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Burka [1]
3 years ago
15

A 65-N box rest on a perfectly smooth surface. The minimum force needed to start the box moving is

Physics
1 answer:
Orlov [11]3 years ago
7 0

when you say "perfectly smooth", I understand that to mean there's no friction.  If that's true, then ANY net horizontal force acting on the box makes it start moving. (D)  Naturally, the greater the force is, the faster the box speeds up.  That's because  F = m A .

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The illustration represents one form of _________________, the process that enables all stars, including our sun, to continuousl
jonny [76]

We have no way to say what the illustration represents, mainly because
you haven't given us a way to see the illustration.

<span>However, the process that all stars, including our sun, use to continuously
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8 0
3 years ago
Read 2 more answers
A softball is thrown from the origin of an x-y coordinate system with an initial speed of 18 m/s at an angle of 35∘ above the ho
podryga [215]
<span>What we need to first do is split the ball's velocity into vertical and horizontal components. To do that multiply by the sin or cos depending upon if you're looking for the horizontal or vertical component. If you're uncertain as to which is which, look at the angle in relationship to 45 degrees. If the angle is less than 45 degrees, the larger value will be the horizontal speed, if the angle is greater than 45 degrees, the larger value will be the vertical speed. So let's calculate the velocities sin(35)*18 m/s = 0.573576436 * 18 m/s = 10.32437585 m/s cos(35)*18 m/s = 0.819152044 * 18 m/s = 14.7447368 m/s Since our angle is less than 45 degrees, the higher velocity is our horizontal velocity which is 14.7447368 m/s. To get the x positions for each moment in time, simply multiply the time by the horizontal speed. So 0.50 s * 14.7447368 m/s = 7.372368399 m 1.00 s * 14.7447368 m/s = 14.7447368 m 1.50 s * 14.7447368 m/s = 22.1171052 m 2.00 s * 14.7447368 m/s = 29.48947359 m Rounding the results to 1 decimal place gives 0.50 s = 7.4 m 1.00 s = 14.7 m 1.50 s = 22.1 m 2.00 s = 29.5 m</span>
6 0
3 years ago
. If you live in a region that has a particular TV station, you can sometimes pick up some of its audio portion on your FM radio
Reil [10]

Answer:

Please see below as the answer is self-explanatory.

Explanation:

The low band of the VHF TV Spectrum, spans channels 2-6, from 54 to 88 Mhz.

In the analog TV, in the Americas, the total bandwidth of any channel is 6 Mhz, with the visual carrier modulated in VSS (Vestigial Side Band) at 1.25 Mhz from the lowest frequency of the channel.

The aural carrier is located at 4.5 Mhz from the visual carrier, and is FM modulated.

For Channel 6, which spans between 82 and  88 Mhz, the visual carrier is at 83.25 Mhz, so the aural carrier is at 87.75 Mhz, which falls within the FM Band, so it is possible to listen the audio part of this channel in a FM radio receiver, even at a lower volume, due to the FM radio has a greater deviation than TV aural carrier.

6 0
3 years ago
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padilas [110]

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8 0
3 years ago
Identify each process labeled in the diagram
marishachu [46]
I am going to need a picture for this question
3 0
3 years ago
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