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3 years ago

Displacement is to velocity as _______ is to acceleration

1 answer:

7 0

Velocity is the rate of change of displacement and both of them are vectors. Acceleration is the rate of change of velocity and both of them are vectors. So the answer is d.

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your softball coach needs to add a new player to the team roster. which of these people would most likely demonstrate the best s

Shkiper50 [21]

It would be easier to answer your question if you attached options. Anyway I remember that the right answer to that question is:<span> marissa, because she has cheered for the team all season long</span>

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3 years ago

Which body system remove carbon dioxide and waste

antiseptic1488 [7]

The excretory system

7 0

3 years ago

Read 2 more answers

11. You want to calculate the displacement of an object thrown over a bridge. Using -10 m/s2 for acceleration due to gravity, wh

IrinaVladis [17]

Answer:

The displacement was 320 meters.

Explanation:

Assuming projectile motion and zero initial speed (i.e., the object was dropped, not thrown down), you can calculate the displacement using the kinematic equation:

$d = \frac{1}{2}gt^2=\frac{1}{2}10\frac{m}{s^2}\cdot 8^2 s^2=320 m$

The displacement was 320 meters.

7 0

3 years ago

A plane travels 350 km south along a straight path with an average velocity of 125 km/h to the south. The plane has a 30 minute

faust18 [17]

Average Velocity = Total Displacement / Total time

1st part of journey, 350 km at velocity 125 km/h

Time = 350 / 125 = 2.8 hours.

2nd part of journey, 220 km at velocity 115 km/h

Time = 220 / 115 = 1.9 hours

Average Velocity = Total Displacement / Total time

= (350 + 220) / (2.8 + 1.9)

= 570 / 4.7 ≈ 121.3 km/hr

Average Velocity ≈ 121 km/hr due south.

Option C.

6 0

3 years ago

At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f

jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp = change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s, 0.65 kg m/s)

The magnitude of this vector is calculated as follows:

$|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s$

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

4 0

3 years ago