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Dominik [7]
1 year ago
11

A 1300 kg car moving at 20 m/s and a 900 kg car moving at 15 m/s in precisely oppositedirections participate in a head-on crash.

If we consider this event to be a perfectlyinelastic collision, what is the speed and direction of the cars after the impact?a
Physics
1 answer:
miskamm [114]1 year ago
5 0

Given

Car 1

m1 = 1300 kg

v1 = 20 m/s

m2 = 900 kg

v2 = -15 m/s

(Negative sign shows that direction of car 2 is opposite to car 1)

Procedure

As per the conservation of linear momentum, "The total momentum of the system before the collision must be equal to the total momentum after the collision". And this applies to the perfectly inelastic collision as well. Then the expression is,

\begin{gathered} m_1v_1+m_2v_2=(m_1+m_2)v \\ v=\frac{m_1v_1+m_2v_2}{m_1+m_2} \\ v=\frac{1300\operatorname{kg}\cdot20m/s-900\operatorname{kg}\cdot15m/s}{1300\operatorname{kg}+900\operatorname{kg}} \\ v=5.681m/s \end{gathered}

Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.

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Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in an elastic, glancing collision. The yello
klemol [59]

Answer:

3.62m/s and 2.83m/s

Explanation:

Apply conservation of momentum

For vertical component,

Pfy = Piy

m* Vof (sin38) - m*Vgf (sin52) = 0

Divide through by m

Vof(sin38) - Vgf(sin52) = 0

Vof(sin38) = Vgf(sin52)

Vof (sin38/sin52) = Vgf

0.7813Vof = Vgf

For horizontal component

Pxf= Pxi

m* Vof (cos38) - m*Vgf (cos52) = m*4.6

Divide through by m

Vof(cos38) + Vgf(cos52) = 4.6

Recall that

0.7813Vof = Vgf

Vof(cos38) + 0.7813 Vof(cos52) = 4.6

0.7880Vof + 0.4810Vof = 4.

1.269Vof = 4.6

Vof = 4.6/1.269

Vof = 3.62m/s

Recall that

0.7813Vof = Vgf

Vgf = 0.7813 * 3.62

Vgf = 2.83m/s

3 0
3 years ago
What human process are carbon sources that are upsetting the balance between CO to update via planets and CO2 released by living
Serhud [2]
The combustion of fossil fuels is releasing more co2 into the atmosphere then what would occur naturally
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3 years ago
The gravitational force of attraction between two students sitting at their desks in physics class is 2.59 × 10−8 N. If one stud
motikmotik
<h2>The distance between students is 2.46 m</h2>

Explanation:

The force of attraction due to Newton's gravitation law is

F = \frac{Gm_1m_2}{r^2}

Here G is the gravitational constant

m₁ is the mass of one student

m₂ is the mass of second student .

and r is the distance between them

Thus r = \sqrt{\frac{Gm_1m_2}{F} }

If we substitute the values in the above equation

r = \sqrt{\frac{6.673x10^-^1^1x31.9x30.0}{2.59x10^-^8} }

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3 0
3 years ago
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Find Vxl and Vyl of a pumpkin launched at a velocity of 55 m/s at an angle of 20 degrees
Vinvika [58]

Answer:

             

Explanation:

is  A projectile is any object on which the only force acting is gravity and air resistance (drag).

Examples of projectiles are:

baseballs and softballs in the air after being hit by the bat

golf balls hit by a club

objects dropped from aircraft, such as people (skydivers), bombs, crates of food being dropped to refugees

objects launched by cannons, such as cannonballs, shells, and circus performers

Once the baseball, softball, golf ball, skydiver, bomb, crate, cannonball, shell, or clown are no longer touching the bat, club, aircraft, or cannon, and are in the air with only gravity and slight air resistance acting on it, then it is a projectile.

Here is an online projectile motion applets to play with, just for fun.

Unless otherwise stated in a particular problem or discussion, we will be ignoring the effects of air resistance.

The key to understanding the motion of projectiles is that the horizontal motion and the vertical motion of the projectile are independent of each other. So we can write separate equations for the displacement of the projectile in the horizontal (x) and vertical (y) directions.

                         

The only common variable between these two equations is t, the time. Because in projectile problems there is usually no acceleration (i.e. we ignore air resistance) in the horizontal direction, we can write

           

The velocity components follow the same equations we used for one-dimensional motion.

                             

Because there is usually no acceleration in the x direction, the x-velocity is constant.

3 0
3 years ago
The light coming out of a concave lens:
torisob [31]
The light coming out of a concave lens will never meet.

So, the answer is A. will never meet.

Happy Studying! ^^
4 0
3 years ago
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