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4 years ago

10

An archer defending a castle is on a 15.5 m high wall. He shoots an arrow straight down at 22.8 m/s. How much time does it take

to reach the ground?

1 answer:

vazorg [7]4 years ago

3 0

Answer: 1.907

Explanation:

I did the math

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3. A 0.35 kg puck slides across the ice with an average force of friction of 0.15 N acting on it. It slides 82 m before coming t

sattari [20]

Answer:

Work done is 12.3 J

Explanation:

We have,

Mass of puck, m = 0.35 kg

Force of friction acting on the puck when it slides is 0.15 N

Distance travelled by the puck is 82 m.

It is required to find the work done on the puck. Finally the puck comes to rest and the force of friction is acting on it. It means the applied force is 0.15 N. Work done is given by

$W=Fd\\\\W=0.15\times 82\\\\W=12.3\ J$

The work done on the puck is 12.3 J.

5 0

4 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

A(n) ________________________ is a molecule that is soluble in both polar and nonpolar solvents.

Alex

Answer: ethanol which will dissolve in both water (a polar solvent) and gasoline (a non-polar solvent)

Explanation:

3 0

3 years ago

Pedro is building a model motor. He would like the motor to be more powerful. Which of the following would best help him increas

Ratling [72]

The answer is B because that that will make it more powerful but less lasting.

7 0

3 years ago

A telephone pole has three cables pulling as shown from above, with F⃗ 1=(300.0iˆ+500.0jˆ) , F⃗ 2=−200.0iˆ , and F⃗ 3=−800.0jˆ .

Ray Of Light [21]

A) Net force in component form: $F=100.0i-300.0j$

B) Magnitude of the net force: 316.2, direction: $-71.6^{\circ}$

Explanation:

A)

The three forces given in this problem are:

$F_1=300i+500j$

$F_2=-200i$

$F_3=-800 j$

The three forces are given in component form, where the components with unit vector i is the component along the x-direction, while the components with unit vector j is the component along the y-direction.

In order to find the net force in component form, we just need to add the components of the three forces along each direction. Therefore:

- Along the x-direction:

$F_x = F_{1x}+F_{2x}+F_{3x}=300+(-200)+0=100$

- Along the y-direction:

$F_y=F_{1y}+F_{2y}+F_{3y}=500+0+(-800)=-300$

So, the net force in component form is

$F=100.0i-300.0j$

B)

The magnitude of a vector F is given by Pythagorean's theorem:

$|F|=\sqrt{F_x^2+F_y^2}$

where in this problem,

$F_x=100$ is the x-component

$F_y=-300$ is the y-component

Substituting,

$|F|=\sqrt{(100)^2+(-300)^2}=316.2$

The direction instead is given by

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{-300}{100})=-71.6^{\circ}$

where the negative sign means the direction is below the positive x-axis.

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

8 0

3 years ago