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4 years ago

12

Through which type of medium does sound travel fastest:: solid, liquid, or gas? Through which type of a medium does light travel

fastest: solid liquid, or gas?

2 answers:

valina [46]4 years ago

7 0

Answer:

Solids;Gases

Explanation:

SOVA2 [1]4 years ago

6 0

Sound travels fastest through solids, while light travels fastest through gases.

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an athlete sprints from 150 m south of the finish line to 65 m south of the finish line in 5.0s what is his average velocity

MatroZZZ [7]

Total displacement=150-65m=85m
Time=5s

$\\ \rm\longrightarrow Avg\:Velocity=\dfrac{Total\:Displacement}{Total\:Time}$

$\\ \rm\longrightarrow Avg\:Velocity=\dfrac{85}{5}$

$\\ \rm\longrightarrow Avg\:Velocity= 17m/s$

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3 years ago

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When an object with a height of 0.10 meter is placed at a distance of 0.20 meter from a convex

seraphim [82]

Answer:

h = 0.1 m = 10 cm

u = 0.2 m = 20 cm

v = 0.06 m = 6 cm

m = h'/h = v/u

h'/10 = 6/20

h' = 60/20 = 3 cm

Therefore image height = 3 cm(0.03 m)

Hope this helps!

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3 years ago

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two forces have the same magnitude F, what is the angle between the two vectors if their sum has a magnitude of (a) 2F? (b) sqrt

kramer

Ok let me help you with this:
<span>
!n case a) Line both up head-to-tail in a straight line... that's the only way to get a sum 2F. In case b) sqrt(2) is the length of the hypotenuse in a 45 degree triangle, so the vectors must be at 90 degrees to each other in the case of c) lined up head to tail, the only way to get 0 is if they point in opposite direction</span>

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4 years ago

A uniform cylindrical grinding wheel of mass 50.0 kg and diameter 1.0 m is turned on by an electric motor. The friction in the b

aliya0001 [1]

Answer:

The torque must be applied to the wheel is 15.7 N-m.

Explanation:

Given that,

Mass of the wheel of cylinder, m = 50 kg

Diameter of the wheel, d = 1 m

Radius, r = 0.5 m

Initial speed off the wheel is 0

Final angular speed of the wheel, 120 rev/min = 12.56 rad/s

Angular displacement, $\theta=20\ rev=125.6\ rad$

The torque is given by :

$\tau=I\alpha \\\\\tau=\dfrac{mr^2}{2}\times (\dfrac{\omega_f^2}{2\theta})\\\\\tau=\dfrac{50\times1^{2}}{2}\times(\dfrac{12.56^{2}}{2\times125.6})\\\\\tau=15.7\ N-m$

So, the torque must be applied to the wheel is 15.7 N-m.

4 0

3 years ago

Please ans both urgent

Tpy6a [65]

Answer:

d I guess not sure

b if down I guess

Explanation:

if worng then plz dont say anything as I am not sure about the ans

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3 years ago

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