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Karolina [17]
3 years ago
9

You have a cat who has a mass of 10kg and is chasing a mouse with an acceleration of 10 m/s2. If it runs into a wall what force

will be exerted by the cat?
Mathematics
1 answer:
V125BC [204]3 years ago
3 0

<u>Given</u><u>:</u>

  • mass of cat ( m ) = 10 kg

  • Acceleration ( a ) = 10 m/s²

<u>To</u><u> </u><u>find</u><u> </u><u>out</u><u>:</u><u> </u>

If it runs into a wall what force will be exerted by the cat?

<u>Formula</u><u> </u><u>used</u><u>:</u>

Force = mass × acceleration

<u>Solution</u><u>:</u>

Force = mass × acceleration

Now, Putting these values of m and a in the above equation.

we get:

F = 10 × 10

= 100 N

Thus, the force will be exerted by the cat is 100 N.

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Answer:

It should 3/10

Step-by-step explanation:

3 0
3 years ago
Can someone help me with this please
hichkok12 [17]
<h3>Answer:  4</h3>

===========================================

Explanation:

For now, ignore the variables. Let's just focus on the coefficients 12 and 16.

The GCF of 12 and 16 is 4 because this is the largest factor they have in common.

12 = 4*3

16 = 4*4

You could create a factor tree to get the prime factorization of each number

12 = 2*2*3

16 = 2*2*2*2

Each factor tree has two copies of '2' leading to the GCF 2*2 = 4.

--------

Now let's go back to the variables. The term 12n and 16w^3 do not have any variables in common. So the final answer won't have any variables in it. If both terms had say a 'w' in them, then w would somehow be involved in the GCF. But that's not the case here. So we just stick to 4 as our answer.

--------

Side note: If you had 12n+16w^3 and you wanted to factor, then you would factor out the GCF 4 to get 12n+16w^3 = 4(3n+4w^3). Use the distribution rule to confirm this.

4 0
3 years ago
A hypothetical square grows so that the length of it's diagonals are increasing at a rate of 8 m/min. How fast is the area of th
guapka [62]

Answer:

32m^{2}/min

Step-by-step explanation:

We let the length of the square be l, the diagonal be z and the area be A.

Then by Pythagoras theorem;

l^{2}+l^{2}=z^{2}\\2l^{2}=z^{2}

The are of a square of length l is given as;

A=l^{2}

Combining these two equations yields;

z^{2}=2A

A=\frac{z^{2} }{2}

Next, we have been given;

\frac{dz}{dt}=8

We are required to find;

\frac{dA}{dt} when z = 4

By chain rule;

\frac{dA}{dt}=\frac{dA}{dz}*\frac{dz}{dt}

Differentiating A=\frac{z^{2} }{2} with respect to z yields;

\frac{dA}{dz}=z

Therefore,

\frac{dA}{dt}=\frac{dA}{dz}*\frac{dz}{dt} = 8z

When z is 4m the area of the square will be increasing at a rate of;

8m/min * 4m = 32m^{2}/min

5 0
3 years ago
Solve for x:<br><br> 5(1 – x) = 20<br><br> x = __
sergey [27]

5(1-x)=20\\5-5x=20\\5x=-15\\x=-3

3 0
2 years ago
Read 2 more answers
ASAP PLEASEEEEEEEE
Tasya [4]

After solving the system of equations,  x= -3 and y=6

Step-by-step explanation:

We need to solve the following system of equations by substitution.

-3x + y = 15\,\,\,(1)\\-8x-7y = -18\,\,\,(2)

Finding value of y from equation(1) and substituting it in equation(2)

y=15+3x\\substituting\,\,in\,\,equation(2)\\-8x-7(15+3x)=-18\\Solving:\\-8x-105-21x=-18\\-8x-21x=-18+105\\-29x=87\\x=\frac{87}{-29}\\x=-3\\

The value of x is x=-3

Now, Finding value of x from eq(1) and substituting in equation(2)

-3x=15-y\\x=\frac{15-y}{-3}\\Substituting\,\,in\,\,equation(2)\\-8(\frac{15-y}{-3})-7y=-18\\\frac{120-8y}{3}-7y=-18\\\frac{120-8y-7y*3}{3}=-18\\\frac{120-8y-21y}{3}=-18\\\frac{120-29y}{3}=-18\\120-29y=-18*3\\120-29y=-54\\-29y=-54-120\\-29y=-174\\y=\frac{-174}{-29}\\y=6

So, Value of y is y=6

After solving the system of equations,  x= -3 and y=6

Keywords: system of equations

Learn more about system of equations at:

  • brainly.com/question/6075514
  • brainly.com/question/7490805
  • brainly.com/question/13168205

#learnwithBrainly

4 0
3 years ago
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