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3 years ago

Please help me I promise will mark u brainest

Physics

1 answer:

zimovet [89]3 years ago

3 0

Answer:

Point A

Explanation:

There are 3 states of equilibrium, which are as follows:

1. Stable Equilibrium

2. Unstable Equilibrium

3. Neutral Equilibrium

1. Stable Equilibrium:

A body is said to be in stable equilibrium, if it comes back to its original position, when it is slightly displaced. In this case the center of mass of the body is raised up as compared to the initial position.

Example: A book lying on a table, a map hanging on the wall, a stone suspended by a spring.

2. Unstable Equilibrium:

A body is said to be in unstable equilibrium if it does not come back to its original position when it is slightly displaced. In this case the center of mass of the body is lowered, as compared to the initial position.

Example: A vertical standing pencil, a stick balanced on finger, a funnel resting on its pointed end.

3. Neutral Equilibrium:

A body is said to be in neutral equilibrium, if on being slightly displaced it does not come back to its original position but occupies a new position similar to its original position. In this case the center of mass of the body remains at same height.

Example: A spherical ball on floor, a cone resting in horizontal position.

So, it is clear from the figure that when we will remove displacement the bus will come back to its original position. Therefore, this is the case of stable equilibrium.

So, the center of mass must have raised up. The original center of mass of the bus must be in the center. So, now the center of mass should have moved above center of bus, vertically.

Therefore, the center of mass of the bus is at point A

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Two cars are facing each other. Car A is at rest while car B is moving toward car A with a constant velocity of 20 m/s. When car

lapo4ka [179]

Answer:

Let's define t = 0s (the initial time) as the moment when Car A starts moving.

Let's find the movement equations of each car.

We know that Car A accelerations with a constant acceleration of 5m/s^2

Then the acceleration equation is:

$A_a(t) = 5m/s^2$

To get the velocity, we integrate over time:

$V_a(t) = (5m/s^2)*t + V_0$

Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:

$V_a(t) = (5m/s^2)*t$

To get the position equation we integrate again over time:

$P_a(t) = 0.5*(5m/s^2)*t^2 + P_0$

Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:

$P_a(t) = 0.5*(5m/s^2)*t^2$

Now let's find the equations for car B.

We know that Car B does not accelerate, then it has a constant velocity given by:

$V_b(t) =20m/s$

To get the position equation, we can integrate:

$P_b(t) = (20m/s)*t + P_0$

This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:

$P_b(t) = (20m/s)*t + 100m$

Now we can answer this:

1) The two cars will meet when their position equations are equal, so we must have:

$P_a(t) = P_b(t)$

We can solve this for t.

$0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0$

This is a quadratic equation, the solutions are given by the Bhaskara's formula:

$t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)} }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}$

We only care for the positive solution, which is:

$t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s$

Car A reaches Car B after 11.48 seconds.

2) How far does car A travel before the two cars meet?

Here we only need to evaluate the position equation for Car A in t = 11.48s:

$P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m$

3) What is the velocity of car B when the two cars meet?

Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s

4) What is the velocity of car A when the two cars meet?

Here we need to evaluate the velocity equation for Car A at t = 11.48s

$V_a(t) = (5m/s^2)*11.48s = 57.4 m/s$

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3 years ago

Suppose you want to use this human engine to lift a 2.35 kg box from the floor to a tabletop 1.30 m above the floor. How much mu

morpeh [17]

Answer:

The increase in the gravitational potential energy is 29.93 joules.

Explanation:

Given that,

Mass of the box, m = 2.35 kg

It is lifted from the floor to a tabletop 1.30 m above the floor, h = 1.3 m

We need to find the increase the gravitational potential energy. Initial it will placed at ground i.e. its initial gravitational potential is equal to 0. The increase in the gravitational potential energy is given by :

$U=mgh$