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2 years ago

13

As a block falls through the air by 40 meter it does work equal to -1800 joule. Determine the mass of a block.

1 answer:

kari74 [83]2 years ago

3 0

Answer:

m = 4.5 kg

Explanation:

w = - 1800 j

Fd = - 1800 j

mgd = - 1800 j

m = - 1800 ÷(gd)

m = - 1800 ÷( 10×-40)

m = 4.5 kg

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A driver of a car traveling at -15m/s applies the brakes, causing a uniform acceleration of +2.0m/s2. If the brakes are applied

klemol [59]

Given :

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To Find :

The velocity of the car at the end of the braking period.

How far has the car moved during the braking period.

Solution :

By equation :

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Now, distance covered by car is :

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3 years ago

A solar collector receives solar radiation at a rate of 0.315 kW per m²and delivers the heat to a storage unit whose temperature

SVEN [57.7K]

Answer:

4370.83044 m²

Explanation:

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$Q_h=0.315A\\\Rightarrow A=\frac{Q_h}{0.315}\\\Rightarrow A=\frac{1376.81159}{0.315}\\\Rightarrow A=4370.83044\ m^2$

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2 years ago

What is the efficiency of an engine that does 288 J of work and exhausts 72 J of heat while taking in 360 J of heat?

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If you're using the machine to heat your workshop, it's 20% efficient.

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3 years ago

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Newton's law of cooling is , where is the temperature of an object, is in hours, is a constant ambient temperature, and is a pos

kirill [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The time taken is $t = 6.89 \ hrs$

Explanation:

From the question we are told that

The value of $k = 0.15hr^{-1}$

The the interior temperature is $u(t = 0) = 70 ^oF$

The external temperature is $T = 11 ^oF$

The required interior temperature is $u_{t} = 32 ^oF$

The newton cooling law is

$\frac{du}{dt} = -k (u -T)$

=> $\frac{du}{u-T} = - kdt$

Now integrate both sides we have

$\int\limits \frac{du}{u-T} =\int\limits - kdt$

$ln (u - T) = -kt + c$

Since c is a constant lnC = c will also give a constant so

$ln (u - T) = -kt + ln C$

=> $\frac{(u -T)}{C} = e^{-kt}$

=> $u = T + Ce^{-kt}$

substituting value

$70 = 11 +Ce^{-0* 0.15}$

$C = 59$

Hence

$u_t = 11 + 59 e^{-0.15 *t}$

$32= 11 + 59 e^{-0.15 *t}$

=> $t = -\frac{ln(\frac{21}{59} )}{0.15}$

$t = 6.89 \ hrs$

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2 years ago