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neonofarm [45]
3 years ago
7

Two cars collide on the parkway. The drivers are following too closely and the driver in the front slows because of traffic to a

velocity of 20.0m/s and the driver behind is still moving at 30.0m/s, both cars traveling in the same direction in a rear end collision. If the cars are stuck together after the collision, how fast will they be traveling?
Physics
1 answer:
musickatia [10]3 years ago
4 0

Answer:

The are moving with a speed of 25 m/s after the collision.

Explanation:

assuming the mass (m) of the cars are equal, let v = 20 m/s be the velocity of the driver in the front and u = 30 m/s be the velocity of the driver behind. let V be the velocity of the cars after the collision.

then, by conservation of linear momentum:

m×(v + u) = V×2×m

(v + u) = V×2

V = (v + u)/2

  = (20 + 30)/2

  = 25 m/s

Therefore, the are moving with a speed of 25 m/s after the collision.

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Science If you want to find the energy quantum of light, you multiply the frequency of the radiation (v) by "h". What is "h"?
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Answer:

"h" signifies Planck's constant

Explanation:

In the equation energy E = h X v

The "h" there signifies Planck's constant

Planck's constant is a value, that shows the rate at which the energy of a photon increases/decreases, as the frequency of its electromagnetic wave changes.

It was named after Max Planck who discovered this unique relationship between the energy of a light wave and its frequency.

Planck's constant, "h" is usually expressed in Joules second

Planck's constant = 6.62607015 \times 10^{-34}  J.s

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A 10-kg cart moving at 5 m/s collides with a 5-kg cart at rest and causes it to move 10 m/s. Which principle explains the result
Over [174]
B) law of conservation of momentum

It states that the total momentum of a system before impact is the same as the total momentum of the system after impact.

In this case total momentum before impact:

10kg*5m/s  + 5kg * 0m/s = 50 kg m/s

After Impact:

10kg*0m/s + 5kg*10m/s = 50 kg m/s

You can see the momentum before and after impact is same as 50 kg m/s  

Of course we assumed that the first cart stopped after the impact, and there are no energy losses.
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3 years ago
A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff
strojnjashka [21]

(a) 6.43\cdot 10^5 J

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

E=K+U

The initial kinetic energy is:

K=\frac{1}{2}mv^2

where m = 58.0 kg is the mass of the projectile and v=140 m/s is the initial speed. Substituting,

K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J

The initial potential energy is given by

U=mgh

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J

So, the initial mechanical energy is

E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J

(b) -1.67 \cdot 10^5 J

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J

while the potential energy is

U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J

So, the mechanical energy is

E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J

And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:

W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J

And the work is negative because air friction is opposite to the direction of motion of the projectile.

(c) 88.1 m/s

The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:

W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J

When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:

E = K

The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:

E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J

And this is only kinetic energy:

E=K=\frac{1}{2}mv^2

So, we can solve to find the final speed:

v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s

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