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3 years ago

Power can be expressed

Physics

1 answer:

Komok [63]3 years ago

5 0

By definition Power is energy per second. As units the Watt, and a Watt is joules/second (2)

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Which of the following has the largest m/mentum?

ASHA 777 [7]

A: The total building of Campbell high school, including the trailers and the construction area

6 0

3 years ago

What is a stream’s discharge rate if it has a width of 10 meters, a depth of 2 meters, and a velocity of 2 meters per second?

julia-pushkina [17]

A stream’s discharge rate if it has a width of 10 meters, a depth of 2 meters, and a velocity of 2 meters per second will be 40 $m^{3}$ /sec .

Discharge rate = velocity * area

= velocity * depth * width

= 2 * 2 * 10 = 40 $m^{3}$ /sec

A stream’s discharge rate if it has a width of 10 meters, a depth of 2 meters, and a velocity of 2 meters per second will be 40 $m^{3}$ /sec .

learn more about discharge rate

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2 years ago

a 25-N net force is applied to a rolling cart and pruduces an acceleration of 5 m/s 2 what’s the cart mass

nignag [31]

Answer:

The mass of the cart is 5 kg

Explanation:

You divide 25 by 5 and get 5. Have a great day! :D

<em>The Equation:</em>

25/5 = 5

8 0

3 years ago

If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu

Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

$\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}$ .

Volume of water:

$V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}$ .

Mass of water:

$m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}$ .

Amount of heat that the 15 L water absorbed:

$Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}$ .

What's the mass of the hot steel tool?

The specific heat of carbon steel is $0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}$ .

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

$Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}$ .

$\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}$ .

$m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}$ .

4 0

3 years ago

Acar accelerates from 4 meters/second to 16 meters/second in 4 seconds. The car's acceleration is

s2008m [1.1K]

To understand this question, you need to understand the concept of acceleration first. Have you ever been in a car and noticed that it was getting faster and faster? That "speeding up" of the car is known as acceleration! Acceleration is essentially the rate at which you speed up.

Okay, so we now know what acceleration is. What are its units? The unit of acceleration is the change in velocity over a period of time: $\frac{∆v}{t}$

If you haven't learned about velocity yet, just think about it as speed for now. The funny-looking triangle, ∆, is a symbol for "the change of." For example, if I started walking at 3 $\frac{feet}{second}$ then sped up to 5 $\frac{feet}{second}$ , then the change in my speed would be 2 $\frac{feet}{second}$ , because I started walking 2 $\frac{feet}{second}$ faster!

Okay, enough with all the explanations. Hopefully, you understand the units now. Let's take a look at the question. A car accelerates from 4 $\frac{meters}{second}$ to 16 $\frac{meters}{second}$ in 4 seconds. What would the acceleration be? Let's set up an equation:

a = $\frac{∆v}{t}$

a is the acceleration, ∆v is the change in velocity, and t is the time elapsed.

Now, let's plug in our values! ∆v is the change in velocity, and to find that we simply have to subtract 16 $\frac{meters}{second}$ by 4 $\frac{meters}{second}$ . That makes sense, right? Back to the equation.

a = $\frac{∆v}{t}$
a = $\frac{16-4}{4}$

(16 - 4 is the change in velocity, and 4 is the number of seconds the car was accelerating)

a = $\frac{12}{4}$

a = 3 ( $\frac{meters}{second^{2}}$ )

We have our answer! The car's acceleration is 3 meters per second^{2}.

(You might be thinking: Wait. Meters per second squared? The reason for that is because acceleration is the rate at which the speed increases! That makes the unit $\frac{\frac{meters}{second}}{second}$ , which can be simplified down to $\frac{meters}{second^{2} }$ )

Let me know if you need clarification on anything I explained here!
- breezyツ

6 0

2 years ago