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4 years ago

The SI unit of power is the

Physics

2 answers:

Oksanka [162]4 years ago

4 0

Watt (W).............

fgiga [73]4 years ago

3 0

Power is measured in watts. A watt is the power that it takes to do one joule ofwork in one second. It can be found using the formula P=Wt. (In this formula, W stands for "work.")Large amounts of energy can be measured in kilowatts (1kW=1×103W), megawatts (1MW=1×106W), or gigawatts (1GW=1×109W). This is helpful This is confusingThe watt is named James Watt, who invented an older unit of power: the horsepower.

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I WILL GIVE BRAINLIEST!!!

astra-53 [7]

Answer:

is that high school work??? cause I don't know it and I'm about to go to high school

7 0

3 years ago

What makes a model not useful?

lina2011 [118]

Answer is b hope this helps

8 0

3 years ago

Place the items in order from the largest wavelength to the shortest wavelength.

Butoxors [25]

From largest to shortest wavelength:

Radio waves, Microwaves, Infrared radiation, Red visible, Orange visible, Yellow visible, Green visible, Blue visible, Violet visible, Ultraviolet, X-rays, Gamma rays

Explanation:

Electromagnetic waves are oscillations of the electric and the magnetic field in a plane perpendicular to the direction of motion the wave.

Electromagnetic waves are the only type of waves able to travel in a vacuum, and in a vacuum they always at the same speed, the speed of light, equal to:

$c=3.0\cdot 10^8 m/s$

Electromagnetic waves are classified into 7 different types, according to their wavelength/frequency. From slongest to shortest wavelength, they are ranked as follows:

Radio waves

Microwaves

Infrared radiation

Visible light

Ultraviolet

X-rays

Gamma rays

Visible light is the only part of the spectrum that the human eye is able to see. Depending on the wavelength of the visible light, we perceive the radiation as a different color. In order from longest to shortest wavelength, colors are:

Red

Orange

Yellow

Green

Blue

Indigo

Violet

Therefore, the correct order from largest to shortest wavelength in the given list is:

Radio waves

Microwaves

Infrared radiation

Red visible

Orange visible

Yellow visible

Green visible

Blue visible

Violet visible

Ultraviolet

X-rays

Gamma rays

Learn more about electromagnetic waves:

brainly.com/question/9184100

brainly.com/question/12450147

#LearnwithBrainly

4 0

3 years ago

In the picture of a planet in orbit around its star area a is double that of area b. What is true of the time the planet takes t

aalyn [17]

The Kepler's laws predict the planetary motion, so there are three laws for this, namely:

1. The orbit of a planet is an ellipse with the Sun (the sun is a star!) at one of the two focus.

2. A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

3. The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.

So, let's use second law. The Sun sweeps out equal areas during equal intervals of time means that if A = B, the time the planet takes to travel A1A2 is equal to the time the planet takes to travel B1B2, but given that A = 2B, then takes twice the time to travel A1A2 compared to B1B2.

8 0

3 years ago

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

8 0

4 years ago