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poizon [28]
3 years ago
10

Problem 18.9 an underwater diver sees the sun 60 ∘ above horizontal. part a how high is the sun above the horizon to a fisherman

in a boat above the diver?
Physics
1 answer:
julsineya [31]3 years ago
3 0

The angles in the equation are not the angles relative to the horizon but are relative to the "normal" which means that the line that is perpendicular to the surface.

The angle under the water is 90 – 60 = 30. 

n1 for water is 1.33, n2 for air is 1 Which you seem to understand. 

(1.33)(sin (30)) = (1.00)(sin (x2)) 

Rearranging the equation above, will give us: x^2 = sin^-1((1.33 sin 30)/1) = 41.68

But remember that that is the angle relative to the normal so you have to deduct it from 90 to get the angle relative to the horizon and you get (90 – 41.68) = 48.32 degrees.

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Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

F_D=\dfrac{1}{2}\rho C_DA v^2

C_D=Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}

Now by putting the values

\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}

\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}

\dfrac{F_D_1}{F_D_2}=0.444

Percentage Change in the drag force

\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100

\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100

\Delta F=\dfrac{1-0.444}{0.444}\times 100

ΔF=125.22 %

Therefore force will increase by 125.22  %.

3 0
3 years ago
Free body diagram definition <br>in physics​
kirill115 [55]

Answer:

In physics and engineering, a free body diagram (force diagram, or FBD) is a graphical illustration used to visualize the applied forces, moments, and resulting reactions on a body in a given condition.

4 0
3 years ago
The cart is initially at rest. Force
Oliga [24]

Answer:

v'=\dfrac{1}{2}v

Explanation:

Given that,

Initial speed of the cart, u = 0

Let F force is applied to the cart for time \Delta t after which the car has speed v. The force on an object is given by :

F = ma

m is the mass of the cart

We need to find the speed of second cart, if the same force is applied for the same time to a second cart with twice the mass. Force becomes,

F=\dfrac{mv'}{t}

v'=\dfrac{F t}{2m}

v'=\dfrac{1}{2}v

So, the speed of second cart is half of the initial speed of first cart. So, the correct option is (b).

5 0
3 years ago
A Red-Tailed Hawk leaves its nest and flies 100.0 meters across the sky in 20 seconds, when it spots a garter snake on the groun
igomit [66]
Speed = distance / time.

Speed of him leaving the nest:
S = 100 / 20sec

5 m/s.

Catching the snake:
S2 = 50 / 5sec

10 m/s.

Average of 5& 10 = 7.5

7.5 m/s has to be the answer.
3 0
3 years ago
An athlete can twirl a baton so that it reaches an angular velocity of 12 rad/s from rest in 0.5 s. If the length of a uniform b
NikAS [45]

Answer:

The torque needed is 46.08 Nm

Explanation:

Given;

angular velocity, ω = 12 rad/s

time of motion, t = 0.5 s

length of the baton, r = 0.8 m

mass of the baton, 0.5 kg

The torque needed to  reach the angular velocity is given by;

τ = F x r

where;

F is the centripetal force of the baton

r is the length of the baton = radius of the circular motion of the baton

F_c =ma_c= \frac{mv^2}{r} = m\omega^2 r

Torque is given by;

\tau = F_c *r\\\\\tau = m \omega^2 r *r\\\\\tau = m \omega^2 r^2 \\\\\tau = (0.5)(12)^2(0.8)^2\\\\\tau = 46.08 \ N m

Therefore, the torque needed is 46.08 Nm

6 0
3 years ago
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