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2 years ago

PLZ HELP do waves transport water with them

Physics

1 answer:

Gnoma [55]2 years ago

5 0

Ocean waves transport energy over vast distances, although the water itself does not move, except up and down

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How much water in units of litre can fill a water tank of 1m3 capacity? Explain plx ans me

mash [69]

1000 litres of water can fill a tank of 1m3 capacity because water has a density of 1000kg/m3 and as 1 kg water = 1 liter ,
1000 liter of water can fill a tank of 1m3

5 0

3 years ago

Neutrons incident on a heavy nucleus with spin J 0 show a resonance at an incident energy ER = 250 eV in the total cross-section

ivolga24 [154]

Answer:

elastic partial width is 2.49 eV

Explanation:

given data

ER E = 250 eV

spin J = 0

cross-section magnitude σ = 1300 barns

peak P = 20ev

to find out

elastic partial width W

solution

we know here that

σ = λ²× W / ( E × π × P ) ...................1

put here all value

σ = (0.286)² × W × $10^{-16}$ / ( 250 × π × 20 )

1300 × $10^{-24}$ = (0.286)² × W × $10^{-16}$ / ( 250 × π × 20 )

solve it and we get W

W = 249.56 × $10^{-2}$

so elastic partial width is 2.49 eV

8 0

3 years ago

Select the correct answer. Ben has $10 to spend on dinner. He can have a pizza delivered, drive to a burger place, or go to the

Rudik [331]

Answer:

A. to eat a healthier meal

Explanation:

5 0

3 years ago

A bowler throws a bowling ball of radius R = 11 cm along a lane. The ball slides on the lane with initial speed vcom,0 = 6.0 m/s

ankoles [38]

Answer:

Explanation:

Radius of the ball is $R=11cm=0.11m$

Initial speed of the ball is $v_{com0}=6.0m/s$

Initial angular speed of the ball is $\omega = 0$

Coefficient of kinetic friction between the ball and the lane

is $\mu =0.35$

Due to the presence of frictional force, ball moves with

decreasing velocity.

(a)

velocity $v_{com0}$ in terms of $\omega$ is

$V_{com0} = -R\omega\\\\=-(0.11m)\omega\\\\= (-0.11\omega)m/s$

(b)

Ball's linear acceleration is given by

$a=-\mu g\\\\=-(0.35) (9.8 m/s^2)\\\\= -3.43m/s^2$

(c)

During sliding, ball's angular acceleration is calculated as

$\alpha=-\frac{\tau}{I}\\\\-\frac{\mu mgR}{(\frac{2}{5}mR^2)}\\\\-\frac{2}{5}\frac{\mu g}{R}\\\\-\frac{2}{5}\frac{(0.35)(9.8)}{0.11}\\\\-77.95rad/s^2$

(d)

The time for which the ball slides is calculated from the

equation of motion is

$V_{cm}= V_{cm0} + at\\\\V_{cm} = V_{cm0} + (-\mu g )t\\\\-0.11\omega=6.0m/s -(3.43m/s^2 )t\\\\-(0.11) (\alpha t) =6.0 m/s - (3.43 m/s^2)\\\\- (0.11)(-77.95 rad/s^2)t = 6.0m/s - (3.43 m/s^2 )t\\\\8.5745t + 3.43t= 6.0\\\\12.0045t = 6.0\\\\t= 0.4998s$