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m_a_m_a [10]
3 years ago
9

A person is lying on a diving board 3.50 m above the surface of the water in a swimming pool. She looks at a penny that is on th

e bottom of the pool directly below her. To her, the penny appears to be a distance of 5.00 m from her.
Physics
1 answer:
sergij07 [2.7K]3 years ago
4 0

Answer:

1.995 m

Explanation:

Distance of penny as seen by the person = 5 m

Height of person from water surface = 3.50 m

Apparent depth of penny = 5 - 3.50 = 1.5 m

refractive index of water, n = 1.33

real depth / apparent depth = n

real depth = 1.33 x 1.5 = 1.995 m

Thus, the actual depth of water at that point is 1.995 m.

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Answer:

Explanation:

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Component of weight acting on spring = mg sinθ

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mg sin17 = k x 2.13

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k = 41.7 N/m

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A crate lies on a plane tilted at an angle θ = 22.5 ∘ to the horizontal, with μk = 0.19.
ValentinkaMS [17]

A) 2.03 m/s^2

Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:

Parallel:

mg sin \theta - \mu_k R = ma (1)

where

m is the mass

g = 9.8 m/s^2 the acceleration of gravity

\theta=22.5^{\circ}

\mu_k = 0.19 is the coefficient of friction

R is the normal reaction

a is the acceleration

Perpendicular:

R-mg cos \theta =0 (2)

From (2) we find

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_k mg cos \theta = ma

Solving for a,

a=g sin \theta - \mu_k g cos \theta = (9.8)(sin 22.5)-(0.19)(9.8)(cos 22.5)=2.03 m/s^2

B) 5.94 m/s

We can solve this part by using the suvat equation

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Here we have

v = ?

u = 0 (it starts from rest)

a=2.03 m/s^2

s = 8.70 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(2.03)(8.70)}=5.94 m/s

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