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2 years ago

What is the average flow rate in cm3/s of gasoline to the engine of a car traveling at 100 km/h if it averages 10. 0 km/l?

Physics

1 answer:

DiKsa [7]2 years ago

4 0

The answer is $2.78 \mathrm{~cm}^{3} / \mathrm{s}$

$Q=\frac{1 \mathrm{~L}}{10.0 \mathrm{~km}} \times \frac{100 \mathrm{~km}}{\mathrm{~h}} \times \frac{1 \mathrm{~h}}{36005} \times \frac{1000 \mathrm{~cm}^{3}}{\mathrm{~L}}$ = $2.78 \mathrm{~cm}^{3} / \mathrm{s}$

Thus, the average flow rate is $2.78 \mathrm{~cm}^{3} / \mathrm{s}$ .

What is average flow rate?

Average flow, which is often represented in cubic feet per second, refers to the stream's average discharge at a certain location.
The flowrate is the rate at which liquid travels through a pipe or into a wellbore from a reservoir.
The volume of fluid that moves across a specific cross-sectional area per unit of time is known as the flow rate.
It is essential to measure flow rates accurately using the right flowmeters to guarantee that fluid management procedures are efficient, safe, and cost-effective.

To learn more about Logistics visit: brainly.com/question/13385366

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Alligators and other reptiles don't use enough metabolic energy to keep their body temperatures constant. they cool off at night

drek231 [11]

To compute for the heat, $Q$ needed to be absorbed or released, we need

$Q = mc\Delta T$

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$Q = (300 kg)(3400 kJ/K)(30-25) = 5100000 J$

Recall that 1 Watt = 1 J/s, thus time needed to absorb radiation from the sun is

$t = \frac{Q}{W}\\ t = 4250 seconds$

That means it takes 4250 seconds for the alligator to warm up to be able to absorb the radiation from the sun.

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4 0

4 years ago

We can compare these two interactions on the basis of impulse (see above), but sometimes, we are more interested in the forces (

kiruha [24]

Complete Question

The complete question is shown on the first uploaded image

Answer:

$I = 476 \ N \cdot s$

$I_1 = 14.21 \ N\cdot s$

$F = 20300 \ N$

Explanation:

Considering the first question

From the question we are told that

The force produced is $F = 3400 \ N$

The duration of the punch is $t = 0.14 \ s$

Generally the impulse delivered is mathematically represented as

$I = F * t$

=> $I = 3400 * 0.14$

=> $I = 476 \ N \cdot s$

Considering the second question

The approaching velocity of the ball is $v_b = 45 \ m/s$

The leaving velocity of the ball is $v_l = -53 \ m/s$

The mass of the ball is $m_b = 0.145 \ kg$

Generally the magnitude of the impulse delivered is mathematically represented as

$I_1 = m* v_b - m * v_l$

=> $I_1 = [0.145 * 45] - [0.145 * -53]$

=> $I_1 = 14.21 \ N\cdot s$

Considering the third question

The duration of the impact of the bat is $t _1 = 0.7 \ ms = 0.7 *10^{-3} \ s$

Generally the average force exerted by the bat is mathematically represented as

$F = \frac{I_1}{t_1}$

=> $F = \frac{14.21 }{0.7 *10^{-3}}$

=> $F = 20300 \ N$

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ASHA 777 [7]

Answer:

B. -40 kgm/s is the answer

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