Match the correct term with each part of the wave

An electric furnace is to melt 40 kg of aluminium/hour. The initial temperature of aluminium is 32°C. Given that aluminium has s

gizmo_the_mogwai [7]

Answer:

Part a)

P = 13.93 kW

Part b)

R = 8357.6 Cents

Explanation:

Part A)

heat required to melt the aluminium is given by

$Q = ms\Delta T + mL$

here we have

$Q = 40(950)(680 - 32) + 40(450 \times 10^3)$

$Q = 24624 kJ + 18000 kJ$

$Q = 42624 kJ$

Since this is the amount of aluminium per hour

so power required to melt is given by

$P = \frac{Q}{t}$

$P = \frac{42624}{3600} kW$

$P = 11.84 kW$

Since the efficiency is 85% so actual power required will be

$P = \frac{11.84}{0.85} = 13.93 kW$

Part B)

Total energy consumed by the furnace for 30 hours

$Energy = power \times time$

$Energy = 13.93 kW\times 30 h$

$Energy = 417.9 kWh$

now the total cost of energy consumption is given as

$R = P \times 20 \frac{Cents}{kWh}$

$R = 417.9 kWh\times 20 \frac{cents}{kWh}$

$R = 8357.6 Cents$

3 0

3 years ago

Different kinds of electromagnetic waves are identified by<br> their -?-

const2013 [10]

Answer:

frequency or wavelength

Explanation:

this is why the electromagnetic spectrum has a decreasing wavelength and increasing frequency

4 0

2 years ago

A 70.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 15.0 m

Levart [38]

Answer:

1.3823 rad/s

20.7345 m/s

28.66129935 m/s²

$a=2.92164g$

2006.29095 N radially outward

Explanation:

r = Radius = 15 m

m = Mass of person = 70 kg

g = Acceleration due to gravity = 9.81 m/s²

Angular velocity is given by

$\omega=13.2\times \dfrac{2\pi}{60}\\\Rightarrow \omega=1.3823\ rad/s$

Angular velocity is 1.3823 rad/s

Linear velocity is given by

$v=r\omega\\\Rightarrow v=15\times 1.3823\\\Rightarrow v=20.7345\ m/s$

The linear velocity is 20.7345 m/s

Centripetal acceleration is given by

$a_c=r\omega^2\\\Rightarrow a_c=15\times 1.3823^2\\\Rightarrow a_c=28.66129935\ m/s^2$

The centripetal acceleration is 28.66129935 m/s²

Acceleration in terms of g

$\dfrac{a}{g}=\dfrac{28.66129935}{9.81}\\\Rightarrow a=2.92164g$

$a=2.92164g$

Centripetal force is given by

$F_c=ma_c\\\Rightarrow F_c=70\times 28.66129935\\\Rightarrow F_c=2006.29095\ N$

The centripetal force is 2006.29095 N radially outward

The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.

3 0

3 years ago

The velocity of a 1.3 kg remote-controlled car is plotted on the graph. The work of segment A is J.

tamaranim1 [39]

Answer: 585 J

Explanation:

We can calculate the work done during segment A by using the work-energy theorem, which states that the work done is equal to the gain in kinetic energy of the object:

$W=K_f -K_i$

where Kf is the final kinetic energy and Ki the initial kinetic energy. The initial kinetic energy is zero (because the initial velocity is 0), while the final kinetic energy is

$K_f =\frac{1}{2}mv^2$