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rjkz [21]
2 years ago
15

What net force is required to accelerate a car at a rate of 2m/s^2 if the car has a mass of 3,000kg? plus explanation with equat

ions? I want to know how to get to the answers.
I've already seen the same question answered, but I want to be able to understand how to get the answer, please.
Physics
1 answer:
Vinvika [58]2 years ago
3 0

Answer:

<h2>6000 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

mass = 3000 kg

acceleration = 2 m/s²

We have

force = 3000 × 2 = 6000

We have the final answer as

<h3>6000 N</h3>

Hope this helps you

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3 years ago
Which type of reaction is this?<br> CH3COOH + H20 2 H+ + CH3C00
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3 years ago
A square is 1.0 m on a side. Point charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners ar
finlep [7]

Answer:

<em>B) 1.0 × 10^5 V</em>

Explanation:

<u>Electric Potential Due To Point Charges </u>

The electric potential produced from a point charge Q at a distance r from the charge is

\displaystyle V=k\frac{Q}{r}

The total electric potential for a system of point charges is equal to the sum of their individual potentials. This is a scalar sum, so direction is not relevant.

We must compute the total electric potential in the center of the square. We need to know the distance from all the corners to the center. The diagonal of the square is

d=\sqrt2 a

where a is the length of the side.

The distance from any corner to the center is half the diagonal, thus

\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}

\displaystyle r=\frac{1}{\sqrt{2}}=0.707\ m

The total potential is  

V_t=V_1+V_2+V_3+V_4

Where V1 and V2 are produced by the +4\mu C charges and V3 and V4 are produced by the two opposite charges of \pm 3\mu\ C. Since all the distances are equal, and the charges producing V3 and V4 are opposite, V3 and V4 cancel each other. We only need to compute V1 or V2, since they are equal, but they won't cancel.

\displaystyle V_1=V_2=k\frac{Q}{r}=9\times 10^9 \frac{4\times 10^{-6}}{0.707}

V_1=V_2=50912\ V

The total potential is

V_t=50912\ V+50912\ V=1\times 10^5\ V

\boxed{V_t=1\times 10^5\ V}

6 0
3 years ago
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