,

I = Current

R = Resistance

V = Voltage (Same that the electromotive force at this case)

In this system we have that the resistance in series of coil and charge measuring device is given by,

$R = R_c + R_d$

And that the current can be expressed as function of charge and time, then

$I = \frac{q}{t}$

Equation Faraday's law and Ohm's law we have,

$V = \epsilon$

$IR = \frac{NBA}{t}$

$(\frac{q}{t})(R_c+R_d) = \frac{NBA}{t}$

Re-arrange for Magnetic Field B, we have

$B = \frac{q(R_c+R_d)}{NA}$

Our values are given as,

$R_c = 58.7\Omega$

$R_d = 45.5\Omega$

$N = 120$

$q = 3.53*10^{-5}C$

$A = 3.21cm^2 = 3.21*10^{-4}m^2$

Replacing,

$B = \frac{(3.53*10^{-5})(58.7+45.5)}{120*3.21*10^{-4}}$

$B = 0.095T$

Therefore the magnetic field in the System is 0.095T

3 0

3 years ago