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pychu [463]
4 years ago
6

What is a watt a unit of? power distance time light

Physics
2 answers:
Pachacha [2.7K]4 years ago
8 0

Answer: Watt is the unit of power

Explanation:

sergey [27]4 years ago
6 0

Answer:

I think the answer is A) Power

Explanation:

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A highly volatile substance has an initial mass of 1200 g and its mass is reduced by 12% each second.
Softa [21]

Answer:

Explanation:

a) 1.00 - 0.12 = 0.88

m = 1200(0.88)^t

b) t = ln(m/1200) / ln(0.88)

c) m = 1200(0.88)^10 = 334.20 g

d) t = ln(10/1200) / ln(0.88) = 37.451... = 37 s

e) t = ln(1/1200) / ln(0.88) = 55.463... = 55 s

4 0
3 years ago
Which of the following is equal to the area under a velocity-time graph
Ksenya-84 [330]

-- The area under a velocity/time graph, between two points in time, is the difference in displacement during that period of time.

-- The area under a speed/time graph, between two points in time, is the distance covered during that period of time.

5 0
3 years ago
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A positively charged atom has
Ulleksa [173]
It has a positive charge
8 0
3 years ago
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Four velcro-lined air-hockey disks collide with each other in a perfectly
Reil [10]

Answer:

The magnitude of the final velocity is approximately 0.526 m/s in approximately the direction of 8.746° East of South

Explanation:

The given collision parameters are;

The kind of collision experienced by the four velcro-lined air-hockey disk = Inelastic collision

The mass of the first disk, m₁ = 50.0 g

The velocity of the first disk, v₁ = 0.80 m/s West = -0.8·i

The mass of the second disk, m₂ = 60.0 g

The velocity of the second disk, v₂ = 2.50 m/s North = 2.5·j

The mass of the third disk, m₃ = 100.0 g

The velocity of the third disk, v₃ = 0.20 m/s East = 0.20·i

The mass of the fourth disk, m₄ = 40.0 g

The velocity of the fourth disk, v₄ = 0.50 m/s South = -0.50·j

Therefore, the total initial momentum of the four velcro-lined air-hockey disk, \Sigma P_{initial} is given as follows;

\Sigma P_{initial} = m₁·v₁ + m₂·v₂ + m₃·v₃ + m₄·v₄ = 50.0×(-0.80·i) + 60.0×(2.50·j) + 100 × (0.20·i) + 40.0 × (-0.50·j)

∴ \Sigma P_{initial} = -40·i + 150·j + 20·i - 20·j = -20·i + 130·j

∴ \Sigma P_{initial} = -20·i + 130·j

By the law of conservation of linear momentum, we have;

\Sigma P_{initial} = \Sigma P _{final} = -20·i + 130·j

Therefore, given that the collision is perfectly inelastic, the disks move as one after the collision and the four masses are added to form one mass, "m", m = m₁ + m₂ + m₃ + m₄ = 50.0 + 60.0 + 100.0 + 40.0 = 250.0

∴ m = 250.0 g

Let, "v" represent the final velocity of the four disks moving as one after the collision

We have;

\Sigma P _{final} = m × v = 250.0 × v = -20·i + 130·j

∴ v = -20·i/250 + 130·j/250 = -0.08·i + 0.52·j

The final velocity of the four disks after collision, v = -0.08·i + 0.52·j

The magnitude of the final velocity, \left | v \right | = √((-0.08)² + (0.52)²) ≈ 0.526

\left | v \right | ≈ 0.526 m/s

The direction of the final velocity, θ = arctan(0.52/(-0.08)) ≈ -81.254°

The direction of the final velocity, θ ≈ -81.254° which is 8.746° East of South

4 0
3 years ago
How do you graph distance and time for an object that moves at a constant speed?
LekaFEV [45]

Answer:

It would be a straight line

Explanation:

On a distance-time graph, an object that moves at constant speed would be represented by a straight line.

In fact, in a distance-time graph, the slope of the line corresponds to the speed of the object. We can demonstrate that. In fact:

- The speed of the object is equal to the ratio between the distance covered (\Delta s) and the time taken (\Delta t):

v=\frac{\Delta s}{\Delta t}

On a distance-time graph, the distance is on the y-axis while the time is on the x-axis. The slope of the line is defined as:

m=\frac{\Delta y}{\Delta x}

But the variation on the y-axis (\Delta y) is equal to the distance covered (\Delta s), while the variation on the x-axis (\Delta x) corresponds to the time taken (\Delta t), so the slope can also be rewritten as

m=\frac{\Delta s}{\Delta t}

which is equal to the speed of the object. Therefore, an object moving at constant speed would be represented by a line with constant slope, which means a straight line.

6 0
3 years ago
Read 2 more answers
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