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3 years ago

The elementary charge equals 1.6 x 10^-9 C. a. True b. False

Physics

1 answer:

yanalaym [24]3 years ago

5 0

Answer:

False

Explanation:

The magnitude of electric charge carried by an electron as well as proton is called elementary charge. Both the electron and proton have the same charge but the difference is that the electrons are negatively charged species and the protons are positively charged.

The elementary charge is equal to $1.6\times 10^{-19}\ C$ . Hence, the given statement "The elementary charge equals $1.6\times 10^{-9}\ C$ " is false.

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Physicists and engineers from around the world have come together to build the largest accelerator in the world, the Large Hadro

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Solution :

Energy of photon, E = 6.7 eV

E = $6.7 \times 1.602 \times 10^{-7}$ joule

Kinetic energy, $K.E. =\frac{1}{2} mv^2 = 1.602 \times 6.7 \times 10^{-7}$

$v^2=\frac{2 \times 1.602 \times 6.7 \times 10^{-7}}{1.6726 \times 10^{-27}}$

$=12.834 \times 10^{-20}$

Kinetic energy at high speeds

$(r-1)\times mc^2 = 6.7 \ eV$

$(r-1)=\frac{6.7 \times 1.602 \times 10^{-7}}{1.6726 \times 10^{-27} \times 9 \times 10^{16}}$

r - 1 = 7130

r = 7130 + 1

r = 7131

$\frac{1}{\sqrt{1-\frac{v^2}{C^2}}}=7131$

$1-\frac{v^2}{C^2} = \left(\frac{1}{7131}\right)^2$

$v^2=C^2\left[1-\left(\frac{1}{7131}\right)^2\right]$

$v=0.99999999017C$

Δ = 1 - 0.99999999017

= 0.00000000933

Relative mass, $m_{rel}=r.m$

$=7131 \times 1.6728 \times 10^{-27}$

$=1.1927 \times 10^{-23}$ kg

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3 years ago

A computer software update involved updating the flash memory of a hardware component. The update failed. A phone technician sai

Alexandra [31]

Answer:

Yes, it's true. Computers do work that way. It's experienced by one of the authors of the book how computers work.

Explanation:

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3 years ago

Two thin wires rings each having a radius R are placed at a distance d apart with their axes coinciding. The charges on the two

tigry1 [53]

Answer:

Q/4πε0 [1/R - 1/√R2+d2]

Explanation:

Q/4πε0 [1/R - 1/√R2+d2] is the answer

explanation is attached.

toppr

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2 years ago

A food packet is dropped from a helicopter during a flood-relief operation from a height of 750 meters. Assuming no drag (air fr

Leviafan [203]

1. Velocity at which the packet reaches the ground: 121.2 m/s

The motion of the packet is a uniformly accelerated motion, with constant acceleration $a=g=9.8 m/s^2$ directed downward, initial vertical position $d=750 m$ , and initial vertical velocity $v_0 = 0$ . We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

$v_f^2 -v_i^2 =2ad$

substituting, we find

$v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s$

2. height at which the packet has half this velocity: 562.6 m

We need to find the heigth at which the packet has a velocity of

$v_f=\frac{121.2 m/s}{2}=60.6 m/s$

In order to do that, we use again the same SUVAT equation substituting $v_f$ with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:

$v_f^2-v_i^2=2ad\\d=\frac{v_f^2}{2a}=\frac{(60.6 m/s)^2}{2(9.8 m/s^2)}=187.4 m$

Which means that the heigth of the packet was

$h=750 m-187.4 m=562.6 m$

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3 years ago

Define orbital velocity

sveta [45]

the Orbital Velocity is the velocity sufficient to cause a natural or artificial satellite to remain in orbit. Inertia of the moving body tends to make it move on in a straight line, while gravitational force tends to pull it down. The orbital path, elliptical or circular, representing a balance between gravity and inertia, and it follows a rue that states that the more massive the body at the centre of attraction is, the higher is the orbital velocity for a particular altitude or distance.

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3 years ago