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2 years ago

Jack is a transgender male who is sexually attracted to males. Based on what you know about sex

Physics

1 answer:

sergey [27]2 years ago

3 0

Answer:

Jack is a homosexual male

Explanation:

Since Jack identifies as male and he is attracted to other males, he would be considered a homosexual male, or gay.

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What is the energy of a photon whose frequency is 6.0 x 10^20?

omeli [17]

Answer:

3.75 MeV

Explanation:

The energy of the photon can be given in terms of frequency as:

E = h * f

Where h = Planck's constant

The frequency of the photon is 6 * 10^20 Hz.

The energy (in Joules) is:

E = 6.63 x10^(-34) * 6 * 10^(20)

E = 39.78 * 10^(-14) J = 3.978 * 10^(-13) J

We are given that:

1 eV = 1.06 * 10^(-19) Joules

This means that 1 Joule will be:

1 J = 1 / (1.06 * 10^(-19)

1 J = 9.434 * 10^(18) eV

=> 3.978 * 10^(-13) J = 3.978 * 10^(-13) * 9.434 * 10^(18) = 3.75 * 10^(6) eV

This is the same as 3.75 MeV.

The correct answer is not in the options, but the closest to it is option C.

6 0

2 years ago

Which of the following describes the relationship between work and power

Pani-rosa [81]

Answer:

Power is the rate which work is done.

Explanation:

<em>Power</em> is the rate which work is done. Power is measured in watts.

<em>Work</em> is the use of force to move an object. Work is measured in joules

7 0

3 years ago

Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

The shear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

b) If the railing edge is 2 ft, the Reynold number is:

$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

The friction drag is equal to:

$F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf$

7 0

2 years ago

An object in static equilibrium has a coefficient of static friction as 0.110. If the normal force acting on the object is 95.0

igomit [66]

Answer:

mass = 9.7 kg

Explanation:

As we know that when object is at rest on the ground of flat base then we will have

$N - mg = 0$

so from here

$N = mg$

now we have

N = 95 Newton

now from above equation we will have

$95 = m\times g$

$95 = m\times 9.8$

$m = 9.7 kg$

7 0

3 years ago

Calculate the efficiency of given pulley System if 40N

jonny [76]

Answer:

20n

Explanation:

60n - 40n = efficiency of pulley system

20n = efficiency of pulley system

4 0

2 years ago