Total momentum before a collision is equal to total momentum after a collision.what does this define

7. Using Analogies Explain how the behavior

klemol [59]

The relative motion of gaseous particles increases with increase in the temperature of the gas molecules just like the motion of popcorn in a popper increases when heat is applied to the popper.

<h3>What is kinetic theory of gas?</h3>

The kinetic theory of gases or matter states that matter consists of tiny particles which are constant motion, colliding with one another and with walls of the containing vessels.

Just like a popcorn in a popcorn popper pops when heat is applied to the popper, gases contained in a cylinder increases their speed when they acquire more kinetic energy as the temperature of the cylinder increases.

Thus, the motion of gas particles depends on the temperature of the containing vessel so also does the random motion of popcorn depends on the temperature of the popper.

Learn more about kinetic theory of gases here: brainly.com/question/11067389

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7 0

7 months ago

A steel beam of mass 1975 kg and length 3 m is attached to the wall with a pin that can rotate freely on its right side. A cable

Nuetrik [128]

Answer:

a) 29062.125 N·m

b) 0 N·m

c) $Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}$

d) T = 11186.02 N

Explanation:

We are given

Beam mass = 1975 kg

Beam length = 3 m

Cable angle = 60° above horizontal

a) We have the formula for torque given as follows;

Torque about the pin = Force × Perpendicular distance of force from pin

Where the force = Force due to gravity or weight, we have

Weight = Mass × Acceleration due to gravity = 1975 × 9.81 = 19374.75 N

Point of action of force = Midpoint for a uniform beam = length/2

∴ Point of action of force = 3/2 = 1.5 m

Torque due to gravity = 19374.75 N × 1.5 m = 29062.125 N·m

b) Torque about the pinned end due to the contact forces between the pin and the beam is given by the following relation;

Since the distance from pin to the contact forces between the pin and the beam is 0, the torque which is force multiplied by perpendicular distance is also 0 N·m

c) To find the expression for the tension force, T we find the sum of the moment forces about the pin as follows

Sum of moments about p is given as follows

∑M = 0 gives;

T·sin(θ) × L= M×L/2×g

Therefore torque due to tension is given by the following expression

$Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}$

d) Plugging in the values in the torque due to tension equation, we have;

$3\times Tsin60 = \frac{1975\times 3\times 9.81}{2} = 29062.125$

Therefore, we make the tension force, T the subject of the formula hence

$T= \frac{29062.125}{3 \times sin(60)} = 11186.02 N$

8 0

3 years ago

Five basket ball teams are playing in a tournament. The teams will play with each other only one time. How many games will be pl

77julia77 [94]

<em>Ten games</em> will be played.

The first team can be any one of 5 . For each of those, the opponent can be any one of the other 4 .

Number of ways to pair them up = (5 x 4) = 20 ways .

<em>BUT</em> ... Whether the Reds play the Blues, or the Blues play the Reds, it's the SAME GAME. Each possible game shows up twice in the list of 20 ways. So there are really only 10 different pairs ==> <em>10 games</em> .

3 0

2 years ago

Why is global warming a cause of concern for the environmentalists?

Svetradugi [14.3K]

Because it can pollute the air and cause wild life to eat stuff that has been contimanted by the air

7 0

2 years ago

A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T

miv72 [106K]

Answer:

C. In the first collision has twice the momentum as when it stays still ( second colllions)

Explanation:

To see which statement is correct, it is best to solve the problem, the momentum is equal to the variation of the moment

I = Δp = m vf - m v₀

I = m (vf -v₀)

Case 1. In car bounces, the initial speed is 0.3 m / s, say that this direction is positive, when the magnitude of the speed bounces it remains constant, but its direction is reversed (vf = -0.3 m / s)

I₁ = m (-0.3 - 0.3)

I₁ = -0.6 m

Case 2. The expensive one that still after the crash so its speed is zero (vf = 0)

I₂ = m (0 - 0.3)

I₂ = -0.3 m

Let's calculate the relationship between the two impulses

I₁ / I₂ = -0.6m / -0.3m

I₂ / I₂ = 2

When it bounces it has twice the momentum as when it stays still

Now let's analyze the answers:

A. False The momentum changes

B. False. The momentum is less in the second collision

C. True. The momentum is double in this collision

D. False. Can be calculated, because the mass is the same throughout the exercise and is eliminated in the equations

E. False. When they say bounces it implies the same speed with the opposite direction

4 0

2 years ago