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Varvara68 [4.7K]
3 years ago
15

You hold glider A of mass 0.125 kg and glider B of mass 0.375 kg at rest on an air track with a compressed spring of negligible

mass between them. When you release the gliders, the spring pushes them apart. Once the gliders are no longer in contact with the spring, glider A is moving to the right at 0.600 m/s. What is the velocity (magnitude and direction) of glider B at this time?
Physics
1 answer:
umka2103 [35]3 years ago
8 0

Answer:

0.2 m/s

Explanation:

m_A = Mass of glider A= 0.125 kg

m_B = Mass of glider B = 0.375 kg

u_A = Initial Velocity of glider A = 0 m/s

u_B = Initial Velocity of glider B = 0 m/s

v_A = Final Velocity of glider A = 0.6 m/s

v_B = Final Velocity of glider B

As linear momentum is conserved

m_{A}u_{A}+m_{B}u_{B}=m_{A}v_{A}+m_{B}v_{B}\\\Rightarrow v_B=\frac{m_{A}u_{A}+m_{B}u_{B}-m_{A}v_{A}}{m_{B}}\\\Rightarrow v_b=\frac{0+0-0.125\times 0.6}{0.375}\\\Rightarrow v_b=-0.2\ m/s

Magnitude of the glider B at this time is 0.2 m/s

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A student falls off a cliff into the lake 54.0 m below. What is the final velocity of the student?
Vladimir79 [104]

Answer:

v_{y} = -32.53 m / s

this velocity is directed downwards

Explanation:

This is a free fall exercise, let's use the expression

         v_{y}^{2} = v_{oy}^{2} + 2 g (y -yo)

where we are assuming that there is friction with the air, as the body falls its initial velocity is zero

         v_{oy} = √ 2g (y - y₀)

let's calculate

         v_{y} = √ (2 9.8 (0-54.0))

         v_{y} = -32.53 m / s

this velocity is directed downwards

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3 years ago
A package is dropped from a helicopter moving upward at 15 m/s
daser333 [38]

The distance the package above the ground when it was released, s ≈ 530 meters

<h3 /><h3>What are kinematic equations?</h3>

The kinematic equation of motion gives the interrelationships of the variables of motion.The correct option for the distance the package above the ground when it was released, is the third option;

It is given that:

The velocity of the helicopter from which the package was dropped = 15 m/s

The time it takes the package to strike the ground = 12 seconds

The required parameter:

The height of the package from the ground when it was dropped

The kinematic equation of motion relating distance, s, time, t, acceleration due to gravity, g, initial velocity, u, and final velocity, v, is applied as follows;

The package continues the upward motion for some time, t₁, given as follows;

Upward motion of the package

v = u - g·t₁

v = 0 at highest point reached by the package;

Therefore;

0 = 15 m/s - 9.81 m/s²  × t₁

t₁ = 15 m/s/(9.81 m/s²) ≈ 1.5295022 seconds

The time the package takes to return to the initial starting point, t₂ = t₁

The time the package falls after returning to the point it was dropped, t₃, is given as follows;

t₃ = t - (t₂ + t₁) = t - 2 × t₁

∴ t₃ = 12 s - 2 × 1.5295022 s ≈ 8.940996 s

From the symmetry of the motion of a projectile, the velocity of the package when returns to its staring point where it was dropped = u (Downwards) = 15 m/s

The distance the package falls, s, which is the distance the package above the ground when it was released, is given as follows;

s = u·t + (1/2)·g·t²

s = 15× 8.940996  + (1/2) × 9.81 × 8.940996² = 526.22755346 ≈ 530

The distance the package falls, s ≈ 530 m = The height of the

The distance the package above the ground when it was released, s ≈ 530 meters

Learn more about the kinematic equations of motion here:

brainly.com/question/16995301

#SPJ4

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choli [55]

F = 52000 N

m = 1060 kg

a= F/m = 52000 N/1060 kg = 49.0566 m/s^2

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