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Lostsunrise [7]
3 years ago
7

A 1/10th scale model of an airplane is tested in a wind tunnel. The reynolds number of the model is the same as that of the full

scale airplane in air. The velocity in the wind tunnel is then
Physics
1 answer:
trasher [3.6K]3 years ago
7 0

Answer:

of the velocity of a full size plane in the air

You might be interested in
A spotlight on the ground shines on a wall 12 meters away. If a man 2 meters high walks away from the spotlight towards the wall
nordsb [41]

Answer: The length of the shadow on the wall is decreasing by 0.6m/s

Explanation:

the specified moment in the problem, the man is standing at point D with his head at point E.

At that moment, his shadow on the wall is y=BC.

The two right triangles ΔABC and ΔADE are similar triangles. As such, their corresponding sides have equal ratios:

ADAB=DEBC

8/12=2/y,∴y=3 meters

If we consider the distance of the man from the building as x then the distance from the spotlight to the man is 12−x.

(12−x) /12=2/y

1− (1 /12x )=2 × 1/y

Let's take derivatives of both sides:

−1 / 12dx = −2 × 1 / y^2 dy

Let's divide both sides by dt:

−1/12⋅dx/dt=−2/y^2⋅dy/dt

At the specified moment:

dxdt=1.6 m/s

y=3

Let's plug them in:

−1/121.6) = - 2/9 × dy/dt

dy/dt = 1.6/12 ÷ 2/9

dy/dt = 1.6/12 × 9/2

dy/dt = 14.4/24 = 0.6m/s

5 0
3 years ago
A 30.0 kg child, initially at rest, slides down a 2.0 m tall slide. The child reaches the bottom of the slide with a speed of 6
HACTEHA [7]

Answer:

Explanation:

Total energy is constant

E = mgh + ½mv² + Fd

At the top of the slide, all energy is potential

E = mgh + 0 + 0

At the bottom of the slide, all potential energy has converted to kinetic and work of friction.

mgh = ½mv² + W

W = mgh - ½mv²

W = 30.0[(9.81)(2.0) - ½6²]

W = 48.6 J

5 0
2 years ago
You want to find out how many atoms of the isotope 65Cu are in a small sample of material. You bombard the sample with neutrons
serious [3.7K]

Answer:

a) number of copper atoms 65 (⁶⁵Cu)  is 7.692 10⁶ atoms

b) m_total Cu = 1.585 10⁹ u = 2.632 10⁻¹⁸ kg

Explanation:

a) For this exercise let's start by using the radioactive decay ratio

           N = N₀  e^{- \lambda t}o e - lambda t

The half-life time is defined as the time it takes for half of the radioactive (activated) atoms to decay, therefore after two half-lives there are

            N = ½ (½ N₀) = ¼ N₀

            N₀ = 4 N

in each decay a photon is emitted so we can use a direct rule of proportions. If an atom emits a photon it has Eo = 1,04 Mev, how many photons it has energy E = 10,000 MeV

          # _atoms = 1 atom (photon) (E / Eo)

          # _atoms = 1 10000 / 1.04

          # _atoms = 9615,4 atoms

          N₀ = 4 #_atoms

          N₀ = 4 9615,4

          N₀=  38461.6  atoms

in the exercise indicates that half of the atoms decay in this way and the other half decays directly to the base state of Zinc, so the total number of activated atoms

          N_activated = 2 # _atoms

          N_activated = 2 38461.6

          N_activated = 76923.2

also indicates that 1% = 0.01 of the nuclei is activated by neutron bombardment

          N_activated = 0.01 N_total

          N_total = N_activated / 0.01

          N_total = 76923.2 / 100

          N_total = 7.692 10⁶ atoms

so the number of copper atoms 65 (⁶⁵Cu)  is 7.692 10⁶

b) the natural abundance of copper is

  ⁶³Cu     69.17%

  ⁶⁵Cu    30.83%

Let's use a direct proportion rule. If there are 7.692 10⁶  ⁶⁵Cu that represents 30.83, how much ⁶³Cu is there that represents 69.17%

                # _63Cu = 69.17%  (7.692 10⁶    / 30.83%)

                # _63Cu = 17.258 10⁶  atom  ⁶³Cu

the total amount of comatose is

              #_total Cu = #_ 65Cu + # _63Cu

              #_total Cu = (7.692 + 17.258) 10⁶

              #_total Cu = 24.95 10⁶

the atomic mass of copper is m_Cu = 63.546 u

          m_total = #_totalCu m_Cu

          m_total = 24.95 10⁶ 63,546 u

          m_total = 1.585 10⁹ u

let's reduce to kg

           m_total Cu = 1.585 10⁹ u (1,66054 10⁻²⁷ kg / 1 u)

           m_total Cu = 2.632 10⁻¹⁸ kg

8 0
2 years ago
Suppose that you have a spring gun that you use to launch a small metal ball. You try the first two settings of the gun. The fir
Ivan

Answer:

The distance s of how far the ball will go at the highest setting = 2.25m

Explanation:

Let consider x to be the representative of the compression and the distance to be s

Recall that:

\dfrac{1}{2}\times K \times  x^2 = mgs +c

By cross multiplying

K \times  x^2 = 2(mgs +c)

K \times  x^2 = 2\times 9.81(ms) +2c

K \times  x^2 = 19.62(ms) +2c

x^2 = A \times  s+B

Thus, for the first setting

x = 1 , s = 0.25

for the second setting

x = 2,   s = 1

1 = 0.25A + B ---  (1)

4 = A + B    ----- (2)

From (1); let B =  1 - 0.25A  and substitute it into (2)

4 = A + 1 - 0.25 A

4 - 1 = A - 0.25 A

3 = 0.75 A

A = 3/0.75

A = 4

From (2)

4 = A + B

4 = 4 + B

B = 4 - 4

B = 0

Therefore, for the highest setting, where x = 3

Then :

x^2 = A \times  s+B will be:

3² =   4s + 0

9 = 4s

s = 9/4

s = 2.25 m

∴

The distance s of how far the ball will go at the highest setting = 2.25m

6 0
3 years ago
what can be done not to let anyone of the country be hungry and clothless discuss in groups and present your findings in the cla
Salsk061 [2.6K]

...get food to people in need when they need it...

8 0
2 years ago
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