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3 years ago

How much power does it take to lift a 30.0 n box 10.0 m high in 5.00 s, if you must apply a 62n force to lift the box?

1 answer:

5 0

Power is defined as the rate at which the body is doing work:
$P=\frac{W}{t}$
Work is defined as displacement done by the force times that displacement:
$W=F\cdot h$
We know that we need 62N to move the box, so when we apply this force along the path of 10m we have done:
$W=62N\cdot10m=620J$
of work.
Now we just divide that by 5s to get how much power is required:
$P=\frac{620J}{5s}=124W$

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a football is thrown upward at a 31° angle to the horizontal.the acceleration of gravity is 9.8m/s^2. To throw the ball a distan

mr Goodwill [35]

The ball's vertical position $y$ in the air at time $t$ is

$y=v_0\sin31^\circ\,t-\dfrac g2t^2$

The ball is at its original height when $y=0$ , which happens at

$v_0\sin31^\circ\,t-\dfrac g2t^2=\dfrac t2\left(2v_0\sin31^\circ-gt)=0$

$\implies t=0\text{ and }t=\dfrac{2v_0\sin31^\circ}g$

Meanwhile, the ball's horizontal position $x$ at time $t$ is

$x=v_0\cos31^circ\,t$

So when the ball reaches its original height a second time, the ball will have traveled a horizontal distance of

$x=\dfrac{2{v_0}^2\sin31^\circ\cos31^\circ}g=\dfrac{{v_0}^2\sin(2\cdot31^\circ)}g$

(which you might recognize as the formula for the range of a projectile)

To reach a distance of $x=77\,\rm m$ , the initial speed $v_0$ would be

$77\,\mathrm m=\dfrac{{v_0}^2\sin62^\circ}{9.8\,\frac{\rm m}{\mathrm s^2}}\implies v_0=29\dfrac{\rm m}{\rm s}$

7 0

3 years ago

Suppose that it takes 0.6 seconds for a mass on a spring to move from its highest position to its lowest position. What is the p

o-na [289]

The mass on the spring is bouncing.

We would call it a wave-like motion, except that it all stays in the same place. But, just like a wave, moving from the highest position to the lowest position
is one-half of a full wiggle.

(The other half consists of moving from the lowest position back up to the
highest position, where it started from.

So, half of the wave-like motion takes 0.6 seconds.

A full cycle of the wave motion ... the actual period of the bounce,
is double that much time . . .
1.2 seconds.

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4 years ago

All are examples of speed except ______

katovenus [111]

Speed is a derived term that is the result when distance or displacement is divided by unit time. Speed is a scalar unit which means it only includes magnitude not direction, thus always positive. The answer thus are the three choices given above.

4 0

4 years ago

The perihelion of the comet TOTAS is 1.69 AU and the aphelion is 4.40 AU. Given that its speed at perihelion is 28 km/s, what is

dybincka [34]

Answer:

The speed at the aphelion is 10.75 km/s.

Explanation:

The angular momentum is defined as:

$L = mrv$ (1)

Since there is no torque acting on the system, it can be expressed in the following way:

$t = \frac{\Delta L}{\Delta t}$

$t \Delta t = \Delta L$

$\Delta L = 0$

$L_{a} - L_{p} = 0$

$L_{a} = L_{p}$ (2)

Replacing equation 1 in equation 2 it is gotten:

$mr_{a}v_{a} =mr_{p}v_{p}$ (3)

Where m is the mass of the comet, $r_{a}$ is the orbital radius at the aphelion, $v_{a}$ is the speed at the aphelion, $r_{p}$ is the orbital radius at the perihelion and $v_{p}$ is the speed at the perihelion.

From equation 3 v_{a} will be isolated:

$v_{a} = \frac{mr_{p}v_{p}}{mr_{a}}$

$v_{a} = \frac{r_{p}v_{p}}{r_{a}}$ (4)

Before replacing all the values in equation 4 it is necessary to express the orbital radius for the perihelion and the aphelion from AU (astronomical units) to meters, and then from meters to kilometers:

$r_{p} = 1.69 AU x \frac{1.496x10^{11} m}{1 AU}$ ⇒ $2.528x10^{11} m$

$r_{p} = 2.528x10^{11} m x \frac{1km}{1000m}$ ⇒ $252800000 km$

$r_{a} = 4.40 AU x \frac{1.496x10^{11} m}{1 AU}$ ⇒ $6.582x10^{11} m$

$r_{p} = 6.582x10^{11} m x \frac{1km}{1000m}$ ⇒ $658200000 km$

Then, finally equation 4 can be used:

$v_{a} = \frac{(252800000 km)(28 km/s)}{(658200000 km)}$

$v_{a} = 10.75 km/s$

Hence, the speed at the aphelion is 10.75 km/s.

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3 years ago

A steam engine absorbs 1.98 × 105 j and expels 1.49 × 105 j in each cycle. assume that all of the remaining energy is used to do

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B) The heat absorbed by the engine is $Q_{in} = 1.98 \cdot 10^5 J$ while the heat expelled is $Q_{out} = 1.49 \cdot 10^5 J$ , therefore the work done by the engine is the difference between the heat absorbed and the heat expelled:
$W=Q_{in} - Q_{out} = 1.98 \cdot 10^5 J - 1.49 \cdot 10^5 J = 0.49 \cdot 10^5 J$

a) The efficiency of the engine is the ratio between the work done by the engine and the heat absorbed, therefore:
$\eta= \frac{W}{Q_{in}} = \frac{0.49 \cdot 10^5 J}{1.98 \cdot 10^5 J}=0.247 = 24.7\%$

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4 years ago