Secondary evidence is the basis for drawing scientific conclusions. <br> a. True<br> b. False

An electron initially has a speed 16 km/s along the x-direction and enters an electric field of strength 27 mV/m that points in

weqwewe [10]

Answer:

a) $t=1.4\times 10^{-5}\ s$

b)S= 46.4 cm

Explanation:

Given that

Velocity = 16 Km/s

V= 16,000 m/s

E= 27 mV/m

E=0.027 V/m

d= 22.5 cm

d= 0.225 m

a)

lets time taken by electron is t

d = V x t

0.225 = 16,000 t

$t=1.4\times 10^{-5}\ s$

b)

We know that

F = m a = E q ------------1

Mass of electron ,m

$m=9.1\times 10^{-31}\ kg$

Charge on electron

$q=1.6\times 10^{-19}\ C$

So now by putting the values in equation 1

$a=\dfrac{E q}{m}$

$a=\dfrac{1.6\times 10^{-19}\times 0.027}{9.1\times 10^{-31}}\ m/s^2$

$a=4.74\times 10^{9}\ m/s^2$

$S= ut+\dfrac{1}{2}at^2$

Here initial velocity u= 0 m/s

$S= \dfrac{1}{2}\times 4.74\times 10^{9}\times (1.4\times 10^{-5})^2\ m$

S=0.464 m

S= 46.4 cm

S is the deflection of electron.

4 0

4 years ago

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A string along which waves can travel is 4.36 m long and has a mass of 222 g. The tension in the string is 60.0 N. What must be

lora16 [44]

Answer:

frequency is 195.467 Hz

Explanation:

given data

length L = 4.36 m

mass m = 222 g = 0.222 kg

tension T = 60 N

amplitude A = 6.43 mm = 6.43 × $10^{-3}$ m

power P = 54 W

to find out

frequency f

solution

first we find here density of string that is

density ( μ )= m/L ................1

μ = 0.222 / 4.36

density μ is 0.050 kg/m

and speed of travelling wave

speed v = √(T/μ) ...............2

speed v = √(60/0.050)

speed v = 34.64 m/s

and we find wavelength by power that is

power = μ×A²×ω²×v / 2 ....................3

here ω is wavelength put value

54 = ( 0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 ) / 2

0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 = 108

ω² = 108 / 7.160 × $10^{-5}$

ω = 1228.16 rad/s

so frequency will be

frequency = ω / 2π

frequency = 1228.16 / 2π

frequency is 195.467 Hz

7 0

4 years ago

How do Newton's laws of motion explain why it is important to keep the ice smooth on a hockey rink so that players can

lord [1]

Answer:

I'm not sure..but please refer to your teacher later.

Answer: Based on Newton's First law of motion (where inertia is involved), smooth ice increases the forceused to accelerate the hockey puck.

Explanation;

smooth ice reduces the resistances between the surface of the figure skates and the ice itself.
based on inertia theory ; the heavier the weight, the larger the inertia.. which explains it takes alot of force to move a heavier object than the lighter ones.. it also hard to *stop* the motion of heavier objects than the lighter ones.
now let's look at the design of the player shoe itself, they have a sharp blade at the bottom of the figure stakes.. which takes us to the law of the force.. the smaller the surface area, the more forces acting on it. So, players force (weight, F= mg) acts on the tip of the blade and on the ice
high inertia (run fast) and high force (attack opponent and pass puck) enables them to perform well in playing hockey
Thus if there's no resistance and the inertia of the player is high then they could run and pass the puck quickly

6 0

3 years ago

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Q 2 Two mirrors meet at right angles. A ray of light is incident on one at an angle of 30°

serg [7]

a ray of light is incident towards a plane mirror at an angles of 30degrees with the mirror surface. what will be the angles of reflection is 60degree.

3 0

3 years ago

Define the fundamental difference between kinematics and dynamics. .

Tcecarenko [31]

In kynematics you describe the motion of particles using vectors and their change in time. You define a position vector r for a particle, and then define velocity v and acceleration a as

$v=\frac{dr}{dt} \\$

$a=\frac{dv}{dt}$

In dynamics Newton's laws predict the acceleration for a given force. Knowing the acceleration, and the kynematical relations defines above, you can solve for the position as a function of time: r(t)

5 0

3 years ago

Secondary evidence is the basis for drawing scientific conclusions. a. True b. False