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rosijanka [135]
3 years ago
8

A 3.00 kg mud ball has a perfectly inelastic collision with a second mud ball that is initially at rest. the composite system mo

ves with a speed equal to one- third the original speed of the 3.00 kg mud ball. what is the mass of the second mud ball?
Physics
1 answer:
viva [34]3 years ago
7 0
Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:

m_{1} * v_{1} +  m_{2} * v_{2} =  m_{1} * v'_{1} +  m_{2} * v'_{2}

In case of perfectly inelastic collision v'1 and v'2 are same.

We have following information:
m₁=3 kg
m₂=? kg
v₁=x m/s
v₂=0 m/s
v'1 = v'2 = 1/3 * v₁

Now we insert given information and solve for m₂:
3*v₁ + 0*? = 3*1/3*v₁ + m₂*1/3*v₁
3v₁ = v₁ + m₂*1/3*v₁
2v₁ = m₂*1/3*v₁
2 = m₂*1/3
m₂= 6kg

Mass of second mud ball is 6kg.
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8 0
3 years ago
A bike travels at 7.5 m/s along a straight road, whereas a car travels at 10.0 m/s along the same road and in the same direction
Ilya [14]

Answer:

t = 8.3 s

Explanation:

As we know that

velocity of bike = 7.5 m/s

velocity of car is 10 m/s

deceleration of car is 0.75 m/s^2

part a)

velocity of bike with respect to car is given as

v_r = 7.5 - 10 = -2.5 m/s

acceleration of bike with respect to car is given as

a_r = 0 - (-0.75) = 0.75 m/s^2

now the distance of the bike with respect to car is given as

d = v_i t + \frac{1}{2}at^2

5 = (-2.5) t + \frac{1}{2}(0.75)t^2

t = 8.3 s

Part b)

3 0
3 years ago
) A coin is tossed 13 times. a) How many different outcomes are possible? 2^8 b) How many different outcomes have exactly 4 head
Svetradugi [14.3K]

Answer:

Explanation:

a) A coin has two sides, therefore the total outcome possible when a coin is tossed is 2 i.e Head (H) and Tail (T)

outcome of two coins will be 4 i.e 2^2

Outcome of three coins will be 8 i.e 2^3 and so on. Since its following a trend, the outcome when 'n' coins is tossed will be 2^n.

Using the general formula, the possible outcome when a coin is tossed 13 times will be "2^13"

b)

3 0
3 years ago
A uniform thin wire is bent into a quarter-circle of radius a = 20.0 cm, and placed in the first quadrant. Determine the coordin
Mashcka [7]

Answer:

r_{cm}=[12.73,12.73]cm

Explanation:

The general equation to calculate the center of mass is:

r_{cm}=1/M*\int\limits {r} \, dm

Any differential of mass can be calculated as:

dm = \lambda*a*d\theta  Where "a" is the radius of the circle and λ is the linear density of the wire.

The linear density is given by:

\lambda=M/L=M/(a*\pi/2)=\frac{2M}{a\pi}

So, the differential of mass is:

dm = \frac{2M}{a\pi}*a*d\theta

dm = \frac{2M}{\pi}*d\theta

Now we proceed to calculate X and Y coordinates of the center of mass separately:

X_{cm}=1/M*\int\limits^{\pi/2}_0 {a*cos\theta*2M/\pi} \, d\theta

Y_{cm}=1/M*\int\limits^{\pi/2}_0 {a*sin\theta*2M/\pi} \, d\theta

Solving both integrals, we get:

X_{cm}=2*a/\pi=12.73cm

Y_{cm}=2*a/\pi=12.73cm

Therefore, the position of the center of mass is:

r_{cm}=[12.73,12.73]cm

5 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!! A IS NOT THE CORRECT ANSWER
patriot [66]

Answer: A

<u>Explanation:</u>

NOTES:

d = 650 meters

t = 10 seconds

**********************************

v = d/t

  = 650 meters/10 seconds

  = 65 meters/second

6 0
4 years ago
Read 2 more answers
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