7. A single-pole GFCI breaker is rated at

During a cold winter day, wind at 50 km/h is blowing parallel to a 4-m-high and 10-m-long wall of a house. If the air outside is

fenix001 [56]

Answer:

to determine the rate of heat loss from that wall by convection = 12780 watts

Explanation:

8 0

3 years ago

Describe the make-up of an internal combustion engine.<br> Pls answer quickly.

Whitepunk [10]

Answer:

The engine consists of a fixed cylinder and a moving piston. The expanding combustion gases push the piston, which in turn rotates the crankshaft. Ultimately, through a system of gears in the power-train, this motion drives the vehicle's wheels.

Explanation:

8 0

3 years ago

Which lists the order of Energy Career Pathways from the source to the customer?

Luden [163]

Answer:

answer is A. EG,Et, and Ed

Explanation:

4 0

3 years ago

Read 2 more answers

The 2-lb block is released from rest at A and slides down along the smooth cylindrical surface. Of the attached spring has a sti

MA_775_DIABLO [31]

Answer:

L = 4.574 ft

Explanation:

Given:

- The weight of the block W = 2 lb

- The initial velocity of the block v_i = 0

- The stiffness of the spring k = 2 lb/ft

- The radius of the cylindrical surface r = 2 ft

Find:

Determine its unstretched length so that it does not allow the block to leave the surface until θ= 60°.

Solution:

- Compute the velocity of the block at θ= 60°. Use Newton's second equation of motion in direction normal to the surface.

F_n = m*a_n

Where, a_n is the centripetal acceleration or normal component of acceleration as follows:

a_n = v^2_2 / r

- Substitute:

F_n = m*v^2_2 / r

Where, F_n normal force acting on block by the surface is:

F_n = W*cos(θ)

- Substitute:

W*cos(θ) = m*v^2_2 / r

v_2 = sqrt ( r*g*cos(θ) )

- Plug in the values:

v^2_2 = 2*32.2*cos(60)

v^2_2 = 32.2 (ft/s)^2

- Apply the conservation of energy between points A and B where θ= 60° :

T_A + V_A = T_B + V_B

Where,

T_A : Kinetic energy of the block at inital position = 0

V_A: potential energy of the block inital position

V_A = 0.5*k*x_A^2

x_A = 2*pi - L ..... ( L is the original length )

V_A = 0.5*2*(2*pi - L)^2 =(2*pi - L)^2

T_B = 0.5*W/g*v_2^2 = 0.5*2 / 32.2 *32.2 = 1

V_B = 0.5*k*x_B^2 + W*2*cos(60)

x_B = 2*0.75*pi - L ..... ( L is the original length )

V_B = 0.5*2*(1.5*pi - L)^2 + 2*1 = 2 + ( 1.5*pi - L )^2

- Input the respective energies back in to the conservation expression:

0 + (2*pi - L)^2 = 1 + 2 + ( 1.5*pi - L )^2

4pi^2 - 4*pi*L + L^2 = 3 + 2.25*pi^2 - 3*pi*L + L^2

pi*L = 1.75*pi^2 - 3

L = 4.574 ft

3 0

3 years ago

An inventor claims to have developed a power cycle operating between hot and cold reservoirs at 1350 K and 295 K, respectively,

Art [367]

Answer:

efficiency = 0.71999 < 0.78148

efficiency = 0.7999 ≈ 0.78148

Explanation:

given data

Hot reservoir Temp Th = 1350 K

Cold reservoir Temp Tc = 295 K

heat transfer rate = 150,000 kJ/h

solution

first we get here maximum possible efficiency that is express as

maximum possible efficiency = 1 - $\frac{Tc}{Th}$ ..........1

put here value and we get

maximum possible efficiency = 1 - $\frac{295}{1350}$

maximum possible efficiency = 0.78148

and

efficiency = $\frac{Wout}{Qin}$ .................2

here Qin = $\frac{150000}{3600}$ = 41.667

so put here value in equation 2 we get

efficiency = $\frac{30}{41.667}$

efficiency = 0.71999 < 0.78148

and

efficiency = $\frac{33.33}{41.667}$

efficiency = 0.7999 ≈ 0.78148

claim might be true or not because there always be looser

5 0

4 years ago