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Dahasolnce [82]

3 years ago

14

a truck is traveling down the highway with a speed of 26 m/s. how long will it take the truck to stop (0m/s) if it's decelerates

at 10 m/s

1 answer:

kogti [31]3 years ago

7 0

Answer: 16 seconds

Explanation: i think

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An open train car rolls along a track while it is being filled with sand. There is negligible friction. As a result of

Angelina_Jolie [31]

Answer:

Explanation:

<h3>that`s a the train car, that you asked the meaning, of that if the train car rolls it`s doing it`s speed, and it`s not ganna fall off the the trail of the train, car.</h3>

3 0

2 years ago

A transverse wave is traveling from right to left. what direction does the medium vibrate?

lisabon 2012 [21]

<span>In transverse waves, particles of the medium vibrate to and from in a direction perpendicular to the direction of energy transport.</span>

5 0

3 years ago

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A ray of light travels across a liquid-to-glass interface. if the indices of refraction for the liquid and glass are, respective

prisoha [69]

When it comes to optics, Snell's law is the basic formula to be used. If you notice, when light hits the water, the light does not travel in the same direction. After, it hits the water, it changes in angle. Light becomes refracted. This is observed when your hands tend to become bigger if you place it underwater. The formula for Snell's Law is

n₁ sin θ₁ = n₂sin θ₂, where n is the index of refraction. This depends on the type of medium. For example, for air, n=1. The parameters θ₁ is the angle of incidence, and θ₂ is the angle of refraction. Critical angle is the incident angle needed so that the refract angle is 90°. So, modifying the equation:

n₁ sin θcrit = n₂sin 90°, since sin 90°=1,
sin θcrit = n₂/n₁
θcrit = sin ⁻¹ (n₂/n₁)

Since liquid comes first before glass, n₁=1.75 and n₂=1.52. Substituting,
θcrit = sin ⁻¹ (1.52/1.75)
θcrit = 60.29°

7 0

3 years ago

Which of the following are true about centripetal force? Check all that apply. A. Without centripetal force, an object cannot ac

Katyanochek1 [597]

B. Friction can be a centripetal force, such as when it keeps a car on the road going around a curve.

C. Gravity can be a centripetal force, such as when it pulls a satellite in its orbit.

Explanation:

The centripetal force is any force that keeps an object moving in circular motion, "pulling" the object towards the centre of the circular trajectory.

Several forces can act as centripetal force. Examples are:

- friction: when a car is going around the curve, is moving by circular motion. The force that keeps the car in circular motion is, in fact, the friction between the tires and the road.

- Gravity: when a satellite moves around the Earth, it is moving by circular motion. The force that keeps the satellite in circular motion is the gravitational attraction between the Earth and the satellite, that pushes the satellite towards the Earth.

The other two options are not correct because:

A) An object can also accelerate if there is no centripetal force (for example, a car speeding up on a straight road is accelerating, but there is no centripetal force since there is no circular motion

D) Centripetal force is not an outward force, since it pushes the object inwards (towards the centre of the trajectory).

3 0

3 years ago

Read 2 more answers

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