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3 years ago

Which scientific term is defined as the rate at which work is accomplished?

1 answer:

7 0

Power is the scientific term defined as the rate at which work is accomplished. Power can be seen everywhere as it is applied to our everyday living. Power is seen when one pushes a heavy cart. It could be seen when a baseball hitter hits the ball.

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The rate (in liters per minute) at which water drains from a tank is recorded at half-minute intervals. Use the average of the l

lana [24]

Answer:

see explanation

Explanation:

You are missing the chart with the rates and time to do this, however, I wll do it with a similar exercise here, and you only need to replace the procedure with your data:

See the attached table.

From the left we have:

r = 1/2 (50 + 48 + 46 + 44 + 42 + 40) = 135 L/min

From the right we have:

r = 1/2 (48 + 46 +44 + 42 + 40 + 38) = 129 L/min.

And this should be the correct answer. Watch your chart and replace if it's neccesary.

3 0

3 years ago

Would a rougher or smoother object absorb thermal energy faster? Why? Give an example.

Nuetrik [128]

Smoother because it will increase energy and when the energy increases it’ll create heat also . Example: A car racing on a smooth road it’ll go faster than a Car speeding on a bumpy and rough road , Hope that helps .

5 0

3 years ago

A spring that is compressed 14.5 cm from its equilibrium position stores 2.99 J of potential energy. Determine the spring consta

strojnjashka [21]

Answer:

284.4233 N/m

Explanation:

k = Spring constant

x = Compression of spring = 14.5 cm

U = Potential energy = 2.99 J

The potential energy of a spring is given by

$U=\dfrac{1}{2}kx^2$

Rearranging to get the value of k

$\\\Rightarrow k=\dfrac{2U}{x^2}\\\Rightarrow k=\dfrac{2\times 2.99}{0.145^2}\\\Rightarrow k=284.4233\ N/m$

The spring constant is 284.4233 N/m

7 0

3 years ago

During the spin cycle of a washing machine, the clothes stick to the outer wall of the barrel as it spins at a rate as high as 1

Darya [45]

To answer the two questions, we need to know two important equations involving centripetal movement:

v = ωr (ω represents angular velocity in radians)

a = $\frac{v^{2}}{r}$

Let's apply the first equation to question a:

v = ωr

v = ((1800*2π) / 60) * 0.26

Wait. 2π? 0.26? 60? Let's break down why these numbers are written differently. In order to use the equation v = ωr, it is important that the units of ω is in radians. Since one revolution is equivalent to 2π radians, we can easily do the conversion from revolutions to radians by multiplying it by 2π. As for 0.26, note that the question asks for the units to be m/s. Since we need meters, we simply convert 26 cm, our radius, into meters. The revolutions is also given in revs/min, and we need to convert it into revs/sec so that we can get our final units correct. As a result, we divide the rate by 60 to convert minutes into seconds.

Back to the equation:

v = ((1800*2π)/60) * 0.26

v = (1800*2(3.14)/60) * 0.26

v = (11304/60) * 0.26

v = 188.4 * 0.26

v = 48.984

v = 49 (m/s)

Now that we know the linear velocity, we can find the centripetal acceleration:

a = $\frac{v^{2}}{r}$

a = $\frac{49^{2}}{0.26}$

a = 9234.6 (m/ $s^{2}$ )

Wow! That's fast!

We now have our answers for a and b:

a. 49 (m/s)

b. 9.2 * $10^{3}$ (m/ $s^{2}$ )

If you have any questions on how I got to these answers, just ask!

- breezyツ

5 0

2 years ago

Using equations, determine the temperature, pressure and density of the air for a aircraft flying at 19.5 km. Is this aircraft s

Viefleur [7K]

Answer:

a) - 72.5°c

b) pressure = 3625.13 Pa

c) density = 0.063 kg/m^3

d) it is a subsonic aircraft

Explanation:

a) Determine Temperature

Temperature at 19.5 km ( 19500 m )

T = -131 + ( 0.003 * altitude in meters )

= -131 + ( 0.003 * 19500 ) = - 72.5°c

b) Determine pressure and density at 19.5 km altitude

Given :

Po (atmospheric pressure at sea level ) = 101kpa

R ( gas constant of air ) = 0.287 KJ/Kgk

T = -72.5°c ≈ 200.5 k

pressure = 3625.13 Pa

hence density = 0.063 kg/m^3

attached below is the remaining part of the solution

C) determine if the aircraft is subsonic or super sonic

Velocity ( v ) = $\sqrt{CRT}$ = $\sqrt{1.4*287*200.5 }$ = 283.8 m/s

hence it is a subsonic aircraft

4 0

3 years ago