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faust18 [17]
3 years ago
13

How much work will a machine with a power rating of 1.1 × 103 watts do in 2.0 hours?

Physics
2 answers:
lyudmila [28]3 years ago
7 0

Answer:

D.

Explanation:

8.1X10 to the power of 6 Joules.

Pie3 years ago
4 0

Answer:

7.92×10^6Joules

Explanation:

Power is defined as the rate of change in work done. Mathematically, Power = Workdone/Time

Given power = 1.1×10³watts

Time = 2.0hours = 2×60×60

= 7,200seconds

Workdone = power × time

Work done = 1.1×10³×7200

Workdone = 7,920,000Joules

= 7.92×10^6Joules

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Can u help me thanks so<br> Much ❤️
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3 years ago
Assume you are in the car and the car is moving at a certain speed to
Setler79 [48]
You are at rest with respect to the car.
You are in motion with respect to the School.

6 0
3 years ago
A wave of water moving up a river, initiated by tidal action and normal resonances within a river estuary, is called a:
IgorC [24]

Answer:

friction solar system

Explanation:

6 0
3 years ago
A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude
Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$.

Now, let us call \lambda the charge per unit length, then we know that

dQ = \lambda Rd\theta;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda  d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda   }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda   }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda   }{R}.$

Now, we know that

\lambda = 3.0*10^{-9}C/m,

k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},

and the radius of the semicircle is

\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9})   }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

7 0
3 years ago
What would happen if you tried to use a prism to disperse a beam that contained only green light?​
Margaret [11]

It is determined by the nature of the green light. Because lasers create light at almost a single frequency, green laser light would appear as a thin line of pure green. Other sources of "green" light emit light at a variety of frequencies, including yellow and blue, resulting in a strong green band in the center that fades into blue-green and yellow-green at the borders.

For example, here’s a graph of the spectrum of a green LED, showing the color range: Attachment #1

and here’s a graph of the transmission spectra of several standard photographic filters, including green: Attachment #2

Learn more about the color spectrum:

  • brainly.com/question/14552374
  • brainly.com/question/20930002

#SPJ2

4 0
2 years ago
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