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brilliants [131]

2 years ago

14

Justin signed a rental agreement for his condo. After he moved out, the owner determined that the condo needed to be cleaned, th

e cost of which totaled $200. How much of the deposit can Justin expect back?

2 answers:

Juli2301 [7.4K]2 years ago

4 0

I looked at the contract and the answer is $800

Sonja [21]2 years ago

3 0

Answer:

$850

Explanation:

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An individual white LED (light-emitting diode) has an efficiency of 20% and uses 1.0 W of electric power. a. How many LEDs must

Neporo4naja [7]

Answer:

8, 8 W

Explanation:

The useful power of 1 Light Emitting Diode is

$0.2\times 1=0.2\ W$

Total power required is 1.6 W

Number of Light Emitting Diodes would be

$n=\dfrac{1.6}{0.2}\\\Rightarrow n=8$

The number of Light Emitting Diodes is 8.

Power would be

$P=8\times 1=8\ W$

The power that is required to run the Light Emitting Diodes is 8 W

7 0

2 years ago

An object is located 13.5 cm in front of a convex mirror, the image being 7.05 cm behind the mirror. A second object, twice as t

goldfiish [28.3K]

Answer:

Second object is located at 42.03 cm in front of mirror

Explanation:

In this question we have given,

object distance from convex mirror ,u=-13.5cm

Image distance from convex mirror,v=7.05cm

let focal length of convex mirror be f

we have to find the distance of second object from convex mirror

we know that u, v and f are related by following formula

$\frac{1}{f} =\frac{1}{v}+ \frac{1}{u}$ .............(1)

put values of u and v in equation (1)

we got,

$\frac{1}{f} =\frac{1}{7.05}+ \frac{1}{-13.5}$

$\frac{1}{f}=\frac{13.5-7.05}{13.5\times 7.05}$

$\frac{1}{f}=\frac{6.45}{13.5\times 7.05}\\f=13.5\times 1.09\\f=14.75$

we have given that

second object is twice as tall as the first object

and image height of both objects are same

it means

$o_{2}=2o_{1}\\i_{1}=i_{2}$ .............(2)

we know that

$\frac{v}{u}=\frac{i}{o}\\i=\frac{o\times v}{u}$

therefore,

$i_{1}=\frac{o_{1}\times v}{u}$ .................(3)

put values of v and u in equation 3

$i_{1}=-\frac{o_{1}\times 7.05}{13.5}$

$i_{1}=-0.52o_{1}$

therefore from equation 2

$i_{2}=-0.52o_{1}$

we know that

$i_{2}=\frac{o_{2}\times V}{U}$ .................(4)

put value of $i_{2}$ and $o_{2}$ in equation 4

$-.52o_{1}=\frac{2o_{1}\times V}{U}$

$U=\frac{2o_{1}\times V}{-.52o_{1}} \\U=-3.85V$

we know that U,V and f are related by following formula

$\frac{1}{f} =\frac{1}{V}+ \frac{1}{U}$ .............(5)

put values of f and U in equation 5

we got

$\frac{1}{14.75} =\frac{1}{V}- \frac{1}{3.85V}$

$\frac{1}{14.75} =\frac{2.85}{3.85V}$

$\frac{1}{14.75} =\frac{2.85}{3.85V}\\V=\frac{2.85\times 14.75}{3.85}\\V=10.91 cm$

Therefore,

U=-10.91\times 3.85

U=-42.03 cm

Second object is located at 42.03 cm in front of mirror

4 0

2 years ago

You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t

Sladkaya [172]

Answer:

0.767

Explanation:

The work done on the truck by the frictional drag force is given by

$W=-Fd$

where

F is the magnitude of the frictional force

d = 38.0 m is the maximum displacement allowed for the truck

The negative sign is due to the fact that the force of friction is opposite to the motion of the truck

The force of friction can also be written as:

$F=\mu mg$

where

$\mu$ is the coefficient of kinetic friction between the truck and the lane

m is the mass of the truck

g is the acceleration of gravity

So we can rewrite the work done as

$W=-\mu mg d$ (1)

According to the work-energy theorem, the work done by friction is equal to the change in kinetic energy of the truck:

$W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (2)

where

v = 0 is the final velocity of the truck

u = 23.9 m/s is the initial velocity of the truck

By combining (1) and (2) we get

$-\frac{1}{2}mu^2 = -\mu mg d$

And solving for $\mu$ , we find the minimum coefficient of kinetic friction able to stop the truck in a distance d:

$\mu = \frac{u^2}{2gd}=\frac{23.9^2}{2(9.8)(38.0)}=0.767$

7 0

2 years ago

Lizette works in her school’s vegetable garden. Every Tuesday, she pulls weeds for 15 minutes. Weeding seems like a never-ending

Gnoma [55]

Answer:

Constant or Controlled variables: Same concentration of vinegar solution, same quantity of vinegar, same type of weed etc

Explanation:

In an experiment, certain variables are kept unchanged or constant for both the experimental group and control group in order not to influence the outcome of the experiment. These variables are called CONTROLLED VARIABLES or CONSTANTS.

In the case of this experiment where Lizette is testing the effect of vinegar on weed, the variable that should be kept the same (controlled variables) for the control group of weeds and the sprayed weeds include Same concentration of vinegar solution, Same quantity of vinegar, same type of weed.

8 0

2 years ago

Read 2 more answers

A toy car of mass 0.15kg accelerates from a speed of 10 cm/s to a speed of 15 cm/s. What is the impulse acting on the car?

OLga [1]

v=v2-v1=15-10=5 cm/s

p=mv=0.15•5=0.75 kg•cm/s

7 0

2 years ago

Read 2 more answers