# How much kinetic energy does a 4 kg cats have while at 9 m/s​

162 Joules

Step-by-step explanation:

Remember, the equation for the kinetic energy of an object is:

where m is the mass of the object in kg (kilograms), v is the velocity of the object in m/s (meters per second), and K is the kinetic energy of the object in J (joules).

We are given that the cat's mass is 4 kg. We will use this for m. We are also given that the cat is moving at 9 m/s. This would be it's speed. We can use this for velocity, or v. Plugging in we get:

So K, or the kinetic energy, would be 162J.

I hope you find my answer and explanation to be helpful. Happy studying.

## Related Questions

PLEASE WRITE THE EQUATION THAT IS WHAT IT IS ASKING!! THE EQUATION IS IN SLOPE INTERCEPT FORM!!

y = -3x-5

Step-by-step explanation:

i am the first person to answer mark me brainliest! :D

Step-by-step explanation:

among a group of students 50 played cricket 50 played hockey and 40 played volleyball. 15 played both cricket and hockey 20 played both hockey and volleyball 15 played cricket and volley ball and 10 played all three. if every student played at least 1 game find the no of students and how many students played only cricket, only hockey and only volley ball

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Explanation:

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

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What is the equivalent of 27/5 in decimal form?

5.4 is equivalent to 27/5.

Step-by-step explanation: 27/5, so 5x5 makes 25 and 2 remaining so 5x0.4=2 so answer is 5+0.4 which equals to 5.4

Two numbers are randomly selected on a number line numbered 1 through 9. The same number can be chosen twice. What is the probability that both numbers are greater than 6?

p= 9/81 = 1/9 thats the real right answer
(3)/(8+7+6+5+4+3+2+1)(no repitition)
=3/36
=1/12

Solve the system of equations. \begin{aligned} & -9x+4y = 6 \\\\ & 9x+5y=-33 \end{aligned} ​ −9x+4y=6 9x+5y=−33 ​

x = -2

y = -3

Step-by-step explanation:

We can use either substitution or elimination for this problem. I will use elimination to solve this problem:

Step 1: Eliminate x by adding the 2 equations together

9y = -27

y = -3

Step 2: Plug in y into one of the original equations to get x

-9x + 4(-3) = 6

-9x -12 = 6

-9x = 18

x = 2

And we have our final answers!