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3 years ago

Two vectors are illustrated in the coordinate plane.

Physics

1 answer:

oksano4ka [1.4K]3 years ago

4 0

Answer: (-4.3, 2.5)

Explanation:

In the second quadrant, magnitude of the vector is 5 and angle is 30° from the negative x-axis.

We can write this in terms of its components:

$-5 cos 30^o \hat i + 5 sin 30^o\hat j$

$-4.3\hat i+2.5 \hat j$

Thus, the components of vector in the second quadrant are (-4.3, 2.5)

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Why does the speed of sound depend on air temperature?

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4 years ago

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The percentage of incident light radiation reflected back to space is termed albedo. True or false?

liubo4ka [24]

Answer:True

Explanation:

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3 years ago

A certain lightning bolt moves 10.0 C of charge. How many charges have been moved if the fundamental unit of charge is 1.6 x 10-

Andrej [43]

To solve this problem, use the ratio given by the total number of electrons or protons that exist as a function of the total charge, and inversely proportional to the value of the fundamental charge. The number of fundamental unit $|q_e|$ that constitutes a charge of 40.0C can be calculated as

$N = \frac{|Q|}{|q_e|}$

Here,

$q_e$ = Value of charge and it is the fundamental charge

Q = Total Charge

N = Total number of electron or protons

The number of fundamental units is calculated as follows

$N = \frac{10.0C}{1.6*10^{-19}C}$

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3 years ago

A parallel-plate capacitor has 2.10 cm × 2.10 cm electrodes with surface charge densities ±1.00×10-6 C/m2. A proton traveling pa

Darya [45]

Answer:

x=0.53 $x10^{-3} m$

Explanation:

Using Gauss law the field is uniform so

E=ζ/ε

Charge densities ⇒ζ=1. $x10x^{-6} \frac{C}{m^{2}}$

ε=8.85 $x10^{-12} \frac{C^{2}}{n*m^{2}}$

$E=\frac{1x10^{-6}\frac{C}{m^{2}}}{8.85x^{-12}\frac{C^{2} }{N*m^{2}}} \\E=0.11299 x10^{-6} \frac{N}{C}$

Force of charge is

$F_{q}=q*E\\F_{q}=1.6x10^{-19}C*0.11299x10^{6}\frac{N}{C} \\F_{q}=1.807x10^{-14} N$

$F_{q}=m*a\\a=\frac{F_{q}}{m}=\frac{1.807x10^{-13}N}{1.67x10^{-27}}\\ a=1.082x0^{14} \frac{m}{s^{2}} \\t=\frac{x}{v}\\ x=2.1cm\frac{1m}{100cm}=0.021m \\v=6.7x10^{6}\frac{m}{s} \\ t=\frac{0.021m}{6.7x10^{6}\frac{m}{s}} \\t=3.13x10^{-9}s$

So finally knowing the acceleration and the time the distance can be find using equation of uniform motion

$x_{f}=x_{o}+\frac{1}{2}*a*t^{2}\\ x_{o}=0\\x_{f}=\frac{1}{2} a*t^{2}=\frac{1}{2}*1.082x10^{14}\frac{m}{s^{2} } *(3.134x^{-9}s)^{2} \\x_{f}=0.53x^{-3}m$

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3 years ago