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irina1246 [14]
3 years ago
13

Answer these questions please and you will get the brain list

Physics
1 answer:
Naddika [18.5K]3 years ago
7 0

I am using the equation F=ma (force equals mass times acceleration) to solve these problems.

1. You are looking for force, and have mass and acceleration. You just plug in the values for mass and acceleration to get the force needed.

F=(15kg)(5m/s^2)

F=75N

2. Again, you are looking for force, and just need to plug in the values for mass and acceleration

F=(3kg)(2.4m/s^2)

F=7.2N

3. In this problem, you have force and mass, but need to find acceleration. To do this, you need to get acceleration alone on one side of the equation - divide each side by m. Your equation will now be F/m=a

a=(5N)/(3.7kg)

a=18.5m/s^2

I did not use significant figures. Let me know if you need to do that and need any help on that. Hope this helps!

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Burning oil and coal adds to the atmosphere.
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Answer:

carbon dioxide

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What do you call any two colors of light that combine to form white light?
Sever21 [200]
Hello there!


We called that complementary colors.

As always, it is my pleasure to help students like you!

4 0
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Which strategy should you use if your research question is too broad for the scope of your project? (1 O narrow the focus of res
Nesterboy [21]

Answer:

"Narrow the focus of research question"

Explanation:

O Narrow the focus of research question

    This is good! You can still use your question, but focus in on something so you have a proper research project.

O Add another research question

    Would adding another question to an already broad question help? No.

O Use the very first source you find for your project

    If your question is too broad, you should not use whatever you see first as it may be incorrect or does not answer the question

O Change the scope of your project​    

    You could, but if you have a set scope for your project (a) you might not be able to change it (b) you don't need to restart

Have a nice day!

    I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

3 0
2 years ago
A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor sto
Fed [463]

Answer:

a)   a = 34.375 m / s²,  b)    v_f = 550 m / s

Explanation:

This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.

a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed

             v_f = \frac{x-x_1}{t}

we substitute the values

             v_f = \frac{ 6600 -x_1}{4}  

The initial part of the movement is carried out with acceleration

             v_f = v₀ + a t

             x₁ = x₀ + v₀ t + ½ a t²

the rocket starts from rest v₀ = 0 with an initial height x₀ = 0

             x₁ = ½ a t²

             v_f = a t

we substitute the values

              x₁ = 1/2  a 16²

              x₁ = 128 a

              v_f = 16 a

let's write our system of equations  

               v_f = \frac{6600 - x_1}{4}

               x₁ = 128 a

               v_f = 16 a

we substitute in the first equation  

               16 a = \frac{6600 -128 a}{4}

               16 4 a = 6600 - 128 a

                a (64 + 128) = 6600

                a = 6600/192

                 a = 34.375 m / s²

b) let's find the time to reach this height

                x = ½ to t²

                t² = 2y / a

                t² = 2 5100 / 34.375

                t² = 296.72

                t = 17.2 s

We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part

               v_f = 16 a

               v_f = 16 34.375

               v_f = 550 m / s

8 0
3 years ago
Find the relation, time velocity graph : acceleration of an object, time position graph : ?
Anit [1.1K]

Answer:

In a time-position graph (s-t graph):

slope = velocity

In a time-velocity graph (v-t graph):

slope = acceleration

area under graph = change in displacement (distance travelled)

In a time-acceleration graph (a-t graph):

area under graph = change in velocity

8 0
3 years ago
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