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ankoles [38]
2 years ago
11

A diver springs upward from a board that is 3.90 m above the water. At the instant she contacts the water her speed is 13.2 m/s

and her body makes an angle of 81.8 ° with respect to the horizontal surface of the water. Determine her initial velocity, both (a) magnitude and (b) direction.
Physics
1 answer:
kodGreya [7K]2 years ago
5 0

Answer:

a) u = 9.88 m/s

b) \theta = 79^0

Explanation:

given,

speed of the diver = 13.2 m/s

Angle made w.r.t horizontal = 81.8 °

The horizontal component of final velocity

v_h = 13.2 cos 81.8 °

The vertical component of final velocity

v_v = 13.2 sin 81.8 °

using equation of motion

initial velocity of vertical component

u_v = \sqrt{v_v^2-2as}

u_v = \sqrt{(13.2 sin 81.8^0)^2-2(-9.8)(-3.9)}

      = 9.70 m/s

the horizontal component of the initial velocity

u_h = v_h = 13.2 cos 81.8 ° = 1.88 m/s

magnitude

u = \sqrt{9.7^2+1.88^2}

u = 9.88 m/s

direction

\theta = tan{-1}(\dfrac{9.7}{1.88})

\theta = 79^0

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C. Evaporating water from the container.

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Water initially at 200 kPa and 300°C is contained in a piston–cylinder device fitted with stops. The water is allowed to cool at
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Answer:

Δu=1300kJ/kg  

Explanation:

Energy at the initial state

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Is saturated vapor at initial pressure we have

p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)

Process 2-3 is a constant volume process

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The overall in internal energy

Δu=u₁-u₃

We replace the values in equation

Δu=u₁-u₃

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What is used to create the sequential order of elements?
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Is it possible for one energy store to transfer into another energy store ?Explained answer
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Yes

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A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistance.
Nookie1986 [14]

Answer:

a. 0.6 m b. 0.385 m c. 3.6 m/s at 287.78° to the horizontal

Explanation:

a. Using s = ut - 1/2gt² for motion under gravity where s = vertical distance = height of table, u = initial vertical velocity of book = 0 m/s, t = time of flight = 0.350 s and g = acceleration due to gravity = 9.8 m/s².

Substituting these these values into s and taking the top of the table as position 0 m, we have.

0 - s = 0t - 1/2gt²

-s = -1/2gt²

s = 1/2gt²

s = 1/2 × 9.8 m/s² × (0.350 s)²

s = 0.6 m

b. Using d = v't where d = horizontal distance from table, v' = horizontal velocity of book = 1.10 m/s and t = time of flight = 0.350 s

d = v't = 1.10 m/s × 0.350 s = 0.385 m

c. Using v² = u² - 2gs where u = initial vertical velocity of book = 0 m/s and g = 9.8 m/s², s = -0.6 m (negative since we are at the bottom and 0 m is at the top)and v = final vertical velocity of book

v² = u² - 2gs

= 0 - 2 × 9.8 m/s² × (-0.6 m)

= 11.76 m²/s²

v = √11.76 m/s

= 3.43 m/s

So, the magnitude of the resultant velocity is V = √(v² + v'²)

= √((3.43 m/s)² + (1.10 m/s)'²)

= √(11.76 m²/s² + 1.21 m²/s²)

= √12.97 m²/s²

= 3.6 m/s

Its direction Ф = tan⁻¹(-v/v') since v is in the negative y direction

= tan⁻¹(-3.43 m/s/1.10 m/s)

= tan⁻¹(-3.1182)

= -72.22°

Ф = -72.22°+ 360 = 287.78° since it is in the third quadrant

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