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pishuonlain [190]

3 years ago

10

If you push on a heavy box that is at rest, you must exert some force to start its motion. 5. however, once the box is in motion

you need a smaller force to maintain that motion. why?

1 answer:

Free_Kalibri [48]3 years ago

5 0

The potential energy has turned into kinetic energy and energy is not lost on the box, some is lost to friction, but most of it is still kept.

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A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the

irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

$x=V_{o} cos \theta t$ (1)

$y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2}$ (2)

Where:

$x$ is the horizontal distance between the cannon and the ball

$V_{o}=23 m/s$ is the cannonball initial velocity

$\theta=0\°$ since the cannonball was shoot horizontally

$t$ is the time

$y=0$ is the final height of the cannonball

$y_{o}=1.96 m$ is the initial height of the cannonball

$g=9.8 m/s^{2}$ is the acceleration due gravity

Isolating $t$ from (2):

$t=\sqrt{-\frac{2(y-y_{o})}{g}}$ (3)

$t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}}$ (4)

$t=0.63 s$ (5)

Substituting (5) in (1):

$x=(23 m/s) cos(0\°) 0.63 s$ (6)

Finally:

$x=14.49 m$

5 0

2 years ago

A red cross helicopter takes off from headquarters and flies 120 km at 70 degrees south of west. There it drops off some relief

fiasKO [112]

Answer:

130 km at 35.38 degrees north of east

Explanation:

Suppose the HQ is at the origin (x = 0, y = 0)

So the coordinates of the helicopter after the 1st flight is

$x_1 = -120cos70^o = -41.04 km$

$y_1 = -120sin70^o = -112.763 km$

After the 2nd flight its coordinate would be:

$x_2 = x_1 - 75sin60^o = -41.04 - 64.95 = -106km$

$y_2 = y_1 + 75cos60^o = -112.763 + 37.5 = -75.263 km$

So in order to fly back to its HQ it must fly a distance and direction of

$s = \sqrt{y_2^2 + x_2^2} = \sqrt{75.263^2 + 106^2} = \sqrt{5664.519169 + 11236} = \sqrt{16900.519169} = 130 km$

$tan\theta = \frac{y_2}{x_2} = \frac{75.263}{106} = 0.71$

$\theta = tan^{-1}0.71 = 0.62 rad \approx 35.38^o$ north of east

3 0

2 years ago

An object is launched with an initial speed of 30 m/s at an angle of 60° above the horizontal . What is the maximum height reach

matrenka [14]

Answer:

H = 34.43 m

Explanation:

Given that,

Initial speed of the object, u = 30 m/s

The angle of projection, $\theta=60^{\circ}$

We need to find the maximum height reached by the object. Let it is H. Using the formula for maximum height reached by the projectile.

$H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(30)^2\times \sin^2(60)}{2\times 9.8}\\\\H=34.43\ m$

So, the maximum height reached by the object is 34.43 m.

6 0

2 years ago

URGENT!!! Which option(s) correctly define an electric circuit? (Select all that apply) a set-up where current flows without a v

ArbitrLikvidat [17]

Answer:

a set up where current flows without a voltage difference

Explanation:

because a circuit is a set up of different components, and throughout the circuit the voltage is the same, even with more components

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3 years ago

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What does altered mean

Elina [12.6K]

Altered means to change something.

5 0

2 years ago

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