1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
zubka84 [21]
3 years ago
9

A computer monitor uses 200 W of power. How much energy does it use in 10 seconds?

Physics
2 answers:
monitta3 years ago
8 0

<u>Answer</u>:

The energy that is consumed by the computer in 10 seconds that uses a power of 200 watt is 5.55 * 10^{-4} \mathrm{KWH}

Explanation:

Given:

Power used by the computer=200W

Time used by the compute=10seconds

To Find:

The Energy Consumed by the computer=?

Solution:

1 W a t t=\frac{1}{1000} W a t t

1Watt=1/1000 Watt

so,

200 Wa t t=\frac{200}{1000} Wa t t

1 \text { second }=\frac{1}{3600} \text { seconds }

So,

10 \text { seconds }=\frac{10}{3600} \text { hours }

Now substituting the values we get ,

Energy = Power * Time

\text { Energy }=\left(\frac{200}{1000}\right) *\left(\frac{10}{3600}\right)

\text {Energy}=(0.2) *(0.00277)

\text {Energy}=5.55 * 10^{-4} \mathrm{KW} \mathrm{H}

Hence the energy consumed is =5.55 * 10^{-4} \mathrm{KW} \mathrm{H}

gavmur [86]3 years ago
3 0

Answer:

<u>The correct answer is 0.556 Watts</u>

Explanation:

The computer monitor uses 200 Watts of power in an hour, that is the standard measure.

If we want to know, how much energy the computer monitor uses in one second, we will have to divide both sides of the equation into 3,600.

1 hour = 60 minutes = 3,600 seconds (60 x 60)

Energy per second = 200/3600

Energy per second = 0.0556 Watts

Therefore to calculate how much energy is used in 10 seconds, we do this:

Energy per second x 10

<u>0.0556 x 10 = 0.556 Watts</u>

<u>The computer monitor uses 0.556 Watts in 10 seconds</u>

You might be interested in
Help Help Help pls...
alexandr1967 [171]

Answer:

1 is B 2 is D 3 is C

Explanation:

3 0
3 years ago
Which of the following is an example of work being done on an object?
tensa zangetsu [6.8K]

Answer:

d. All of these

Explanation:

work is said to be done when a force is applied to an object through a certain distance. the SI unit of workdone is joules or newton per meter

mathematically

workdone = force x distance.

from the answers,  work is being done because there is force applied in a certain distance.

  • from wagon is used to carry vegetables from a garden.
  • pulley is used to get water from a well.
  • hammer is used to remove a nail from a wall.

4 0
3 years ago
Read 2 more answers
A top fuel dragster covers the quarter mile in 4.5 seconds. What is it's speed in miles per hour
gladu [14]
<span>Distance in Miles (.25) x 3600 (seconds in an hour) / time in seconds = 200 MPH. Drag racing calculates the top speed via a speed trap starting 66 feet from the finish line.</span>
6 0
3 years ago
Read 2 more answers
A car has two horns, one emitting a frequency of 199 hz and the other emitting a frequency of 203 hz. what beat frequency do the
andriy [413]
Beat frequency, fb = |f2-f1|

That is, beat frequency is the absolute difference between two frequencies. Is is as a results of destructive and constructive inferences.

Therefore, in this case:

fb = 203 - 199 = 4 Hz
3 0
2 years ago
The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the
MakcuM [25]

Answer:

Explanation:

Half-life problems are modeled as exponential equations.  The half-life formula is P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}} where P_o is the initial amount, k is the length of the half-life, t is the amount of time that has elapsed since the initial measurement was taken, and P is the amount that remains at time t.

P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}

<u>Deriving the half-life formula</u>

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, y=a b^{t}, where t is still the amount of time, and y is the amount remaining at time t.  The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

y=a b^{t}

(14.2)=ab^{(0)}

14.2=a *1

14.2=a

So, a=14.2, which represents out initial amount of the substance, and our equation becomes: y=14.2 b^{t}

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}

\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}

0.5=b^{8.0252}

\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}

\sqrt[8.0252]{\frac{1}{2}}=b

Recalling exponent properties, one could find that  \left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b, which will give the equation identical to the half-life formula.  However, recalling this trivia about exponent properties is not necessary to solve this problem.  One can just evaluate the radical in a calculator:

b=0.9172535661...

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

y=14.2* (0.9172535661)^t  or y=14.2*(0.5)^{\frac{t}{8.0252}} where y represents the amount remaining at time t.

<u>Solving for the amount remaining</u>

With the equation set up, substitute the amount of time it takes to cross the Pacific to solve for the amount remaining:

y=14.2* (0.9172535661)^{(31.8)}          y=14.2*(0.5)^{\frac{(31.8)}{8.0252}}

y=14.2* 0.0641450581                    y=14.2*(0.5)^{3.962518068}

y=0.9108598257                              y=14.2* 0.0641450581

                                                        y=0.9108598257

Since both the initial amount of Iodine, and the amount of time were given to 3 significant figures, the amount remaining after 31.8days is 0.911g.

8 0
1 year ago
Other questions:
  • The human ear canal is about 2.9 cm long and can be regarded as a tube open at one end and closed at the eardrum. What is the fu
    13·1 answer
  • If a rigid body rotates about point O, the sum of the moments of the external forces acting on the body about point O equals whi
    11·1 answer
  • 3. A train has broken through the wall of a train station. During the collision, what can be said about the force exerted by the
    15·1 answer
  • The term ampacity is defined as the _____ current, in amperes, that a conductor can carry continuously under conditions of use w
    9·1 answer
  • 1. A ski-plane with a total mass of 1200 kg lands towards the west on a frozen lake at 30.0
    15·1 answer
  • An open train car moves with speed 18.5 m/s on a flat frictionless railroad track, with no engine pulling the car. It begins to
    6·1 answer
  • PLSS HELP!! Are these True or False questions
    9·1 answer
  • What does a negative acceleration mean if the object has positive velocity?
    12·2 answers
  • What are the two groups of planets based on composition?
    7·1 answer
  • If a lever has an input arm of 80 cm and an output arm of 20 cm, what is its ideal mechanical advantage?
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!