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2 years ago

Acceleration Practice

Physics

2 answers:

Trava [24]2 years ago

8 0

Answer:

0.85 m/s²

Explanation:

Acceleration is change in velocity over change in time. In mathematically, it can be expressed as:

$\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{\vec v_2 - \vec v_1}{t_2-t_1}}$

Our final velocity is given to be 17 m/s in 20 seconds. Initial velocity is at starting point which is 0 m/s in 0 second. Therefore:

$\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{17-0}{20-0}}\\\\\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{17}{20}}\\\\\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = 0.85 \ \, \sf{m/s^2}}$

Therefore, the acceleration of a horse from starting point to 17 m/s in 20 seconds is 0.85 m/s²

just olya [345]2 years ago

4 0

Answer:

0.85 m/s^2

Explanation:

Acceleration = (final velocity - starting velocity) / time

Final velocity = 17

Starting velocity = 0

Time = 20

1. Substitute:

Acceleration = (starting velocity - final velocity) / time --> Acceleration = ( 17 - 0) / 20

2. Solve:

Acceleration = ( 17 - 0) / 20 -->

Acceleration = 0.85

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Answer:

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Explanation:

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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

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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.

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3 years ago

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emmainna [20.7K]

Explanation:

It is given that,

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3 years ago