A transformer is used to convert mechanical energy to electric energy

A barbell consists of two small balls, each with mass m at the ends of a very low mass rod of length d. The barbell is mounted o

sveta [45]

The total angular momentum of the system about point B is $L=m_1r_1\omega_1+m_2r_2\omega_2$

Angular momentum, also known as moment of momentum or rotational momentum, is the rotating counterpart of linear momentum.

A rigid object's angular momentum is defined as the product of its moment of inertia and its angular velocity. If there is no external torque on the object, it is analogous to linear momentum and is subject to the fundamental constraints of the conservation of angular momentum principle. The vector quantity angular momentum It is derived from the expression for a particle's angular momentum.

Given,

mass of ball 1 = m1

m₂ mass of ball 2=m2

v₁ is the velocity of ball=r₁ω₁

v₂ is the velocity of ball 2=r₂ω₂

The total angular momentum is given as;

$V_{total}=r_1\omega_1+r_2\omega_2\\\\L=m_1r_1\omega_1+m_2r_2\omega_2$

Hence the total angular momentum will be $L=m_1r_1\omega_1+m_2r_2\omega_2$

To learn more about angular momentum refer here

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6 0

1 year ago

The force an ideal spring exerts on an object is given by Fx = -kx, where x measures the displacement of the object from its equ

shusha [124]

Answer:

The work done by this force can be found via the following formula

$W = \int{F(x)} \, dx = \int\limits^0_{-20} {(-kx)} \, dx = \frac{-kx^2}{2}\left \{ {{x=0} \atop {x=-20}} \right. = \frac{-60*(-20)^2}{2} \\W = -12000J$

Explanation:

Alternatively, the work done by the object is equal to the elastic potantial energy done by the spring.

$U = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2 =0 - \frac{1}{2}60(-20)^2 = -12000J$

6 0

3 years ago

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and

nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}$

k = 1.4

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K$

Work done is given as;

$W = \frac{1}{2} *m*(v_i^2 - v_e^2)$

inlet velocity is negligible;

$v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s$

Therefore, the exit velocity is 629.41 m/s

6 0

3 years ago

Heat is added to an open pan of water at 100.0°c, vaporizing the water. the expanding steam that results does 43.0 kj of work, a

vampirchik [111]

Heat = change in internal energy + Work done The internal energy of a system = heat added and mechanical work done by the system, i.e. U = Q + W rearranging the formula above, will give us: Q = deltaU + W
Q = U - W = 604 kJ - 43.0 kJ = 561,000 J would be the answer.

6 0

3 years ago

Read 2 more answers

Which of the following statements applies to deductive reasoning? 1 Deductive reasoning applies specific knowledge. 2 Deductive

rodikova [14]

Answer:

3 deductive reasoning begins with a general statement hope .

Explanation:

I'mI'm wrong hope I'm not wrong if I'm wrong you can correct me right away thank you

6 0

3 years ago