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3 years ago

An object has a velocity of 8 m/s and a kinetic energy of 480 J. What is the mass of the object? (Formula: )

Physics

1 answer:

andrezito [222]3 years ago

6 0

Answer:

1/2mv^2 a(7.5 b(15 kg c(60 kg d(120 kg. nevermind i found the answer its (15 kg) because to solve for m its m= K2/v squared.

Explanation:

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What is mechanical waves

LuckyWell [14K]

Mechanical waves are those waves that require a material medium for propagation.

<h3>What are mechanical waves?</h3>

Generally, we define a wave as a disturbance along a medium which transfers energy. It then follows that waves move energy from one point to another.

Waves can be classified as;

Mechanical waves
Electromagnetic waves

Mechanical waves are those waves that require a material medium for propagation such as sound, and waves on a strings.

Learn more about mechanical waves: brainly.com/question/9242091

6 0

2 years ago

Read 2 more answers

Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is

saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

$\dot{m_{i}} = \rho_{i} v_{i}A_{i}$

Where:

$v_{i}$ : is the initial velocity = 20 m/s

$A_{i}$ : is the inlet area of the nozzle = 60 cm²

$\rho_{i}$ : is the density of entrance = 2.21 kg/m³

$\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s$

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

$\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}$

$0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}$

$A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}$

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!

5 0

3 years ago

Read 2 more answers

An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to v

maxonik [38]

Answer:

a) The final pressure is 1.68 atm.

b) The work done by the gas is 305.3 J.

Explanation:

a) The final pressure of an isothermal expansion is given by:

$T = \frac{PV}{nR}$

$T_{i} = T_{f}$

$\frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR}$

Where:

$P_{i}$ : is the initial pressure = 5.79 atm

$P_{f}$ : is the final pressure =?

$V_{i}$ : is the initial volume = 420 cm³

$V_{f}$ : is the final volume = 1450 cm³

n: is the number of moles of the gas

R: is the gas constant

$P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm$

Hence, the final pressure is 1.68 atm.

b) The work done by the isothermal expansion is:

$W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J$

Therefore, the work done by the gas is 305.3 J.

I hope it helps you!

3 0

3 years ago

A constant magnetic flux through a closed loop of wire induces an emf in that loop. True or false?

miskamm [114]

Answer: False

Explanation: In order to explain this problem we have to use the Faraday law, which say

dФm/dt=-ε it means that the variation of the magnetic field flux with time is equal to the emf ( electromotive force). In our case the magnetic flux is constant then there is not a emf induced in a wire closed loop.

6 0

3 years ago

2. An electron and a proton are separated by 5 cm:

Diano4ka-milaya [45]

Answer:

a) the charge of an electron is equivalent to the magnitude of the elementary charge but barring a negative sign since the side of the elementary charge is roughly 1.602 * 10 - 19 Columbus then the charge of the electronic is-1.602 * 10 - 19

b) b=2T on the electron moving in the magnetic field

7 0

2 years ago