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ozzi
1 month ago
10

A cart of mass M is attached to an ideal spring that can stretch and compress equally well. The cart and spring rest on a smooth

horizontal track. The cart is pulled to position A and released. The cart then moves toward position E, where it reverses direction and returns again to position A.
a) show with the sketch of a position diagram how the velocity of the cart changes between the points indicated
b) The dots below represent the cart at different labeled positions. Draw free body diagrams showing and labeling all the forces exerted on the cart at each labeled position. Draw the relative lengths of all vectors to reflect the relative magnitude of all forces. You will create one set of sketches for when the cart is moving to the right and a second set for when the car is moving to the left.
c) do the diagrams above indicate whether the cart is moving left or right? Justify your claim with evidence.
d) at each position, compare the direction of the net force exerted by the spring on the cart and the carts displacement from equilibrium when at that position. Note that this question is not asking whether the cart is moving right or left. Use these results to briefly explain why your claim in part C makes sense.

Physics
1 answer:
Maru [3.3K]1 month ago
4 0

It’s important to note that ‘the cart or spring is situated on a frictionless horizontal track’, which means there are no frictional forces acting on them.

Explanation:

a) Per the scenario, when the cart is drawn to position A and then let go, its initial speed at A (at time t=0) is 0 m/s. The cart travels toward position E, reaches a point where it reverses direction, and then heads back to position A. During this second phase, as the cart travels from A to E, its speed increases until it becomes zero again at point E, at which point it changes direction again, thus

( File has been attached)

b) Let’s assume the distance between two successive points is x meters and the spring constant is k N/m

c) ( File has been attached)

d) Moving to the right

e) Moving to the left

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A rock is dropped from the top of a tall building. The rock's displacement in the last second before it hits the ground is 46 %
inna [3103]

The height measures 69.68 m

Explanation:

Given data

Before striking the ground =  46 % of the total distance

To establish

the height

Solution

We know here acceleration and displacement, which is

d = (0.5)gt²..............1

Here d is the distance, g is the acceleration, and t is time

So, when an object falls it will be

h = 4.9 t²....................2

For the first part of the inquiry

The falling objects account for

54 % of the total distance

0.54 h = 4.9 (t-1)²...................3

Thus,

Now we possess two equations with unknown variables

We can equate both equations

The first equation already solves for h

Substituting h in the second equation allows us to find t

0.54 × 4.9 t² = 4.9 (t-1)²  

t = 0.576 s and  3.771 s

We choose here 3.771 s since 0.576 s is negligible; the distance covered in the last second before it impacts the ground is 46 % of the entire fall.

Thus, selecting t = 3.771 s

Then h from equation 2

h = 4.9 t²

h = 4.9 (3.771)²

h =  69.68 m

Thus, the height is 69.68 m

6 0
1 month ago
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
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Answer:

The mass of the child is 14.133 kg

Explanation:

According to the principle of linear momentum conservation, we get;

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

The negative sign signifies that their velocities were directed oppositely

Considering both the child and the ball are initially at rest, we have;

v₁ = 0 m/s and v₂= 0 m/s

Thus;

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

Substituting the known values gives us;

(m₁ + 2.4) × 0.45 = 2.4 × 3.1

(m₁ + 2.4) = 16.533

This implies m₁ + 2.4 = 16.533

Thus, m₁ = 16.533 - 2.4 = 14.133 kg

The mass of the child equals 14.133 kg.

3 0
1 month ago
slader A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of
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Response: 800N

Clarification:

Provided data:

Ball mass = 0.8kg

Contact duration = 0.05 seconds

Final and initial speed = 25m/s

The average force exerted by the ball on the wall can be calculated using the following relationship:

Force (F) = mass (m) * average acceleration (a)

a= (initial velocity (u) + final velocity (v))/t

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t = contact time of the ball = 0.05s

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Hence,

The average force magnitude (F)

F=ma

m = ball mass = 0.8

a = 1000m/s^2

F = 0.8 * 1000

F = 800N

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