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4 years ago

Which equation is most likely used to determine the acceleration from a velocity vs:time graph?

2 answers:

7 0

Acceleration, a = (v - u)/t

where v is the final velocity, u is the initial velocity, and t is the time.

This formula on a velocity time graph represents the slope of the graph.

Mars2501 [29]4 years ago

3 0

Answer: Hello there!

in this case, you have a graph of velocity vs time, then in this graph is represented the function v(t).

The acceleration is defined as the integral with respect to the time of the velocity v(t), then we could calculate the mean acceleration between two times in a next way:

$a = \frac{v(t2) - v(t1))}{t2 -t1}$

this is the slope between the times t2 and t1, where t2>t1 and V(t2) is the velocity at the time t2 (the final velocity) and v(t1) is the velocity at the time t1 (the initial velocity). In this case, a is the mean acceleration between these times.

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Which of the following employs the use of an external combustion engine?

just olya [345]

The answer should be ''all the above''

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4 years ago

Help with physical science please

alex41 [277]

1. Elastic potential energy (D. EEl)

In this situation, the spring is compressed with the toy on top of it. The toy is stationary, so it does not have kinetic energy. However, the spring is compressed, so it does have elastic potential energy, given by:

$E_{EL}=\frac{1}{2}kx^2$

where k is the spring constant and x is the compression of the spring.

2. Gravitational potential energy (C. Eg)

In this situation, the spring has been released, so it returns to its natural position, so its elastic potential energy is zero. The toy is also stationary, since it is at its top position, where its velocity is zero, so its kinetic energy is also zero. However, the toy is now at a certain height h above the spring, so it has gravitational potential energy given by:

$E_g = mgh$

where m is the mass of the toy and g is the gravitational acceleration.

3. Gravitational potential and kinetic energy (A. Eg and EK)

In this situation, the toy is falling: so, it is moving with a certain speed v, so it has kinetic energy given by

$E_k = \frac{1}{2}mv^2$

Also, since it is at a certain height above the spring, it still has some gravitational potential energy, as in the previous point.

4. Gravitational potential energy (C. Eg)

The jumper is standing on the bridge, so it has gravitational potential energy given by its height h above the ground:

$E_g=mgh$

where m is the mass of the jumper.

5. This exercise has the same text of the previous one.

8 0

3 years ago

A test rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward

Ksenya-84 [330]

I think th<span>e upward acceleration of the rocket during the burn phase is 7.</span>8 m/s2

6 0

4 years ago

During a race, a runner runs at a speed of 6 m/s. Four seconds later, she is running at a speed of 10 m/s. what is the runners a

Georgia [21]

<span>a = (v2 - v1)/t = (10 - 6)/2 = 2 m/sec/sec (average acceleration)</span>

8 0

4 years ago

In an economy, the demand for labor is given by the equation W = 15 - (1/200) L and the supply of labor is given by the equation

mr_godi [17]

Answer:

the equilibrium wage rate is 10 and the equilibrium quantity of labor is 1000 workers

Explanation:

The equilibrium wage rate and the equilibrium quantity of labor are found as the point where the equation of demand intercepts the equation of supply, so the equilibrium quantity of labor is:

$W_{Demand} = W_{Supply}$

15 - (1/200) L = 5 + (1/200) L

15 - 5 = (1/200) L + (1/200) L

10 = (2/200) L

(10*200)/2 = L

1000 = L

Then, the equilibrium wage rate is calculated using either the equation of demand for labor or the equation of supply of labor. If we use the equation of demand for labor, we get:

W = 15 - (1/200) L

W = 15 - (1/200) 1000

W = 10

Finally, the equilibrium wage rate is 10 and the equilibrium quantity of labor is 1000 workers

7 0

3 years ago