If a person is walking at 1.2 m/s and 60 seconds later the person is running at 10 m/s, what was the acceleration rate?

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?

DiKsa [7]

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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4 0

2 years ago

AGREE OR DISAGREE: All light can be seen.

Tema [17]

Answer:

Disagree

Explanation:

We cannot see infared lights.

3 0

3 years ago

The law of reflection says that the angle of incidence is

nekit [7.7K]

The law of reflection states that the angle of incidence is equal to the angle of reflection. Furthermore, the law of reflection states that the incident ray, the reflected ray and the normal all lie in the same plane.

hope this helps :)

4 0

3 years ago

If Kim’s egg traveled a distance of 10m in 5 seconds, calculate the speed. (Speed=distance/time)

Crazy boy [7]

Answer:

2 meters a second or 2 m/s

8 0

3 years ago

Read 2 more answers

A 0.100 kg ball hangs from a spring of negligible mass. When the ball is hung on the spring and is at rest, the spring is stretc

anzhelika [568]

Answer:

a) 4.9 N/m

b) 1.4 m/s

c) 0.225 s

Explanation:

Hooke's law states that

F = k * Δx

Where

F: force applied to a spring

k: constant of the spring

Δx: elongation of the spring

The force applied in this case is the weight of the ball, this is

P = m * g = 0.1 kg * 9.81 m/s^2 = 0.981 N

Rearrainging Hooke's law:

k = F / Δx

k = 0.981 / 0.2 = 4.9 N/m

If the ball is pulled down the spring will acquire some potential energy, when it is released, the potential energy will be released as kinetic energy on the ball

$Ec = \frac{1}{2} * m * v^2$

Elastic potential energy is:

$U = \frac{1}{2} * k * \Delta x^2$

The energy gained from the 0.2m pull will be turned into kinetic energy

Ec = U

Therefore:

$\frac{1}{2} * m * v^2 = \frac{1}{2} * k * \Delta x^2$

Rearranging:

$v^2 = \frac{k}{m} * \Delta x^2$

$v = \Delta x * \sqrt{\frac{k}{m}}$

$v = 0.2 * \sqrt{\frac{4.9}{0.1}} = 1.4 m/s$

After being released the ball will oscillate at the natural frequency of the system, which is

$f = \frac{1}{2 * \pi} * \sqrt{\frac{k}{m}}$

And the period will be:

$T = 2 * \pi * \sqrt{\frac{m}{k}}$

The period in this case is:

$T = 2 * \pi * \sqrt{\frac{0.1}{4.9}} = 0.9 s$

The ball will move up and down taking T time to complete a cycle, the movement from the stretched position to the equilibrium position takes T/4 = 0.225 s

8 0

3 years ago

If a person is walking at 1.2 m/s and 60 seconds later the person is running at 10 m/s, what was the acceleration rate?​

If a person is walking at 1.2 m/s and 60 seconds later the person is running at 10 m/s, what was the acceleration rate?