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3 years ago

This illustration shows two opposing forces pulling on a wagon. Which description best describes how the wagon will move?

Physics

1 answer:

Talja [164]3 years ago

8 0

Answer:

The wagon will move to the right.

Explanation:

From the question given above, the following data were obtained:

Force applied to the left (Fₗ) = 10 N

Force applied to the right (Fᵣ) = 30 N

Direction of the wagon =.?

To determine the direction in which the wagon will move, we shall determine the net force acting on the wagon. This can be obtained as follow:

Force applied to the left (Fₗ) = 10 N

Force applied to the right (Fᵣ) = 30 N

Net force (Fₙ) =?

Fₙ = Fᵣ – Fₗ

Fₙ = 30 – 10

Fₙ = 20 N to the right

From the calculations made above, the net force acting on the wagon is 20 N to the right. Hence the wagon will move to the right.

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3 years ago

How much work must be done to bring three electrons from a great distance apart to 5.0×10^−10 m from one another (at the corners

Inessa05 [86]

Answer:

1.38 x 10^-18 J

Explanation:

q = - 1.6 x 10^-19 C

d = 5 x 10^-10 m

the potential energy of the system gives the value of work done

The formula for the potential energy is given by

$U =\frac{Kq_{1}q_{2}}{d}$

So, the total potential energy of teh system is

$U =\frac{Kq_{1}q_{2}}{d}+\frac{Kq_{2}q_{3}}{d}+\frac{Kq_{1}q_{3}}{d}$

As all the charges are same and the distance between the two charges is same so the total potential energy becomes

$U =3\times \frac{Kq^{2}}{d}$

K = 9 x 10^9 Nm^2/C^2

By substituting the values

$U =3\times \frac{9\times 10^{9}\times \ 1.6 \times 1.6 \times 10^{-38}}{5\times 10^{-10}}$

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4 years ago

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klasskru [66]

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3 years ago

31. If you threw a baseball straight out at 45 m/s from a height of 1.5 meters (A) how long would it be in the air? B) How far o

coldgirl [10]

Answer:

A) t = 0.55 s

B) x = 24.8 m

Explanation:

A) We can find the time at which the ball will be in the air using the following equation:

$y_{f} = y_{0} + v_{0y}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ is the final height= 0

$y_{0}$ is the initial height= 1.5 m

$v_{0y}$ is the component of the initial speed in the vertical direction = 0 m/s

t: is the time =?

g: is the gravity = 9.81 m/s²

$0 = 1.5 m - \frac{1}{2}9.81 m/s^{2}t^{2}$

By solving the above equation for t we have:

$t = \sqrt{\frac{2*1.5 m}{9.81 m/s^{2}}} = 0.55 s$

Hence, the ball will stay 0.55 seconds in the air.

B) We can find the distance traveled by the ball as follows:

$x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}$

Where:

a: is the acceleration in the horizontal direction = 0

$x_{f}$ is the final position =?

$x_{0}$ is the initial position = 0

$v_{0x}$ is the component of the initial speed in the horizontal direction = 45 m/s

$x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}$

$x_{f} = 0 + 45 m/s*0.55 s + 0 = 24.8 m$

Therefore, the ball will travel 24.8 meters.

I hope it helps you!

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