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kolbaska11 [484]
3 years ago
10

Which best describes a relationship among the vectors? Vectors P and Py are components of vector Px. Vectors Px and Py are compo

nents of vector P.
Physics
2 answers:
Semmy [17]3 years ago
4 0

Answer:

D - Vectors Px and Py are components of vector P.

Explanation:

mafiozo [28]3 years ago
3 0
If we have a vector P, then we name it's components by the axis we project it onto.

In 2D Cartesian coordinate system we have one x axis and one y axis.

Vector P can then be represented with two components: Px and Py.

Px is the component of vector P on the x axis and Py is the component of the vector P on the y axis. 
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Which is true of the greenhouse effect?
Aleksandr [31]

Answer:

c is correct option thanks to brainly

3 0
3 years ago
A parallel-plate capacitor is constructed from two aluminum foils of 1 square centimeter area each placedon both sides of a rubb
Svet_ta [14]

Answer:

The voltage will be 0.0125V

Explanation:

See the picture attached

4 0
3 years ago
Unpolarized light with intensity I0I0I_0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal po
zvonat [6]

Answer:

0.293I_0

Explanation:

When the unpolarized light passes through the first polarizer, only the component of the light parallel to the axis of the polarizer passes through.

Therefore, after the first polarizer, the intensity of light passing through it is halved, so the intensity after the first polarizer is:

I_1=\frac{I_0}{2}

Then, the light passes through the second polarizer. In this case, the intensity of the light passing through the 2nd polarizer is given by Malus' law:

I_2=I_1 cos^2 \theta

where

\theta is the angle between the axes of the two polarizer

Here we have

\theta=40^{\circ}

So the intensity after the 2nd polarizer is

I_2=I_1 (cos 40^{\circ})^2=0.587I_1

And substituting the expression for I1, we find:

I_2=0.587 (\frac{I_0}{2})=0.293I_0

5 0
2 years ago
a bubble of air of volume 1cm^3 is released by a deep sea diver at a depth where the pressure is 4.0 atmospheres. assuming its t
lys-0071 [83]

Answer:

hope this helps!

Explanation:

Volume of the air bubble, V1=1.0cm3=1.0×10−6m3

Bubble rises to height, d=40m

Temperature at a depth of 40 m, T1=12oC=285K

Temperature at the surface of the lake, T2=35oC=308K

The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa 

The pressure at the depth of 40 m: P1=1atm+dρg

Where,

ρ is the density of water =103kg/m3

g is the acceleration due to gravity =9.8m/s2

∴P1=1.103×105+40×103×9.8=493300Pa

We have T1P1V1=T2P2V2

Where, V2 is the volume of the air bubble when it reaches the surface.

V2=

8 0
2 years ago
If you wish to warm 100 kg of water by 20°C for your bath, how much heat is required? (Give your answer in calories and joules.)
taurus [48]
Q = mcθ

Where m = mass of water in kg.
c = specific heat capacity in kJ/kg⁰C, c for water = 4200 kJ/kg⁰C
θ = temperature rise in ⁰C

Q = 100*4200* 20    Note here the temperature rise is 20 ⁰C
Q = 8 400 000 J

In calories,  4.2 J = 1 Calorie
=  8 400 000 / 4.2   = 200 000

Q = 200 000 Calories
4 0
2 years ago
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