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3 years ago

13

an incandescent lightbulb has an efficiency of 2.1% and a power of 60 w. how much light energy does the lighbulb produce in 1 se

cond

1 answer:

AveGali [126]3 years ago

5 0

Explanation:

Power output of the bulb:

0.021 × 60 W = 1.26 W

Energy produced by the bulb in 1 second:

E = Pt

E = (1.26 W) (1 s)

E = 1.26 J

Round as needed.

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What Determines the amount of TE of an object

Luden [163]

Answer:

To find the volume of a rectangular object, measure the length, width and height. Multiply the length times the width and multiply the result by the height. The result is the volume. Give the result in cubic units, such as cubic centimeters.

Explanation:

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2 years ago

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Suppose the magnitude of the proton charge differs from the magnitude of the electron charge by a mere 1 part in 109

Oksana_A [137]

Answer:

A) F = 1.09 10 5 N, b) Yes

Explanation:

Part A

For this exercise we need the number of free electrons in copper, as the valence of copper +1 there is a free electron for each atom. Let's use the concept of density to find the mass of copper in the sphere

ρ = m / V

.m = ρ V = ρ 4/3 π r³

The radius is half the diameter

r = 1.9 10⁻² / 2 = 0.95 10⁻² m

ρ = 8960 kg / m3

m = 8960 4/3 π (0.95 10⁻²)³

m = 3.2179 10⁻² kg

The molecular weight of copper is 63,546 g / mol which has 6,022 10²³ atoms

With this we can use a rule of proportions to enter the number of atom is this mass

#_atom = 6.022 10²³ 3.2179 10⁻² / 63.546 10⁻³

#_atom = 3,049 10²³

Therefore there is the same number of electrons, as they indicate that the charge of the protone and the electon differs by 1/10⁹ the total charge for each spherical is

q = e / 10⁹ #_atom

q = e / 10⁹ 3,049 1023

q = 3,049 10⁴ (-1.6 10⁻¹⁹)

q = -4,878 10-5 C

Electric force is

F = k q₁q₂ / r²

F = k q² / r²

Let's calculate

F = 8.99 10⁹ (4.878 10⁻⁵)²2 / (1.4 10⁻²)²

F = 1.09 10 5 N

This is a force of repulsion.

Part B

The magnitude of this force is in very easy to detect

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3 years ago

Give an example in which a small force exerts a large torque. give another example in which a large force exerts a small torque.

goldenfox [79]

<span>Since the torque involves the product of force times lever arm, a small force can exert a greater torque than a larger force if the small force has a large enough lever arm.

With a large force exerts a small torque is a gate, hinged in its vertical line (axis). When pushed from a point near to the hinge, a very large amount is needed to open the gate.
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</span>

3 0

3 years ago

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How much power is needed to lift the 200-N object to a height of 10 m in 4 s?

harkovskaia [24]

Answer:

500 watts

Explanation:

Recall that the definition of power is the amount of energy delivered per unit of time.

In our case, the energy delivered is potential energy which we can estimate as the product of the weight of the object times the distance it is lifted above ground:

200 N x 10 m = 2000 Nm

then the power is the quotient of this potential energy divided the time it took to lift the object to that position:

Power = 2000 / 4 Nm/s = 500 Nm/s = 500 watts

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3 years ago

Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

The shear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

b) If the railing edge is 2 ft, the Reynold number is:

$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

The friction drag is equal to:

$F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf$

7 0

3 years ago