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vesna_86 [32]
3 years ago
12

Place your hands on the side of your head, with your palms facing forward. This should look like you have given yourself large e

ars. Now completely rotate the Earth (your head) relative to the Sun (the light). What did you just model?
Physics
1 answer:
nirvana33 [79]3 years ago
7 0

Answer:

the solar system

Explanation:

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Ignoring air resistance, if a 20 kg ball and a 400 kg crate were both dropped from the
masya89 [10]
The crate’s acceleration is gravity= 9.81m/s^2
The ball’s acceleration is also gravity=9.81m/s^2


8 0
3 years ago
Yes or no
Tju [1.3M]

Answer:

1. No, it is a colloid

2. Yes

Explanation:

8 0
2 years ago
An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and
Andre45 [30]

Answer:

change in entropy is 1.44 kJ/ K

Explanation:

from steam tables

At 150 kPa

specific volume

Vf = 0.001053 m^3/kg

vg = 1.1594 m^3/kg

specific entropy values are

Sf = 1.4337 kJ/kg K

Sfg = 5.789 kJ/kg

initial specific volume is calculated as

v_1 = vf + x(vg - vf)

      = 0.001053 + 0.25(1.1594 - 0.001053)

v_1 = 0.20964  m^3/kg

s_1 = Sf + x(Sfg)

     = 1.4337 + 0.25 \times 5.7894 = 2.88 kJ/kg K

FROM STEAM Table

at 200 kPa

specific volume

Vf = 0.001061 m^3/kg

vg = 0.88578 m^3/kg

specific entropy values are

Sf = 1.5302 kJ/kg K

Sfg = 5.5698 kJ/kg

constant volume  sov_1 -  v_2  = 0.29064 m^3/kg

v_2 = v_1 = vf + x(vg - vf)

       =0.29064 = x_2(0.88578 - 0.001061)

x_2 = 0.327

s_2 = 1.5302 + 0.32 \times 5.5968 = 3.36035 kJ/kg K

Change in entropy \Delta s = m(s_2 - s_1)

              =3( 3.36035 - 2.88) =  1.44 kJ/kg

8 0
3 years ago
which of the following are vector quantities? check all that apply. a. force b. acceleration c. displacement d. mass
Ratling [72]
Force, acceleration, and Displacement are all vector quantities.
4 0
3 years ago
A laser pulse of duration 25 ms has a total energy of 1.4 J. The wavelength of this radiation is
SpyIntel [72]

Answer:

n = 4 x 10¹⁸ photons

Explanation:

First, we will calculate the energy of one photon in the radiation:

E = \frac{hc}{\lambda}\\\\

where,

E = Energy of one photon = ?

h = Plank's Constant = 6.625 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of radiation = 567 nm = 5.67 x 10⁻⁷ m

Therefore,

E = \frac{(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{5.67\ x\ 10^{-7}\ m}

E = 3.505 x 10⁻¹⁹ J

Now, the number of photons to make up the total energy can be calculated as follows:

Total\ Energy = nE\\1.4\ J = n(3.505\ x\ 10^{-19}\ J)\\n = \frac{1.4\ J}{3.505\ x\ 10^{-19}\ J}\\

<u>n = 4 x 10¹⁸ photons</u>

8 0
3 years ago
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