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2 years ago

For every force there is one that is _ in size but ____ in direction.

Physics

2 answers:

Vaselesa [24]2 years ago

5 0

Explanation:

For every force there is one that is <em><u>equal</u></em> in size but <em><u>opposite</u></em> in direction.

Karolina [17]2 years ago

4 0

Answer:

<h3>» equal , opposite</h3>

Explanation:

For every force there is one that is <u>equal</u><u> </u>

in size but <u>opposite</u> in direction.

<h3>because , </h3>

<h3>Newton's Third law states that</h3>

<h3>For every force, there is another force of equal magnitude, acting in the opposite direction.</h3>

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A 0.20-kg particle moves along the x axis under the influence of a conservative force. The potential energy is given by

anastassius [24]

Answer:

e) 11 m/s

Explanation:

For a particle under the action of a conservative force, its mechanical energy at point 0 is must be equal to its mechanical energy at point 1:

$K_1+U_1=K_0+U_0\\\\\frac{mv_1^2}{2}+[(8.0\frac{J}{m^2})(x_1)^2+(2.0\frac{J}{m^4})(x_1)^4]=\frac{mv_0^2}{2}+[(8.0\frac{J}{m^2})(x_0)^2+(2.0\frac{J}{m^4})(x_0)^4]$

In $x_1=1.0m$ the speed is given, so $v_1=5.0\frac{m}{s}$ and $x_0=0$ . Replacing:

$\frac{mv_1^2}{2}+[(8.0\frac{J}{m^2})(1m)^2+(2.0\frac{J}{m^4})(1m)^4]=\frac{mv_0^2}{2}+[(8.0\frac{J}{m^2})0^2+(2.0\frac{J}{m^4})(0)^4]\\\frac{mv_1^2}{2}+8.0J+2.0J=\frac{mv_0^2}{2}\\\frac{mv_0^2}{2}=\frac{mv_1^2}{2}+10J\\v_0=\sqrt{\frac{2}{m}(\frac{mv_1^2}{2}+10J)}\\v_0=\sqrt{\frac{2}{0.2kg}(\frac{(0.2kg)(5\frac{m}{s})^2}{2}+10J)}\\v_0=11.18\frac{m}{s}$

7 0

3 years ago

Select the correct answer.

lawyer [7]

I believe the answer is C

5 0

3 years ago

Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person’s voice w

abruzzese [7]

Answer:

3.85*10^8m

Explanation:

We know that

Speed v = distance/time t

So distance = v x t

So = 3*10^8 x 2.56s

= distatance = 7.7*10^8m

However this is double the distance from the earth to the moon, so to get distance from earth to the moon we divide by 2

7.7*10^8/2= 3.85*10^8m

8 0

2 years ago

A box is being dragged with a horizontal force of 65 N for 12 meters. If there is a force of friction acting on it

asambeis [7]

Answer:

A. 780 J

B. 120 J

C. 660 J

Explanation:

From the question given above the following data were obtained:

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

A. Determination of the work done by the dragging force.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Workdone (Wd) by dragging force =?

Wd = Fₔ × s

Wd = 65 × 12

Wd = 780 J

Therefore, the work done by the dragging force is 780 J

B. Determination of the work done by friction.

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Workdone (Wd) by friction =?

Wd = Fբ × s

Wd = 10 × 12

Wd = 120 J

Therefore, the work done by friction is 120 J

C. Determination of the net work done on the box.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Net work done (Wd) =?

Next, we shall determine the net force acting on the box. This can be obtained as follow:

Dragging force (Fₔ) = 65 N

Force of friction (Fբ) = 10 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fբ

Fₙ = 65 – 10

Fₙ = 55 N

Thus, the net force acting on the box is 55 N

Finally, we shall determine the net work done on the box as follow:

Distance (s) = 12 m

Net force (Fₙ) = 55 N

Net work done (Wd) =?

Wd = Fₙ × s

Wd = 55 × 12

Wd = 660 J

Therefore, the net work done on the box is 660 J

4 0

2 years ago

What is point c called?

FromTheMoon [43]

Then every line segment has one and only one mid-point.

（っ＾▿＾）

5 0

2 years ago

Read 2 more answers

For every force there is one that is ___________ in size but ______________ in direction.

For every force there is one that is _ in size but ____ in direction.