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MArishka [77]
2 years ago
5

John rode 3,150 m at an average speed of 350 m/min. If he

Physics
1 answer:
Vikentia [17]2 years ago
6 0

Answer:

mabye 8

Explanation:

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You haven't said how much power the stereo uses.  It matters !

Whatever that number is, the maximum hours per month is

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Which metal in Period 5 is very reactive and has two valence electrons in each atom?
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3 years ago
The maximum energy a bone can absorb without breaking is surprisingly small. For a healthy human of mass 60 kg60 kg, experimenta
netineya [11]

Answer:

<em>the maximum height a man can jump from and land rigidly upright on both feet without breaking his legs is 0.34 m</em>

<em></em>

Explanation:

Mass of a healthy man = 60 kg

energy the bone can take without breaking = 200 J

If a healthy man jumps from a height 'h', he falls with an energy equal to the potential energy due to his initial height above the ground.

initial potential energy of the healthy man = mgh

where m = mass of the man

g = acceleration due to gravity = 9.81 m/s^2

h = the height above ground

==> PE = 60 x 9.81 x h = 588.6h

If we assume that all energy is absorbed in the leg bones in a rigid landing, then we can safely say that this calculated PE for a healthy man is equal to the energy his bone can absorb in the jump without breaking.

equating, we have

200 = 588.6h

<em>the maximum height a man can jump from without breaking his legs = 200/588.6 = 0.34 m</em>

When people jump from a height, the sudden deceleration to zero can impact a big force on the leg bones, shattering them. If the time spent in decelerating to zero is increased, the overall force on the leg bones is reduced greatly.

<em>Bending the knees gradually on landing from a jump from a height, and then rolling increases the time spent decelerating, and reduces the impact force on the legs due to the landing</em>. If you observe carefully you will see that this is what professional stunts men and acrobats do when they jump from a height.

5 0
2 years ago
In a Young's double-slit experiment, 610-nm-wavelength light is sent through the slits. The intensity at an angle of 2.95° from
Sliva [168]

Answer:

spacing between the slits is 405.32043 ×10^{-9}  m

Explanation:

Given data

wavelength = 610 nm

angle = 2.95°

central bright fringe = 85%

to find out

spacing between the slits

solution

we know that spacing between slit is

I = 4I_{0} × cos²∅/2

so

I/4I_{0}  = cos²∅/2

here I/4I_{0} is 85 % = 0.85

so

0.85 = cos²∅/2

cos∅/2 = √0.85

∅ = 2 ×cos^{-1} 0.921954

∅  = 45.56°

∅  = 45.56° ×π/180 = 0.7949 rad

and we know that here

∅  = 2π d sinθ / wavelength

so

d = ∅× wavelength /  ( 2π  sinθ )

put all value

d = 0.795 × 610×10^{-9} / ( 2π  sin2.95 )

d = 405.32043 ×10^{-9}  m

spacing between the slits is 405.32043 ×10^{-9}  m

7 0
2 years ago
A concave lens has a focal length of 25cm. it's power in diaptor is​
IgorLugansk [536]

As we know that :

\begin{gathered}\large{\boxed{\sf{P \: = \: \dfrac{1}{f}}}} \\ \\ \rightarrow {\sf{P \: = \: \dfrac{1}{-25}}}\end{gathered}

Power, is in Meter. So divide focal length by 100

\begin{gathered}\rightarrow {\sf{P \: = \: \dfrac{1}{\dfrac{-25}{100}}}} \\ \\ \rightarrow {\sf{P \: = \: \dfrac{-100}{25}}} \\ \\ \rightarrow {\sf{P \: = \:- 4}} \\ \\ \underline{\sf{\therefore \: Power \: of \: Concave \: lens \: is \: - \: 4D}}\end{gathered}

8 0
1 year ago
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