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sergeinik
3 months ago
15

Whale sharks swim forward while ascending or descending. They swim along a straight-line path at a shallow angle as they move fr

om the surface to deep water or from the depths to the surface. In one recorded dive, a shark started 50 m below the surface and swam at 0.85 m/s along a path tipped at a 13 ∘ angle above the horizontal until reaching the surface.
Physics
1 answer:
Keith_Richards [3.2K]3 months ago
4 0
Although you didn't provide the question, I'm here to clarify the issue and guide you in locating the necessary information.

1) The shark rises at an angle of 13°, calculated as follows:

Vertical ascending speed = 0.85m/s * sin(13°)

Horizontal speed = 0.85m/s * cos(13°)

2) The distance the shark covers while ascending is determined as follows

Vertical ascent length = 50 m

Horizontal distance, y:

\frac{y}{50} = \frac{0.85sin(13)}{0.85cos(13)}

This gives y = 50 * tan(13°)

=> y = 11.54 m.

3) In conclusion:

1) The shark traveled 50 m upward vertically and 11.54 m horizontally.

2) To find the length of the path taken by the shark, we can use Pythagoras' theorem:

hypotenuse^2 = (50m)^2 + (11.54m)^2 = 2633.25m^2

hypotenuse = 51.35m

Consequently, the shark swam 51.35 m to surface.

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An airplane flies with a velocity of 55.0 m/s [35o N of W] with respect to the air (this is known as air speed). If the velocity
ValentinkaMS [3465]
V - wind speed;
53° - 35° = 18°
v² = 55² + 40² - 2 · 55 · 40 · cos 18°
v² = 3025 + 1600 - 2 · 55 · 40 · 0.951
v² = 440.6
v = √440.6
v = 20.99 ≈ 21 m/s
Conclusion: The wind speed calculates to 21 m/s.  
5 0
2 months ago
A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro
Ostrovityanka [3204]

Answer:

a) v_{o} =16m/s

b) v=9.8m/s

c) \beta =-35.46º

Explanation:

According to the problem, the distance from the building where the ball hits is 16m, and its final elevation exceeds the initial height by 8m.

With this information, we can compute the ball’s starting speed.

a) Let's first assess the horizontal trajectory.

x=v_{ox}t

x=v_{o}cos(60)t

v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)} (1)

This gives us our initial equation.

Next, we need to examine the vertical trajectory.

y=y_{o}+v_{oy}t+\frac{1}{2}gt^2

y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2

Utilizing v_{o} in our first equation (1)

8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2

\frac{1}{2}(9.8)t^2=16tan(60)-8

Now let’s solve for t.

t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s

The ball takes two seconds to reach the adjacent building, allowing us to compute its initial speed.

v_{o}=\frac{16m}{(2s)cos(60)}=16m/s

b) To determine the velocity magnitude just before impact, we must calculate both x and y components.

v_{x}=v_{ox}+at=16cos(60)=8m/s

v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s

The computed velocity magnitude is:

v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s

c) The ball's angle is:

\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º

4 0
4 months ago
A water jet that leaves a nozzle at 60 m/s at a flow rate of 120 kg/s is to be used to generate power by striking the buckets lo
Ostrovityanka [3204]

Response:

216000 W or 216 kW

Details:

Power: This refers to the rate at which energy is utilized or consumed. The standard unit for power is the Watt (W).

In general,

Power = Energy/time

P = E/t........................ Equation 1.

However,

E = 1/2mv²..................... Equation 2

with m representing mass, and v indicating velocity.

Substituting equation 2 into equation 1,

P = 1/2mv²/t...................... Equation 3

Assuming flow rate (Q) = m/t,

Q = m/t................ Equation 4

Insert equation 4 into equation 3,

P = Qv²/2........................ Equation 5

Where Q stands for flow rate, v is velocity, and P denotes power.

Given: Q = 120 kg/s, v = 60 m/s

Plug these values into equation 5,

P = 120(60)²/2

P = 60(60)²

P = 60×3600

P = 216000 W.

Thus, the potential power generation of the water jet is 216000 W or 216 kW.

5 0
4 months ago
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