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3 years ago

If a cart of 10 kg mass has a force of 5 newtons exerted on it, what is its acceleration? m/s2

Physics

2 answers:

kherson [118]3 years ago

6 0

Answer:

.5 m/s2

Explanation:

Elza [17]3 years ago

4 0

Answer:

= 0.5 m/s²

Explanation:

According to Newton's second law of motion, the resultant force is directly proportion to the rate of change of linear momentum.

Therefore; F = ma , where F is the Force, m is the mass and a is the acceleration.

Thus; a = F/m

but; F = 5 N, and m = 10 kg

 a = 5 /10

= 0.5 m/s²

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3 years ago

If the density of the fuel is 0.821 g/ml, what mass of fuel has the airplane taken on?

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Mass of fuel has the airplane took on = 167.89 kg

$\boxed{\boxed{\bold{Further~explanation}}}$

Density is a quantity derived from the mass and volume

density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than objects with a smaller type of mass

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7 0

3 years ago

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A force of 20 N acts on a 4 kg cart for 10 s . How far will it go (using the second law) ?

Elden [556K]

This took me a short while to figure out, but I am still not entirely sure if this is correct, this is just from my basic understanding of Newtons Second Law of Motion.

You have a 4kg cart with a force of 20N acting on it.

The formula for working out the acceleration is.
a=Fnet÷mass

Substitute in the information.
a=20N÷4kg

Now you solve it to give you.
a=5m/s

So now what you should be able to do is figure out that after 10 seconds the cart travelling at 5m/s would have travelled 10 metres.

This is achieved by finding out how many 5's go into 10 which is 2.
So you do 5×2 which equals 10.

The 4kg cart has travelled 10 meters in 10 seconds with a force of 20N acting upon it.

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3 years ago

Two forces P and Q act on an object of mass 15.0 kg with Q being the larger of the two forces. When both forces are directed to

Rom4ik [11]

Answer:

P=6.25N and Q=16.25N

Explanation:

In order to solve this problem we must first draw a free body diagram for both situation, (see attached picture).

Now, we need to analyze the two free body diagrams. So let's analyze the first diagram. Since the body is accelerated, then the sum of forces is equal to mass times acceleration, so we get:

$\Sigma F=ma$

We can assume there will be only the two mentioned forces P and Q, so

the sum of forces will be:

P+Q=ma

$P+Q=(15kg)(1.50m/s^{2})$

P+Q=22.5N

We can do the same analysis for the second free body diagram:

$\Sigma F=ma$

$Q-P=(15kg)(0.7m/s^{2})$

Q-P=10.5N

so now we have a system of equations we can solve by elimination:

Q+P=22.5N

Q-P=10.5N

Now, we can add the two equations together so the P force is eliminated, so we get:

2Q=32.5N

now we can solve for Q:

$Q=\frac{32.5N}{2}$

Q=16.25N

Now we can use any of the equations to find P.

Q+P=22.5N

P=22.5N-Q

when substituting for Q we get:

P=22.5N-16.25N

P=6.25N

5 0

4 years ago