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3 years ago

What is the standard unit of velocity

Physics

2 answers:

Bond [772]3 years ago

7 0

Answer:

Explanation:

Meters per second

juin [17]3 years ago

5 0

Answer:

m/s (meters per second)

Explanation:

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A plane wave with a wavelength of 500 nm is incident normally ona single slit with a width of 5.0 × 10–6 m.Consider waves that r

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To solve this exercise it is necessary to use the concepts related to Difference in Phase.

The Difference in phase is given by

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A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of

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Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

$h=ut-\dfrac{1}{2}gt^2$

$h=(u\sin60)\times t-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2$

$h=41.6\ m$

(b). We need to calculate the maximum height of the ball

Using formula of height

$h_{max}=\dfrac{(u\sin\theta)^2}{2g}$

Put the value into the formula

$h=\dfrac{(33\sin60)^2}{2\times 9.8}$

$h=41.67\ m$

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

$v=u-gt$

$v=u\sin\theta-gt$

Put the value into the formula

$v_{y}=33\times\sin 60-9.8\times3.0$

$v_{y}=-0.82\ m/s$

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=33\times\cos60$

$v_{x}=16.5\ m/s$

We need to calculate the ball's impact speed

Using formula of velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(16.5)^2+(-0.82)^2}$

$v=16.52\ m/s$

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

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2 years ago