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3 years ago

a car is traveling at 26 m/s starts to decelerate steadily. It comes to a complete stop in 6 seconds. What is the acceleration?

Physics

2 answers:

torisob [31]3 years ago

5 0

I can’t see if anyone have gave you the answer yet

ad-work [718]3 years ago

4 0

-0.5 m/s^2

Hope you enjoy!

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A skydiver reaches terminal velocity. Then he opens his parachute.

schepotkina [342]

Answer:

it is 0.9999.10896

Explanation:

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3 years ago

Read 2 more answers

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. the heavie

Doss [256]

By conservation of momentum,
Distance lighter fragment slide= 7*6.7=46.9m

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3 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

Which statement describes the horizontal acceleration of a projectile? It is –9.8 m/s2. It is 9.8 m/s2. It is constant. It is ze

lina2011 [118]

In a projectile, the horizontal acceleration is zero. The velocity remains constant at all times. However, the <u>vertical acceleration</u> is -9.81m/s^2.

Hope this helps!

6 0

3 years ago

Read 2 more answers

The space shuttle is accelerated off its launch pad to a velocity of 525 m/s in 18.0 seconds.

Eva8 [605]

Answer: 29.17m/s^2

Explanation:

Given the following :

Velocity = 525 m/s

Time = 18 seconds

Acceleration = change in Velocity with time

Using the motion equation:

v = u + at

Where v = final Velocity

u = Initial Velocity and t = time

Plugging our values

525 = 0 + a × 18

525 = 18(a)

a = 525 / 18

a = 29.166666

a = 29.17 m/s^2

8 0

3 years ago