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2 years ago

Describe and name the different types of collision. In which are the linear momentum and kinetic energy conserved

Physics

1 answer:

ipn [44]2 years ago

8 0

Answer:

1. Elastic collision

2. Inelastic collision

Explanation:

Elastic collision: collision is said to be elastic if total kinetic energy is not conserved and if there is a rebound after collision

the collision is described by the equation bellow

$m1U1+ m2U2= m1V1+m2V2$

Inelastic collision: this type of collision occurs when the total kinetic energy of a body is conserved or when the bodies sticks together and move with a common velocity

the collision is described by the equation bellow

$m1U1+ m2U2= V(m1+m2)$

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Which of the following is not a reason fluorescent lamps are advantageous over incandescent lamps?

cricket20 [7]

Answer;

B. Fluorescent lamps operate at a higher temperature than incandescent lamps.

Explanation;

-A fluorescent lamp, is a type of electric light (lamp) that uses ultraviolet emitted by mercury vapor to excite a phosphor, which emits visible light.

-A fluorescent lamp produces less heat, thus, it is much more efficient. A fluorescent bulb can produce between 50 and 100 lumens per watt. This makes fluorescent bulbs four to six times more efficient than incandescent bulbs.

-Fluorescent lamps operate best around room temperature. At much lower or higher temperatures, efficacy decreases.

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2 years ago

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What are the principles of electrodynamic

Sveta_85 [38]

Answer:

the branch of mechanics concerned with the interaction of electric currents with magnetic fields or with other electric currents.

Explanation:

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2 years ago

Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit

Ede4ka [16]

Answer:

$386.2^{\circ}F$

Explanation:

We are given that

$P_1=200lbf/in^2$

$P_2=60lbf/in^2$

$v_1=200ft/s$

$v_2=1700ft/s$

$T_1=500^{\circ}F$

$Q=0$

$C_p=1BTU/lb^{\circ}F$

We have to find the exit temperature.

By steady energy flow equation

$h_1+v^2_1+Q=h_2+v^2_2$

$C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}$

$1BTU/lb=25037ft^2/s^2$

Substitute the values

$1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}$

$500+1.598=T_2+115.4$

$T_2=500+1.598-115.4$

$T_2=386.2^{\circ}F$

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2 years ago

7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N

artcher [175]

Answer:

20.4 $m/s^{2}$

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 $m/s^{2}$

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2 years ago

A trip is taken that passes through the following points in order

riadik2000 [5.3K]

Answer:

35, I got you bro, i got you

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2 years ago