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Kazeer [188]
3 years ago
8

The efficiency of a device such as a lamp can be calculated using this equation:

Physics
2 answers:
loris [4]3 years ago
6 0

efficiency = (useful energy transferred ÷ energy supplied) × 100

It's easy to use this formula, but we have to know both the useful energy and the energy supplied.  The drawing doesn't tell us the useful energy, so we have to find a clever way to figure it out.  I see two ways to do it:

<u>Way #1:</u>

We all know about the law of conservation of energy.  So we know that the total energy coming out must be  250J, because that's how much energy is going in.  The wasted energy is 75J, so the rest of the 250J must be the useful energy . . . (250J - 75J) = 175J useful energy.

(useful energy) / (energy supplied) =  (175J) / (250J) = <em>70% efficiency</em>

================================

<u>Way #2: </u>

How much of the energy is wasted ? . . . 75J wasted

What percentage of the Input is that 75J ? . . . 75/250 = 30% wasted

30% of the input energy is wasted.  That leaves the other <em>70%</em> to be useful energy.

jek_recluse [69]3 years ago
3 0

Answer:

70%

Explanation:

I did the USA Test Prep.

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If a liquid has a density of 1.67 g/cm^3, what is the volume of 45g of the liquid?
Aloiza [94]

Answer:

V = 26.95 cm³

Explanation:

Density is given by the formula :

ρ = m÷V

Density = mass ÷ Volume

Given both density and mass we rearrange, substitute and solve for Volume :

Rearranging the equation to make Volume the subject :

ρ = m÷V

ρV = m

V = m÷ ρ

Now substitute :

V = 45 ÷ 1.67

V = 26.9461077844

Take 2 decimal places as the density is 2 decimal places :

V = 26.95

Units will be cm³ as it is volume

Hope this helped and have a good day

8 0
2 years ago
You wad up a piece of paper and throw it into the wastebasket. How far will
vitfil [10]

The range of the piece of paper is C) 1.4 m

Explanation:

The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:

- A uniform motion along the horizontal direction, with constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, g=9.8 m/s^2)

From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

d=\frac{u^2 sin 2\theta}{g}

where

u is the initial velocity

\theta is the angle of projection

g is the acceleration of gravity

For the piece of paper in this problem,

u = 4.3 m/s

\theta=65^{\circ}

Substituting,

d=\frac{(4.3)^2 sin(2\cdot 65^{\circ})}{9.8}=1.45 m \sim 1.4 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0
3 years ago
Read 2 more answers
Substances released into the air are known as
irakobra [83]

Your answer is Emissions

3 0
3 years ago
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Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.300 m and carries a current
Marrrta [24]

Answer:

Explanation:

Since the wires attract each other , the direction of current will be same in both the wires .

Let I be current in wire which is along x - axis

force of attraction per unit length between the two current carrying wire is given by

\frac{\mu_0}{4\pi}  x \frac{2 I_1\times I_2}{d}

where I₁ and I₂ are currents in the wires and d is distance between the two

Putting the given values

285 x 10⁻⁶ = 10⁻⁷ x \frac{2\times25.5\times I_2}{.3}

I₂ = 16.76 A

Current in the wire along x axis is 16.76 A

To find point where magnetic field is zero due the these wires

The point will lie between the two wires  as current is in the same direction.

Let at y = y , the neutral point lies

k 2 x  \frac{16.76}{y} = k 2 x \frac{25.5}{.3-y}

25.5y = 16.76 x .3 - 16.76y

42.26 y = 5.028

y = .119

= .12 m

3 0
3 years ago
A fisherman is fishing from a bridge and is using a "44.0-N test line." In other words, the line will sustain a maximum force of
avanturin [10]

Answer:

(a) W= 44N

(b)W= 31.65 N

Explanation:

Data

T=44 N : Maximum force that the rope can withstand without breaking

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

(a)  We apply the formula (1) at constant speed , then, a=0

W: heaviest fish that can be pulled up vertically

∑F = 0

T-W =0  

W = T

W= 44N

(b)  We apply the formula (1) , a= 1.26 m/s²

W: heaviest fish that can be pulled up vertically

W= m*g

m= W/g

g= 9.8 m/s² : acceleration due to gravity

∑F = 0

T-W = m*a

T= W+(W/g)*a

44=W*(1+1/9.8)* (1.26 )

44= W* 1.39

W= 44/1.39

W= 31.65 N

7 0
3 years ago
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