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2 years ago

14

In the ENGR 10 lab (E391), there are 50 long light bulbs (P=100 W) and 30 regular bulbs (P=60 W). How much energy is consumed li

ghting the lab during the 3 hour lab period. Calculate the amount of energy saved if all bulbs are replaced by CFL (spiral) bulbs (P=25W).

1 answer:

Alenkinab [10]2 years ago

6 0

Answer:

Total energy saving will be 0.8 KWH

Explanation:

We have given there are 50 long light bulbs of power 100 W so total power of 50 bulb = 100×50 = 5000 W = 5 KW

30 bulbs are of power 60 W

So total power of 30 bulbs = 30×60 = 1800 W = 1.8 KW

Total power of 80 bulbs = 1.8+5 = 6.8 KW

Total time = 3 hour

We know that energy $E=power\times time=6.8\times 3=20.4KWH$

Now power of each CFL bulb = 25 W

So power of 80 bulbs = 80×25 = 2000 W = 2 KW

Energy of 80 bulbs = 2×3 = 6 KWH

So total energy saving = 6.8-6 = 0.8 KWH

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A ball of mass 0.4 kg is initially at rest on the ground. It is kicked and leaves the kicker's foot with a speed of 5.0 m/s in a

yawa3891 [41]

Answer:

the answer the correct is 3

Explanation:

Let's use the relationship between momentum and momentum

I = Δp

I = m $v_{f}$ - m v₀

Let's calculate

I = 0.4 5.0 - 0

I = 2.0 N s

By Newton's law of action and reaction the force on the ball is equal to the force that the ball exerts on the foot, therefore the impulse on the foot of equal magnitude, but in the opposite direction

I = 2.0 Ns with 60°

When reviewing the answer the correct is 3

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2 years ago

16. A 7500 kg 18-wheeler traveling at 20 m/s exits onto the runaway truck ramp on the freeway.

miskamm [114]

Answer:

<em>765,000 Joule</em>

Explanation:

<u>Principle of Conservation of Energy </u>

The total energy in an isolated system cannot be created or destroyed, but transformed. Moving objects have kinetic energy, objects placed in some height above a reference level have gravitational potential energy. When they change their motion variables, one energy converts into the other, but if the numbers don't fit, we know there was some other type of energy acting into the system. The most common reason for energy 'losses' is the thermal energy, produced when objects move in rough surfaces or take friction from the air.

The 7,500 kg truck is originally traveling at 20 m/s to a certain height we'll set to 0. Thus, its total energy is

$\displaystyle E_1=\frac{mv^2}{2}$

$\displaystyle E_1=\frac{7,500\ 20^2}{2}$

$E_1=1,500,000\ Joule$

When it comes to a stop, its speed is 0 and its height is 10 m higher than before. It means all the kinetic energy was transformed into other types of energy. The gravitational potential energy is

$U=mgh=(7,500)(9.8)(10)=735,000\ Joule$

Since this number is not equal to the previous value of the energy, the difference is due to thermal energy dissipated by friction

$E_t=1,500,000\ Joule-735,000\ Joule=765,000\ Joule$

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1 year ago

A densly wound cylindrical coil has 210 turns per meter, a 5 cm radius, and carries 38 mA. What is the magnitude of the uniform

kaheart [24]

Answer:

(a). The magnitude of the uniform magnetic field is 10.02 μT

(b). The the magnitude of the total magnetic field is 10.78 μT.

Explanation:

Given that,

Number of turns = 210

Radius = 5 cm

Current = 38 mA

Current in the wire = 500 mA

We need to calculate the magnetic field inside a coil

Using formula of magnetic field

$B=\mu_{0} ni$

Put the value into the formula

$B_{c}=4\pi\times10^{-7}\times210\times38\times10^{-3}$

$B_{c}=0.0000100\ T$

$B_{c}=10.02\times10^{-6}\ T$

Now a straight wire is inserted into the coil that carries 500 mA. It travels down the central axis of the coil

Distance $d=\dfrac{5}{2}$

$d=2.5\ cm$

We need to calculate the magnetic field from the straight wire

Using formula of magnetic field

$B_{w}=\dfrac{\mu_{0}I}{2\pi d}$

Put the value into the formula

$B_{w}=\dfrac{4\pi\times10^{-7}\times500\times10^{-3}}{2\times\pi\times2.5\times10^{-2}}$

$B_{w}=4.0\times10^{-6}\ T$

This field is perpendicular to the wire.

The magnitude of magnetic field is

$B=\sqrt{B_{c}^2+B_{w}^2}$

$B=\sqrt{(10.02\times10^{-6})^2+(4.0\times10^{-6})^2}$

$B=0.00001078\ T$

$B=10.78\times10^{-6}\ T$

$B=10.78\ \mu\ T$

Hence, (a). The magnitude of the uniform magnetic field is 10.02 μT

(b). The the magnitude of the total magnetic field is 10.78 μT.

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1 year ago

A uniform Rectangular Parallelepiped of mass m and edges a, b, and c is rotating with the constant angular velocity ω around an

Sonbull [250]

Answer:

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Explanation:

The moment of parallel pipe rotating about it's axis is given by the formula;

I = $\frac{M}{3} (a^{2} +b^{2} )$ ---------------------------------1

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k = $\frac{1}{2} Iw^{2}$ --------------------------------2

Putting equation 1 into equation 2, we have;

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k = $\frac{Mw^{2} }{6} (a^{2} +b^{2} )$

(b) The angular momentum is given by the formula;

τ = Iw -----------------------3

Putting equation 1 into equation 3, we have

τ = $\frac{Mw}{3} (a^{2} +b^{2} )$

But

τ = dτ/dt = $\frac{M}{3} (a^{2} +b^{2} )\frac{dw}{dt}$ ------------------4

where

dw/dt = angular acceleration =∝

Equation 4 becomes;

τ = $\frac{M}{3} (a^{2} +b^{2} )$ ∝

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1 year ago

Nc-1 has the same dimension as

lisabon 2012 [21]

Answer:

Explanation:

The answer is electric field intensity. Electric field intensity is the force per unit positive charge which the charge exerts at any point.

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2 years ago