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3 years ago

10

The word “virtual” refers to something that exists in effect but not in actual fact. How does this definition relate to the virt

ual image you see of yourself in a plane mirror?

2 answers:

atroni [7]3 years ago

8 0

Because you see yourself the opposite way in a mirror. So yes your “seeing” yourself but not how everyone else sees you.

givi [52]3 years ago

5 0

Answer:

The image seems to be behind the mirror, but nothing is really there.

Explanation:

Sample response edge 2021

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•• Your roommate is working on his bicycle and has the bike upside down. He spins the 60-cm-diameter wheel, and you notice that

Y_Kistochka [10]

Answer:

$v=0.57\frac{m}{s}$

$a_c=10.83\frac{m}{s^2}$

Explanation:

We have an uniform circular motion, therefore, the pebble’s speed is given by the distance traveled in a revolution $(2\pi r)$ and the period (T), since this is the time pebble’s takes to complete a revolution:

$v=\frac{2\pi r}{T}$

The period is inversely proportional to the frequency:

$T=\frac{1}{f}$

So, we have:

$v=\frac{2\pi r}{\frac{1}{f}}\\v=2\pi rf\\$

Recall that the radius is the half of the diameter and one revolution per is equal to one Hz:

$v=2\pi (30*10^{-2}m)(3Hz)\\v=0.57\frac{m}{s}$

The centripetal acceleration is defined as:

$a_c=\frac{v^2}{r}\\a_c=\frac{(0.57\frac{m}{s})^2}{30*10^{-2}m}\\\\a_c=10.83\frac{m}{s^2}$

6 0

3 years ago

Isla made ice by putting water in the freezer. Freezing is an example of (2 points)

SSSSS [86.1K]

Answer:

a physical change

6 0

3 years ago

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PLEASE HELP, evaporation,liquids,solids, and gases involved

sukhopar [10]

Yes they are involved

5 0

3 years ago

in a lab, resonance tubes are used to determine experimentally the speed of sound. using the data given, evaluate the best appro

jek_recluse [69]

The approximate speed of sound is about 761 mph or 335.28 meters/second.

<h3>What's the approximate speed of sound?</h3>

If we consider the normal atmosphere at sea level, the speed of sound is about 761 mph or 335.28 meters/second. The speed of sound is different in different mediums such as solid and liquid.

So we can conclude that the approximate speed of sound is about 761 mph or 335.28 meters/second.

Learn more about speed here: brainly.com/question/4931057

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2 years ago

A 2000-kg truck is being used to lift a 400-kg boulder B that is on a 50-kg pallet A. Knowing the acceleration of the rear-wheel

anzhelika [568]

Answer:

(a) reaction at each front wheel is 5272N (upward)

(b) force between boulder and pallet is 4124N (compression)

Explanation:

Acceleration of the truck $a_{t$ = 1 m/ $s^{2}$ (to the left)

when the truck moves 1 m to the left, the boulder is B and pallet A are raised 0.5 m, then,

$a_{A}$ = 0.5 m/ $s^{2}$ (upward) , $a_{B}$ = 0.5 m/ $s^{2}$ (upward)

Let T be tension in the cable

pallet and boulder: ∑fy = ∑(fy)eff = 2T- ( $m_{A}$ + $m_{B}$ )g = ( $m_{A}$ + $m_{B}$ ) $a_{B}$

= 2T- (400 + 50)*(9.81 m/ $s^{2}$ ) = (400 + 50)*(0.5 m/ $s^{2}$ )

T = 2320N

Truck: $M_{R}$ = ∑( $M_{R}$ )eff: = $-N_{f}$ (3.4m) + $m_{T}$ (2.0m) - T (0.6m)= $m_{T} a_{T}$ (1.0m)

Nf = (2.0m)(2000 kg)(9.81 m/ $s^{2}$ )/3.4m - (0.6 m)(2320 N)/3.4m + (1.0 m)(2000 kg)(1.0 m/ $s^{2}$ ) = 11541.2N - 409.4N - 588.2N = 10544N

∑fy (upward) = ∑(fy)eff: $N_{f}$ + $N_{R}$ - $m_{T}$ g = 0

10544 + $N_{R}$ - (2000kg)(9.81 m/ $s^{2}$ ) = 0

$N_{R}$ = 9076N

∑fx (to the left) = ∑(fx)eff: $F_{R}$ - T = $m_{T} a_{T}$

$F_{R}$ = 2320N + (2000kg)(9.81 m/ $s^{2}$ ) = 4320N

(a) reaction at each front wheel:

1/2 $N_{f}$ (upward): 1/2 (10544N) = 5272N (upward)

(b) force between boulder and pallet:

∑fy (upward) = ∑(fy)eff: $N_{B}$ + $M_{B}$ g - $m_{B}$ $a_{B}$

$N_{B}$ = (400kg)(9.81 m/ $s^{2}$ ) + (400kg)(0.5 m/ $s^{2}$ ) = 4124N (compression)

3 0

4 years ago

Read 2 more answers