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3 years ago

Unlike a longitudinal wave, the disturbance in a transverse wave is

Physics

2 answers:

slamgirl [31]3 years ago

4 0

Answer:

The disturbance of molecules are perpendicular to wave motion

Explanation:

Transverse waves: in transverse waves all the medium moles vibrates perpendicular to the wave motion such that the net displacement of all molecules are zero.

While in longitudinal waves all the molecules will oscillate in the direction of wave propagation. All molecules transfer the energy from one point to other point while oscillating parallel to the wave propagation direction.

leva [86]3 years ago

3 0

perpendicular to the wave motion.

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A 65 kilogram astronaut weighs 638 newtons at the surface of earth what is the mass of the astronaut at the surface of the moon,

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An injured monkey sits perched on a tree branch 3.0 m above the ground, while a wildlife veterinarian is kneeling down in the bu

Irina18 [472]

here in the given situation if monkey starts free fall at the same instant when veterinarian shoots towards it then we know that vertical component of motion of monkey and the dart will be same as under gravity

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tia_tia [17]

Answer:

A:7.2 $ms^{-1}$

B:14.25 $ms^{-1}$

C:1.45 $sec$

D:10.3 $m$

E:2.9 $sec$

F:20.88 $m$

Explanation:

Let $v$ be the velocity and $\alpha$ be the angle between the velocity and ground.

Question A:

Horizontal component of velocity is given by $vcos(\alpha )$ .

So,horizontal component is $16\times cos(63)=16\times 0.45=7.2ms^{-1}$

Question B:

Vertical component of velocity is given by $vsin(\alpha )$ .

So,vertical component is $16\times sin(63)=16\times 0.89=14.25ms^{-1}$

Question C:

Time required is given by $\frac{\text{vertical component of velocity}}{g}}=\frac{14.25}{9.8}=1.45 seconds$

Question D:

Maximum height is given by $\frac{\text{vertical component of velocity}^{2}}{2g}}=\frac{203.06}{19.6}=10.3m$

Question E:

Time of flight is twice the time required to reach maximum height= $2\times 1.45=2.9 seconds$ .

Question F:

The distance between the player and ball after landing is called range and is given by $\text{horizontal component of velocity}\times \text{time of flight}=$ $7.2\times 2.9=20.88m$