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2 years ago

If an object has an initial velocity of 2.0 m/s and its rate of change in velocity is 3.0 m/s2

Physics

1 answer:

erica [24]2 years ago

6 0

Answer:

14m/s2

Explanation:

initial velocity=2m/s

final velocity =?

time=4s

acceleration=3.0m/s

v=u+at

v=(2)+(3*4)

v=14m/s

by substituting the values in the equation we get the value 14m/s.

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Andreas93 [3]

High school???
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3 years ago

A truck traveling with an initial velocity of 22m/s comes to a stop in 17.32 secs. What is the acceleration of the truck?

Tomtit [17]

Explanation:

Given:

v₀ = 22 m/s

v = 0 m/s

t = 17.32 s

Find: a

v = at + v₀

(0 m/s) = a (17.32 s) + (22 m/s)

a = -1.270 m/s²

Round as needed.

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3 years ago

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Svetradugi [14.3K]

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3 years ago

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A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.

I hope it helps you!

3 0

3 years ago

g Two masses are involved in a collision on an axis (one dimensional). One mass is six times the mass of the second. Both masses

statuscvo [17]

Answer:

v₁f = 0.5714 m/s (→)

v₂f = 2.5714 m/s (→)

e = 1

It was a perfectly elastic collision.

Explanation:

m₁ = m

m₂ = 6m₁ = 6m

v₁i = 4 m/s

v₂i = 2 m/s

v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i + ((2m₂) / (m₁ + m₂)) v₂i

v₁f = ((m – 6m) / (m + 6m)) * (4) + ((2*6m) / (m + 6m)) * (2)

v₁f = 0.5714 m/s (→)

v₂f = ((2m₁) / (m₁ + m₂)) v₁i + ((m₂ – m₁) / (m₁ + m₂)) v₂i

v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)

v₂f = 2.5714 m/s (→)

e = - (v₁f - v₂f) / (v₁i - v₂i) ⇒ e = - (0.5714 - 2.5714) / (4 - 2) = 1

It was a perfectly elastic collision.

8 0

3 years ago