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prisoha [69]
3 years ago
6

When an object is balanced about a pivot, the total clockwise moment must be equal to the total __________ __________. What two

words complete this sentence?
Physics
1 answer:
Keith_Richards [23]3 years ago
5 0

Answer:

When an object is balanced, about a pivot, the total clockwise moment must be equal to the total anticlockwise moment about that pivot.

Hope that helps.

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Match the type of heat transfer with its description
VikaD [51]

Answer:

1. Convection

2. Radiation

3. Conduction

Hope this helps!

Explanation:

5 0
3 years ago
Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 64.7 N64.7 N , Ji
WARRIOR [948]

Answer:

(a) Magnitude of force is 262.51 N

(b) Angle with East direction is -14.75^{o}

Explanation:

Force by Jack in vector form

\overrightarrow F _1} = 64.7{\rm{ N}}\left( {\hat i} \right)  

Force by Jill in Vector form is given by

\begin{array}{c}\\{\overrightarrow F _2} = 86.5{\rm{ N }}\cos {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\hat i} \right) + 86.5{\rm{ N }}\sin {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\widehat j} \right)\\\\ = 61.16{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\widehat j} \right)\\\end{array}

Force by Jane is

\begin{array}{c}\\{\overrightarrow F _3} = 181{\rm{ N }}\cos {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\hat i} \right) + 181{\rm{ N }}\sin {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( { - \widehat j} \right)\\\\ = 128{\rm{ N}}\left( {\hat i} \right) + 128{\rm{ N}}\left( { - \widehat j} \right)\\\end{array}

Net force is:

\overrightarrow F = {\overrightarrow F _1} + {\overrightarrow F _2} + {\overrightarrow F _3}

Hence

\begin{array}{c}\\\overrightarrow F = 64.7{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\widehat j} \right) + 128{\rm{ N}}\left( {\hat i} \right) + 128{\rm{ N}}\left( { - \widehat j} \right)\\\\ = 253.86{\rm{ N}}\left( {\hat i} \right) - 66.84{\rm{ N}}\left( {\widehat j} \right)\\\end{array}

The net force will be given by

F = \sqrt {{{\left( {{F_x}} \right)}^2} + {{\left( {{F_y}} \right)}^2}

Since F_{x}=253.86N and F_{y}=-66.84N

\begin{array}{c}\\F = \sqrt {{{\left( {253.86{\rm{ N}}} \right)}^2} + {{\left( { - 66.84{\rm{ N}}} \right)}^2}} \\\\ = {\bf{262.51 N}}\\\end{array}

The direction of net force is:

\theta = {\tan ^{ - 1}}\left {\frac{{{F_y}}}{{{F_x}}}}

Since F_{x}=253.86N and F_{y}=-66.84N  

\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{ - 66.84{\rm{ N}}}}{{253.86{\rm{ N}}}}} \right)\\\\ = {\tan ^{ - 1}}\left( { - 0.2633} \right)\\\\ = {\bf{ - 14}}{\bf{.7}}{{\bf{5}}^{\bf{o}}}\\\end{array}

The angle with East direction is -14.75^{o}

Net force exerted on the donkey is in the south-east direction. So, the angle of net force from the east direction is -14.75^{o} and it is 14.75^{o} from the south.

5 0
3 years ago
Угол между осями поляризатора и
Yakvenalex [24]
Nolur acil lütfen yalvarırım yalvarırım
7 0
3 years ago
a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficien
Law Incorporation [45]

Answer:

<u>The magnitude of the friction force is 8197.60 N</u>

Explanation:

Using the definition of the centripetal force we have:

\Sigma F=ma_{c}=m\frac{v^{2}}{R}

Where:

  • m is the mass of the car
  • v is the speed
  • R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

F_{f}=m\frac{v^{2}}{R}

F_{f}=2100\frac{18^{2}}{83}

F_{f}=8197.60 \: N

<u>Therefore the magnitude of the friction force is 8197.60 N</u>

I hope it helps you!

7 0
3 years ago
A 2.0   kg 2.0kg2, point, 0, start text, k, g, end text cart moving right at 5.0   m s 5.0 s m ​ 5, point, 0, start fraction, st
butalik [34]
Yes así le tienes que hacer para aprender
7 0
3 years ago
Read 2 more answers
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