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ira [324]
3 years ago
15

Two vectors are presented as a=3.0i +5.0j and b=2.0i+4.0j find (a) a x b, ab (c) (a+b)b and (d) the component of a along the dir

ection of
b.
Physics
1 answer:
Svet_ta [14]3 years ago
3 0
Let's ask this question step by step:
 Part A) 
 a x b = (3.0i + 5.0j) x (2.0i + 4.0j) = (12-10) k = 2k
 ab = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
 Part (c)
 (a + b) b = [(3.0i + 5.0j) + (2.0i + 4.0j)]. (2.0i + 4.0j)
 (a + b) b = (5.0i + 9.0j). (2.0i + 4.0j)
 (a + b) b = 10 + 36
 (a + b) b = 46
 Part (d)
 comp (ba) = (a.b) / lbl
 a.b = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
 lbl = root ((2.0) ^ 2 + (4.0) ^ 2) = root (20)
 comp (ba) = 26 / root (20)
 answer
 2k
 26
 46
 26 / root (20)
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When temperature decreases but actual water vapor content remains the same, what happens to relative humidity?
ruslelena [56]

Answer:

d. Relative humidity increases.

Explanation:

The expression of relative humidity in terms of absolute humidity, absolute pressure and saturation pression at measured temperature is:

\phi = \frac{\omega \cdot P}{(0.622+\omega)\cdot P_{sat}}

When temperature decreases, the saturation pressure decreases also and, consequently, relative humidity increases. Therefore, the right answer is option D.

8 0
3 years ago
Read 2 more answers
two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel .calculate the equivalent series resist
Novosadov [1.4K]

Explanation:

Given that,

Two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel.

For series combination,

R_{eq}=R_1+R_2

For parallel combination,

\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}

When 6 ohm and 3 ohm are in series,

R_s=6+3\\\\R_s=9\ \Omega

When 6 ohm and 3 ohm are in paralle,

\dfrac{1}{R_p}=\dfrac{1}{6}+\dfrac{1}{3}\\\\R_p=2\ \Omega

So, the equivalent resistance in series combination is 9 ohms and in parallel combination it is 2 ohms.

6 0
2 years ago
A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w
spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

  1. directly proportional to its length i.eR\propto l
  2. inversely proportional to its cross section area i.eR\propto \frac{1}{A}

Therefore

R=\rho\frac{l}{A}

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2

R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }

\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }......(1)

Again for tungsten:

R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }

\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }........(2)

Given that R_1=R_2   and    l_1=l_2

Dividing the equation (1) and (2)

\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}

\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}   [since R_1=R_2   and    l_1=l_2]

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}

\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

8 0
3 years ago
3. When two things are the same temperature:
JulijaS [17]
The answer is definitely C.) their molecules move at the same average speed
3 0
3 years ago
For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is 28.0 cmcm long)
loris [4]

This question is incomplete, the complete question is;

Two plane mirrors intersect at right angles. A laser beam strikes the first of them at a point 11.5 cm from their point of intersection, as shown in the figure.

For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is 28.0 cmcm long) after reflecting from the first mirror

Answer: angle of incidence is 39.4°

Explanation:

Given that;

two plain mirrors intersect at right angle (90°)

distance d = 11.5 cm

S = 28.0 cm

Now the angle that the reflection ray males with first the mirror equal theta  (∅)

so

tan∅ = (S/2) / d

tan∅ = (28/2) / 11.5

tan∅ = 14 / 11.5

tan∅ = 1.2173

∅ = tan⁻¹ (1.2173)

∅ = 50.6°

so angle of incidence = 90° - ∅

= 90° - 50.6°

= 39.4°

Therefore angle of incidence is 39.4°

5 0
3 years ago
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