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3 years ago

You accelerate from 2m/s to 6m/s while traveling a distance of 2m; what was your acceleration?

Physics

1 answer:

nata0808 [166]3 years ago

7 0

Answer:

The acceleration is 8 m/s²

Explanation:

The given parameters are;

The initial velocity, u = 2 m/s

The final velocity, v = 6 m/s

The distance the acceleration took place, s = 2 m

The acceleration, a, can be found from the following kinematic equation;

v² = u² + 2·a·s

By substituting the values, we have;

6² = 2² + 2 × a × 2

6² - 2² = 2 × a × 2

32 = 4·a

a = 32/4 = 8 m/s²

The acceleration, a, of the given motion = 8 m/s².

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andrew-mc [135]

Answer:

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Explanation:

use f=ma

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3 years ago

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Calculate the elastic potential energy stored in a spring if it has a force constant of 150 N/m. the spring is extended to a len

Alenkinab [10]

Answer:

6.75J

Explanation:

U=1/2KΔx²

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3 years ago

1-A car moves toward east 12km is represented as A and it turns towards south 16km is represented as B. What is the resultant ve

hjlf

1. A-20 km south east

The car's displacement consists of two components into two different directions. Using a system of coordinates in which x represents the east direction and y represents the south direction, the two displacements are:

$d_x = 12 km$ east

$d_y = 16 km$ south

Since the two components are orthogonal to each other, we can find the resultant displacement by using Pythagorean's theorem:

$d=\sqrt{d_x^2+d_y^2}=\sqrt{(12 km)^2+(16 km)^2}=\sqrt{400}=20 km$

and the direction is between the two original directions, so south-east.

2. D. 10 m/s

First of all, we need to calculate the total time the stone took to hit the ground. Since the vertical distance covered is S = 78.4 m, and since the motion is an accelerated motion with constant acceleration g=9.8 m/s^2, we have

$S=\frac{1}{2}gt^2$

From which we find the total time of the fall, t:

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(78.4 m)}{9.8 m/s^2}}=4 s$

Now we can consider the horizontal motion of the stone: we know that the stone travels for d = 40 m in a time of t = 4 s, therefore the horizontal velocity of the stone is

$v=\frac{d}{t}=\frac{40 m}{4 s}=10 m/s$

3. B=32.32 m

As in the previous problem, we have to calculate the total time it takes for the stone to reach the river first. Since the vertical distance covered is S = 20 m, we have

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(20 m)}{9.8 m/s^2}}=2.0 s$