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Harlamova29_29 [7]
3 years ago
15

A 600‑kg car accelerates at a rate of 3 m/s2. How much net force is acting on the car to cause this acceleration?

Physics
1 answer:
finlep [7]3 years ago
4 0
600(kg) x 3(m/s^2) = 1800N (newtons)
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If mars has an orbital period of 1.84 years how far from the sun is it and how
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<span>141.6 million mi,and idk what u mean by how</span>
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3 years ago
The table represents the speed of a car in a northern direction over several seconds. A 2-column table with 5 rows. The first co
Oksanka [162]

Answer:

Column 1 should be titled “Time,” and Column 2 should be titled

“Velocity.” ⇒ 1st answer

Explanation:

* <em>Lets revise some definitions of graphs</em>

A velocity-time graph represents the speed and direction an object

travels over a period of time.

Velocity-time graph is called speed-time graph.

The vertical axis of a velocity-time graph represents the velocity of the

object and horizontal axis represents the time from the initial position

In an acceleration time graph y-axis represents the acceleration, and

x-axis represents the time

The table represents the <em>speed of a car in a northern</em> direction over

several seconds

That means the data in the table represent the <em>velocity</em> of the car in a

certain time

Time (s)             Velocity (m/s)  

   0                            5

   2                            10

   4                            15

   6                            20

   8                            25

   10                           30

Column 1 would be on the x-axis

x-axis represents the time

Column 2 would be on the y-axis

y-axis represents the velocity

Column 1 should be titled “Time,” and Column 2 should be titled

“Velocity.”

3 0
4 years ago
Read 2 more answers
A point charge A of charge +4micro coloumb and another B of -1 micro coloumb are placed at a distance in air 1m apart then the d
andrew11 [14]

Answer:

Explanation:

Given that,

A point charge is placed between two charges

Q1 = 4 μC

Q2 = -1 μC

Distance between the two charges is 1m

We want to find the point when the electric field will be zero.

Electric field can be calculated using

E = kQ/r²

Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.

Then, the magnitude of the electric at point x is zero.

E = kQ1 / r² + kQ2 / r²

0 = kQ1 / x²  - kQ2 / (1-x)²

kQ1 / x² = kQ2 / (1-x)²

Divide through by k

Q1 / x² = Q2 / (1-x)²

4μ / x² = 1μ / (1 - x)²

Divide through by μ

4 / x² = 1 / (1-x)²

Cross multiply

4(1-x)² = x²

4(1-2x+x²) = x²

4 - 8x + 4x² = x²

4x² - 8x + 4 - x² = 0

3x² - 8x + 4 = 0

Check attachment for solution of quadratic equation

We found that,

x = 2m or x = ⅔m

So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.

5 0
3 years ago
What is the torque τb about axis b due to the force f⃗ ? (b is the point at cartesian coordinates (0,b), located a distance b fr
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Check the attached file for the solution.

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Two objects gravitationally attract with a force of 70N. If the masses of both of the objects are increased by 15x and the dista
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(B is the answer!!
I took the test and I got a 100! Hope this helps
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3 years ago
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