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ch4aika [34]
2 years ago
14

What is the velocity of a quarter dropped from a tower after 10 seconds?

Physics
1 answer:
Goryan [66]2 years ago
3 0
U = 0, the initial vertical velocity

Assume g = 9.8 m/s² and ignore air resistance.

Use the formula
v = u + gt
where v =  vertical velocity after t seconds.
That is,
v = 0 + (9.8 m/s²)*(10 s) = 98.0 m/s

Answer: 98.0 m/s
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An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below
grin007 [14]
<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let P be the pressure at a point.

Let p be the density fluid at a point.

Let v be the velocity of fluid at a point.

Bernoulli's equation states that P+\frac{1}{2}pv^{2}+pgh=constant for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let p_{up} be the pressure of a point just above the wing.

Let p_{do} be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}

So,p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa

Force is given by the product of pressure difference and area.

Given that area is 27ms^{2}.

So,lifting force is 16398.48\times 27=442758.96N

6 0
3 years ago
A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting
rusak2 [61]

Answer:

The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is \sqrt{725} ft/ sec

Explanation:

Given:

h(t) =  25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec . 5 = 125 ft

x(5) = 10 ft/sec . 5 = 50 ft

Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

D(t) = \sqrt{h^2 + x^2}

Lets find the derivative of distance with respect to time

\frac{dD}{dt} (t)  = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}

Substituting the values of h(t) and  x(t) and simplifying we get,

\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}

\frac{dh}{dt} = 25ft/sec

\frac{dx}{dt} = 10 ft/sec

\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}  = \frac{725}{\sqrt{725}}  = \sqrt{725} ft / sec

5 0
3 years ago
If you were given distance and period of time, what can you calculate?
antiseptic1488 [7]

If you were given distance & period of time, you would be able to calculate the speed.

Hope this helps!

3 0
3 years ago
James and John dive from an overhang into the lake below. James simply drops straight down from the edge. John takes a running s
liraira [26]

Answer:

Both of them reach the lake at the same time.

Explanation:

We have equation of motion s = ut + 0.5at²

Vertical motion of James : -

          Initial velocity, u = 0 m/s

         Acceleration, a = g

         Displacement, s = h

    Substituting,

                  s = ut + 0.5 at²

                 h = 0 x t + 0.5 x g x t²

                 t_{James}=\sqrt{\frac{2h}{g}}

Vertical motion of John : -

          Initial velocity, u = 0 m/s

         Acceleration, a = g

         Displacement, s = h

    Substituting,

                  s = ut + 0.5 at²

                 h = 0 x t + 0.5 x g x t²

                 t_{John}=\sqrt{\frac{2h}{g}}

So both times are same.

Both of them reach the lake at the same time.

3 0
3 years ago
What is an electronic signal
rosijanka [135]

where are the answer choises

6 0
3 years ago
Read 2 more answers
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