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pashok25 [27]
2 years ago
14

Anyone know this question?

Physics
1 answer:
rodikova [14]2 years ago
7 0

Answer:

The answer is Gravitational pull

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What is the difference between a series circuit and a parallel circuit?
Angelina_Jolie [31]
Parallel has more than one circuit or form of energy

series has only one form of energy circuit
6 0
3 years ago
Calculate the amount of heat (kcal) released when 50.0g of steam at 100*c hits the skin, condenses, and cools to a body temperat
luda_lava [24]
As the steam touches the skin, it undergoes a phase change and releases latent heat due to the phase change. As it reaches equilibrium, it releases sensible heat. We calculate as follows:

Q = latent heat + sensible Heat
Q = 2.26 kJ / g (50.0 g) + 50.0 g ( 4.18 J / g C) (37 C - 100 C) ( 1 kJ / 1000 J)
Q = 99.833 kJ
6 0
3 years ago
Read 2 more answers
A force vector F1 points due east and has a magnitude of 200N. A second force F2 is added to F1. The resultant of the two vector
PilotLPTM [1.2K]

Answer:

The second vector \vec{F_2} points due West with a magnitude of 600N

Explanation:

The original vector \vec{F_1} points with a magnitude of 200N due east, the Resultant vector \vec{R} points due west (that's how east/west direction can be interpreted, from east to west) with a magnitude of  400N. If we choose East as the positive direction and West as the negative one, we can write the following vectorial equation:

\vec{F_1}+\vec{F_2}=\vec{R}\implies\vec{F_2}=\vec{R}-\vec{F_1}=-400N-200N=-600N

With the negative sign signifying that the vector points west.

3 0
3 years ago
A modern compact fluorescent lamp contains 1.4 mg of mercury (Hg). If each mercury atom in the lamp were to emit a single photon
Reika [66]

Answer:

A. 1.64 J

Explanation:

First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:

n=\frac{m}{M_m}

where

n is the number of moles

m = 1.4 mg = 0.0014 g is the mass of mercury

Mm = 200.6 g/mol is the molar mass of mercury

Substituting, we find

n=\frac{0.0014 g}{200.6 g/mol}=7.0\cdot 10^{-6} mol

Now we have to find the number of atoms contained in this sample of mercury, which is given by:

N=n N_A

where

n is the number of moles

N_A=6.022\cdot 10^{23} mol^{-1} is the Avogadro number

Substituting,

N=(7.0\cdot 10^{-6} mol)(6.022\cdot 10^{23} mol^{-1})=4.22\cdot 10^{18} atoms

The energy emitted by each atom (the energy of one photon) is

E_1 = \frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda=508 nm=5.08\cdot 10^{-7}nm is the wavelength

Substituting,

E_1 = \frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{5.08\cdot 10^{-7} m}=3.92\cdot 10^{-19} J

And so, the total energy emitted by the sample is

E=nE_1 = (4.22\cdot 10^{18} )(3.92\cdot 10^{-19}J)=1.64 J

4 0
3 years ago
12. Which of the following statements about
aliya0001 [1]
A seems to be correct
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2 years ago
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