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3 years ago

5

A hammer slides down a roof that makes a 30.0°angle with the horizontal. What are the magnitudes of the components of the hammer

's velocity at the edge of the roof if it is moving at a speed of 8.25 m/s

1 answer:

dlinn [17]3 years ago

7 0

<h2>Answer:7.14 $ms^{-1}$ ,4.125 $ms^{-1}$ </h2>

Explanation:

Whenever an object is moving in a 2D frame,its motion can be analysed as if it is travelling in two independent 1D frames.

One of such independent 1D frames are along horizontal and another along vertical.

Let $v$ be the total velocity.

Given that, $v=8.25ms^{-1}$

We call the horizontal velocity as $v_{h}$ and the vertical velocity as $v_{v}$ .

$v_{h}$ = $vCos\alpha$

$v_{v}=vSin\alpha$

where $\alpha$ is the angle between the object and horizontal.

It is given that $\alpha =30^{0}$

$v_{h}=8.25\times Cos(30^{0})=7.14ms^{-1}$

$v_{v}=8.25\times Sin(30^{0})=4.125ms^{-1}$

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A bolt of lightning discharges 9.7 C in 8.9 x 10^-5 s. What is the average current during the discharge?

Anastaziya [24]

Answer: $1.089\times 10^5\ A$

Explanation:

Given

Charge discharged $Q=9.7\C$

time taken $t=8.9\times 10^{-5}\ s$

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$\Rightarrow I=\dfrac{Q}{t}\\\\\Rightarrow I=\dfrac{9.7}{8.9\times 10^{-5}}\\\\\Rightarrow I=1.089\times 10^5\ A$

Therefore, the average current is $1.089\times 10^5\ A$

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2 years ago

A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,

kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

$\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}$

so that the <em>x</em>-component of $\vec v_{\rm ave}$ is

$\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}$

and its <em>y</em>-component is

$\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}$

Solve for $x_2$ and $y_2$ , which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

$x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}$

$y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}$

Note that I'm reading the given details as

$x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}$

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

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3 years ago

A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 5100 MJ with respect to Eart

serious [3.7K]

To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,

$E_{initial} = E_{final}$

$PE_{initial}+KE_{initial} = PE_{final}+KE_{final}$

Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that

$KE_{final} = (PE_{initial}+KE_{initial} )-PE_{final}$

$KE_{final} = (5100+4200)-5700$

$KE_{final} = 3600MJ$

Therefore the final kinetic energy is 3600MJ

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3 years ago

A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t

Evgen [1.6K]

Answers:

a) $65.075 kgm/s$

b) $10.526 s$

c) $61.82 N$

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

$I=\Delta p=p_{2}-p_{1}$ (1)

Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

<h3>b) Time </h3>

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :

$V_{2}=V_{1}+at$ (6)

$V_{2}^{2}=V_{1}^{2}+2ad$ (7)

Where:

$a$ is the acceleration

$d=0.01 m$ is the length the ball was compressed

$t$ is the time

Finding $a$ from (7):

$a=\frac{V_{2}^{2}-V_{1}^{2}}{2d}$ (8)

$a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)}$ (9)

$a=90.25 m/s^{2}$ (10)

Substituting (10) in (6):

$57 m/s=-38 m/s+(90.25 m/s^{2})t$ (11)

Finding $t$ :

$t=1.052 s$ (12)

<h3>c) Force applied to the ball by the bat </h3>

According to Newton's second law of motion, the force $F$ is proportional to the variation of momentum $\Delta p$ in time $\Delta t$ :

$F=\frac{\Delta p}{\Delta t}$ (13)

$F=\frac{65.075 kgm/s}{1.052 s}$ (14)

Finally:

$F=61.82 N$

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3 years ago

Read 2 more answers

A person carries a plank of wood 2 m long with one hand pushing down on it at one end with a force F1 and the other hand holding

sveta [45]

Answer:

F1= 588 N

F2= 784 N

Explanation:

Please see the attached file.

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3 years ago