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ratelena [41]
2 years ago
13

In a collision, a 5 kg mass moving at 2m/s transfers all of its momentum to a 1 kg mass. What is the velocity of the 1kg mass af

ter the collision?
Will give brainliest
Mathematics
2 answers:
statuscvo [17]2 years ago
6 0
Answer:
velocity = 10 m/sec in the same direction as the first body did

Explanation:
The momentum of the body can be calculated as follows:
momentum = mass * velocity

For the first body, we have:
mass = 5 kg
velocity = 2 m/sec
momentum = mass * velocity
momentum = 5 * 2 = 10 kg.m/sec

We know that this momentum is transferred completely to the second body

For the second bode, we have:
momentum = 10 Kg.m/sec
mass = 1 kg
momentum = mass * velocity
10 = 1 * velocity 
velocity = 10/1
velocity = 10 m/sec

Finally, we should get the direction of the motion:
Both the velocity of the first and second bodies have positive values. Therefore, the second body is moving in the same direction as the first body did.

Hope this helps :)
Len [333]2 years ago
5 0

Answer:

3.67m/s

Step-by-step explanation:

According to the law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of momentum of bodies after collision. This bodies moves with a common velocity after collision.

Momentum = mass × velocity

For the 5kg ball with velocity of 2m/s;

Its momentum = 5×2 = 10kgm/s²

For the 1kg body;

It's momentum = 10kgm/s (since the 5kg mass transfers all of its momentum to a 1 kg mass.)

Momentum after collision of the bodies will give;

Momentum = (5kg+1kg)v where v is their common velocity.

Momentum after collision = 6v

Using the law of conservation of momentum;

10+10 = 6v

20 = 6v

v = 20/6

v = 3.67m/s

Since both masses have the same velocity after collision, the velocity of the 1kg mass after the collision will also be 3.67m/s

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The bases of a trapezoid are 24 feet and 16 feet. The height of the trapezoid is 12 feet. What is the area of the trapezoid?
stepan [7]
The formula to use is A=(b1+b2)h if one base is 24 and the other one is 16 which equals to 40 when you add then you still have H which is 12 so you could either divide 12 by two to get half or multiply by 12 and then divide it
40•12=480
480 divided by 2= 240
Your final answer should be 240ft
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2 years ago
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vodomira [7]
Hello
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Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose
Furkat [3]

Answer:

a) 0.164 = 16.4% probability that a disk has exactly one missing pulse

b) 0.017 = 1.7% probability that a disk has at least two missing pulses

c) 0.671 = 67.1% probability that neither contains a missing pulse

Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the binomial distribution(for item c).

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}


In which

x is the number of sucesses


e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Poisson mean:

\mu = 0.2

a. What is the probability that a disk has exactly one missing pulse?

One disk, so Poisson.

This is P(X = 1).

P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164


0.164 = 16.4% probability that a disk has exactly one missing pulse

b. What is the probability that a disk has at least two missing pulses?

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1)

In which

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}&#10;

P(X = 0) = \frac{e^{-0.2}*0.2^{0}}{(0)!} = 0.819

P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164&#10;

P(X < 2) = P(X = 0) + P(X = 1) = 0.819 + 0.164 = 0.983

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.983 = 0.017

0.017 = 1.7% probability that a disk has at least two missing pulses

c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

Two disks, so binomial with n = 2.

A disk has a 0.819 probability of containing no missing pulse, and a 1 - 0.819 = 0.181 probability of containing a missing pulse, so p = 0.181

We want to find P(X = 0).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{2,0}.(0.181)^{0}.(0.819)^{2} = 0.671

0.671 = 67.1% probability that neither contains a missing pulse

8 0
2 years ago
Help quick!
damaskus [11]
I thought that was all one equation lol this is very simple just plug in -3 for x 
which is -3^2-3 which is -12 
3 0
2 years ago
0.04,4/10,0.14,44/100 from least to greatest
garik1379 [7]
0.04, 0.14, 4/10, 44/100
5 0
3 years ago
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