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3 years ago

8

How is this possible?

1 answer:

stellarik [79]3 years ago

6 0

Explanation:

this is how I understand it!!

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The wavelength of light visible to the human eye is on the order of 5 x 10-7 m. if the speed of light in air is 3 x 108 m/s, fin

juin [17]

The frequency of light wave is 0.6*10^15 Hz.

The relation between the speed of light, frequency, and wavelength is given as

c=λν

Plugging the values in the above equation

3*10^8=5*10^(-7)*ν

ν=0.6*10^15 Hz.

Therefore the frequency of the light wave is 0.6*10^15 Hz.

6 0

4 years ago

I need help on this pls

Alexandra [31]

1. First answer
2. Troposphere
3. Rotation
4. Humidity
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5 0

3 years ago

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A block is given a very brief push up a 20.0 degree frictionless incline to give it an initial speed of 12.0 m/s.(a) How far alo

Orlov [11]

Explanation:

(a) Net force acting on the block is as follows.

$F_{net} = -mg Sin (\theta)$

or, ma = -mg Sin (\theta)[/tex]

a = $-g Sin (\theta)$

= $-9.8 \times Sin (20^{o})$

= -3.35 $m/s^{2}$

According to the kinematic equation of motion,

$v^{2} - v^{2}_{o} = 2as$

Distance traveled by the block before stopping is as follows.

s = $\frac{v^{2} - v^{2}_{o}}{2a}$

= $\frac{(0)^{2} - (12.0)^{2}_{o}}{2 \times -3.35}$

= 21.5 m

According to the kinematic equation of motion,

v = $v_{o} + at$

0 = $12.0 m/s + \frac{1}{2} \times -3.35 m/s^{2} \times t$

$t_{1}$ = 7.16 sec

Therefore, before coming to rest the surface of the plane will slide the box till 7.16 sec.

(b) When the block is moving down the inline then net force acting on the block is as follows.

$F_{net} = -mg Sin (\theta)$

ma = $mg Sin (\theta)$

a = $g Sin (\theta)$

= $9.8 m/s^{2} \times Sin (20^{o})$

= 3.35 $m/s^{2}$

Kinematics equation of the motion is as follows.

s = $v_{o}t + \frac{1}{2}at^{2}$

21.5 m = $0 + \frac{1}{2} \times 3.35 m/s^{2} \times t^{2}$

$t_{2}$ = $\sqrt{\frac{2 \times 21.5 m}{3.35 m/s^{2}}}$

= 3.58 sec

Hence, total time taken by the block to return to its starting position is as follows.

t = $t_{1} + t_{2}$

= 7.16 sec + 3.58 sec

= 10.7 sec

Thus, we can conclude that 10.7 sec time it take to return to its starting position.

3 0

4 years ago

You are on a train that is traveling at 3.0 m/s along a level straight track. Very near and parallel to the track is a wall that

loris [4]

Questions Diagram is attached below

Answer:

$T=2.08s$

Explanation:

From the question we are told that:

Speed of Train $V=3.0m.s$

Angle $\theta=12\textdegree$

Height of window $h_w=0.90m$

Width of window $w_w=2.0m$

The Horizontal distance between B and A from Trigonometric Laws is mathematically given by

$b=\frac{0.9}{tan12}$

$b=4.23$

Therefore

Distance from A-A

$d_a=2.0+4.23$

$d_a=6.23$

Therefore

Time Required to travel trough d is mathematically given as

$T=\frac{d_a}{v}$

$T=\frac{6.23}{3}$

$T=2.08s$

5 0

3 years ago

An imaginary line perpendicular to a reflecting surface is called _________.

n200080 [17]

<span>An imaginary line perpendicular to a reflecting surface is called "a normal" (principle line)

So, Your Answer would be Option B

Hope this helps!</span>

5 0

3 years ago

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