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2 years ago

EARTH SCIENCE HELP FAST PLEASE

Physics

1 answer:

xxTIMURxx [149]2 years ago

8 0

I think its C not sure tho

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What is the logical oreder for these words : verse, interlude,verse,refrain, introduction,refrain,coda,refrain,verse

Nikolay [14]

Sorry to say but I know that t(e introduction is first and the coda is last

3 0

3 years ago

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

2 years ago

Imagine imagining an imagination.

ozzi

I’m imagining imagining imagining an imagination...

4 0

3 years ago

A circular horizontal surface having an area of 1020.5 mm2that is completely immersed in a fluid experiences a uniform force of

Bingel [31]

Answer:

Faya vei Abhikesh brass hahahahahaha

Explanation:

3 0

2 years ago

A horizontal 2.00\ m2.00 m long, 5.00\ kg5.00 kg uniform beam that lies along the east-west direction is acted on by two forces.

Sunny_sXe [5.5K]

Answer: $240\ rad/s^2$

Explanation:

Given

Length of beam $l=2\ m$

mass of beam $m=5\ kg$

Two forces of equal intensity acted in the opposite direction, therefore, they create a torque of magnitude

$\tau =F\times l=200\times 2=400\ N.m$

Also, the beam starts rotating about its center

So, the moment of inertia of the beam is

$I=\dfrac{ml^2}{12}=\dfrac{5\times 2^2}{12}\\\\I=\dfrac{5}{3}\ kg.m^2$

Torque is the product of moment of inertia and angular acceleration

$\Rightarrow \tau=I\alpha\\\\\Rightarrow 400=\dfrac{5}{3}\times \alpha\\\\\Rightarrow \alpha =240\ rad/s^2$

7 0

2 years ago