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Mars2501 [29]
2 years ago
8

An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass

1.2 kg is attached to the free end of the spring. The glider is pulled toward the right along a frictionless air track, and then released. Now the glider is moving in simple harmonic motion with amplitude 0.045 m. The motion is horizontal (one-dimensional).
Suddenly, Slimer holding an apple flies in and approaches the glider. Slimer drops the apple vertically onto the glider from a very small height. The apple sticks to the glider. The mass of the apple is 0.18 kg.
Recall that the total mechanical energy is E = 1/2 mv^2 + 1/2 kx^2 = 1/2 kA^2 = constant
(a) Calculate the new amplitude of the motion of the glider with apple if the apple is dropped at the moment when the glider passes through its equilibrium position, x = 0 m.
Hints: The total energy of the glider just before the collision is E = 1/2mglider v^2 = 1/2KA^2
The apple sticks to the glider in a completely inelastic collision. The glider is now moving with the apple but at a lower speed. The linear momentum is conserved. Write a corresponding equation.
Also, assume that the collision is very short, so just before the collision the glider is at x = 0 m, and just after the collision the glider and apple are still at x = 0 m. Therefore, the total energy of the glider just after the collision is Enew = 1/2mglider + applev^2new = 1/2kA^2new
(b) Calculate the period of the motion of the glider and the period of the motion of the glider with apple.
Hint: it's a very simple question.

Physics
1 answer:
solniwko [45]2 years ago
4 0

Answer:

A) The new amplitude = 0.048 m

B) Period T = 0.6 seconds

Explanation: Please find the attached files for the solution

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An airplane flying at 115 m/s due east makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to
ankoles [38]

Answer:

The magnitude of the centripetal acceleration during the turn is a=12.04\ m/s^2.

Explanation:

Given :

Speed to the airplane in circular path , v = 115 m/s.

Time taken by plane to turn , t= 15 s.

Also , the plane turns from east to south i.e. quarter of a circle .

Therefore, time taken to complete whole circle is , T=t\times 4=60\ s.

Now , Velocity ,

v=\dfrac{2\pi r}{T}\\\\115=\dfrac{2\times 3.14\times r}{60}\\\\r=1098.73\ m.

Also , we know :

Centripetal acceleration ,

a=\dfrac{v^2}{r}

Putting all values we get :

a=12.04\ m/s^2.

Hence , this is the required solution .

5 0
1 year ago
Terminal velocity. A rider on a bike with the combined mass of 100kg attains a terminal speed of 15m/s on a 12% slope. Assuming
Firlakuza [10]

Answer:

0.9378

Explanation:

Weight (W) of the rider = 100 kg;

since 1 kg = 9.8067 N

100 kg will be = 980.67 N

W = 980.67 N

At the slope of 12%, the angle θ is calculated as:

tan \ \theta = \dfrac{12}{100} \\ \\  tan \ \theta = 0.12 \\ \\  \theta = tan^{-1}(0.12) \\\\ \theta = 6.84^0

The drag force D = Wsinθ

\dfrac{1}{2}C_v \rho AV^2 = W sin \theta

where;

\rho = 1.23 \ kg/m^3

A = 0.9 m²

V = 15 m/s

∴

Drag coefficient C_D = \dfrac{2 *W*sin \theta}{\rho *A *V^2}

C_D =\dfrac{2 *980.67*sin 6.84}{1.23 *0.9 *15^2}

C_D =0.9378

8 0
1 year ago
Water is formed when two hydrogen atoms bond to an oxygen atom. The hydrogen and the oxygen in this example are different A) com
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2 years ago
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A child on a 2.4 kg scooter at rest throws a 2.2 kg ball. The ball is given a speed of 3.1 m/s and the child and scooter move in
kykrilka [37]

Answer:

The child's mass is 14.133 kg

Explanation:

From the principle of conservation of linear momentum, we have;

(m₁ + m₂) × v₁ + m₃ × v₂ = (m₁ + m₂)  × v₃ - m₃ × v₄

We include the negative sign as the velocities were given as moving in the opposite directions

Since the child and the ball are at rest, we have;

v₁ = 0 m/s and v₂= 0 m/s

Hence;

0 = m₁ × v₃ - m₂ × v₄

(m₁ + m₂)× v₃ = m₃ × v₄

Where:

m₁ = Mass of the child

m₂ = Mass of the scooter = 2.4 kg

v₃ = Final velocity of the child and scooter = 0.45 m/s

m₃ = Mass of the ball = 2.4 kg

v₄ = Final velocity of the ball = 3.1 m/s

Plugging the values gives;

(m₁ + 2.4)× 0.45 = 2.4 × 3.1

(m₁ + 2.4) = 16.533

∴ m₁ + 2.4 = 16.533

m₁ = 16.533 - 2.4 = 14.133 kg

The child's mass = 14.133 kg.

3 0
2 years ago
A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point loc
Dafna11 [192]

Answer:

(a) 18.75 rad/s²

(b) 14920.78 rev

Explanation:

(a)

First we find the acceleration of the centrifuge using,

a = (v-u)/t......................... Equation 1

Where v = final velocity, u = initial velocity, t = time.

Given: v = 150 m/s,  u = 0 m/s ( from rest), t = 100 s

Substitute into equation 1

a = (150-0)/100

a = 1.5 m/s²

Secondly we calculate for the angular acceleration using

α = a/r..................... Equation 2

Where α = angular acceleration, r = radius of the centrifuge

Given: a = 1.5 m/s², r = 8 cm = 0.08 m

substitute into equation 2

α = 1.5/0.08

α = 18.75 rad/s²

(b)

Using,

Ф = (ω'+ω).t/2........................... Equation 3

Where Ф = number of revolution of the centrifuge, ω' = initial angular velocity, ω = Final angular velocity.

But,

ω = v/r and ω' = u/r

therefore,

Ф = (u/r+v/r).t/2

where u = 0 m/s (at rest),  = 150 m/s, r = 0.08 m, t = 100 s

Ф = [(0/0.08)+(150/0.08)].100/2

Ф = 93750 rad

If,

1 rad = 0.159155 rev,

Ф = (93750×0.159155) rev

Ф = 14920.78 rev

6 0
2 years ago
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