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3 years ago

The energy of a wave is directly proportional to the square of the ___. D.Amplitude c. crest b. wavelength a.frequency

Physics

1 answer:

IRISSAK [1]3 years ago

4 0

The energy of a wave is directly proportional to the square of the D.Amplitude

the energy of a wave is given as

E = (0.5) (μ v t ) w² A²

where t = time

μ = mass per unit length of the string

v = wave propagation of velocity.

w = angular frequency

A = amplitude

E = energy of wave

From the equation , we see that

the energy of wave "E" is directly proportional to A².

hence the correct choice is D. Amplitude.

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3 years ago

Which type of molecule is added to this genetic sequece might create a insertion mutation? A-G-A-T-G-C-A-G-A-G-T-T-A-C-G-G

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4 years ago

A ball is tossed with a velocity of 10 m/s directed vertically upward from a window located 20 m above the ground. Determine the

marusya05 [52]

Answer:

Explanation:

Given

Initial velocity of ball $u=10\ m/s$

height of window $h=20\ m$

Using Equation of motion

$y=ut+\frac{1}{2}at^2$

where u=initial velocity

t=time

a=acceleration

As ball is already is at a height of 20 m so

$Y=ut+\frac{1}{2}at^2+20$

$Y=10\times t+0.5\times (-9.8)t^2+20$

$Y=-4.9t^2+10t+20$

(b)highest point is obtained at v=0

$v^2-u^2=2as$

where

v=final velocity

u=initial velocity

a=acceleration

s=displacement

$(0)-10^2=2\times (-9.8)\times s$

$s=\frac{100}{19.6}$

$s=5.102\ m$

Highest Point will be $s+20=25.102\ m$

(c)Time taken when the ball hit the ground i.e. at Y=0

$-4.9t^2+10t+20=0$

$t=3.28\ s$

impact velocity $v=\sqrt{2\times 9.8\times 25.102}$

$v=22.181\ m/s$

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3 years ago

Explain why surface tension occurs

Fynjy0 [20]

Hey there! The cohesive forces between liquid molecules are responsible for the phenomenon known as surface tension. The molecules at the surface do not have other like molecules on all sides of them and consequently they cohere more strongly to those directly associated with them on the surface. Hope this helps! :)

5 0

3 years ago

A heavy rope 6.00 m long and weighing 29.4 N is attached at one end to a ceiling and hangs vertically. A 0.500-kg mass is suspen

aivan3 [116]

Answer:

a) $v=3.1252\ m.s^{-1}$

b) $v=39.0672\ m.s^{-1}$

c) $v=8.2685\ m.s^{-1}$

d) No,

No.

Explanation:

Given:

length of rope, $l=6\ m$

weight of the rope, $w=29.4\ N$

mass suspended at the lower end of the rope, $M=0.5\ kg$

Now the mass of the rope:

$m=\frac{w}{g}$

$m=\frac{29.4}{9.8}$

$m=3.01\ kg$

So the linear mass density of rope:

$\mu=\frac{m}{l}$

$\mu=\frac{3.01}{6}$

$\mu=0.5017\ kg.m^{-1}$

We know that the speed of wave in a tensed rope is given as:

$v=\sqrt{\frac{F_T}{\mu} }$

where:

$F_T=$ tension force in the rope

At the bottom of the hanging rope we have an extra mass suspended. So the tension at the bottom of the rope:

$F_T=M\times g$

$F_T=0.5\times 9.8$

$F_T=4.9\ N$

Therefore the speed of the wave at the bottom point of the rope:

$v=\sqrt{\frac{4.9}{0.5017} }$

$v=3.1252\ m.s^{-1}$

Tension at a point in the middle of the rope:

$F_T=M\times g+\frac{w}{2}$

$F_T=0.5\times 9.8+\frac{29.4}{2}$

$F_T=19.6\ N$

Now wave speed at this point:

$v=\sqrt{\frac{19.6}{0.5017} }$

$v=39.0672\ m.s^{-1}$

Tension at a point in the top of the rope:

$F_T=M\times g+w$

$F_T=0.5\times 9.8+29.4$

$F_T=34.3\ N$

Now wave speed at this point:

$v=\sqrt{\frac{34.3}{0.5017} }$

$v=8.2685\ m.s^{-1}$

Tension at the middle of the rope is not the average tension of tension at the top and bottom of the rope because we have an extra mass attached at the bottom end of the rope.

Also the wave speed at the mid of the rope is not the average f the speeds at the top and the bottom of the ropes because it depends upon the tension of the rope at the concerned points.

7 0

3 years ago