What is (9.8) / (6.75x10-6)

An electron in a television tube is accelerated uniformly from rest to a speed of 8.4\times 10^7~\text{m/s}8.4×10 7 m/s over

stich3 [128]

Answer:

P=3.42×10^-6 J/s

Explanation:

From the kinematics of motion with constant acceleration we know that :

vf^2=vi^2+2*a(xf-xi)

Where :

• vf , vi, are the the final and the initial velocity of the electron

• a is the acceleration of the electron

• xf , xi are the final and the initial position of the electron .

Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.

Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m

vf^2 =vi^2+2*a(xf-xi)

vf^2-vi^2=2*a(xf-xi)

2*a(xf-xi)= vf^2-vi^2

a = (vf^2-vi^2)/2(xf-xi)

Pluging known information to get :

a = (vf^2-vi^2)/2(xf-xi)

= 1.411 × 10^17

From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m

so,

vf^2 =vi^2+2*a(xf-xi)

vf^2 =5.312× 10^7

From the following Eq. we can calculate the time elapsed in this motion .

xf =xi+vi*t+1/2*a*t

t=√2(xf-xi)/a

t=3.765×10^-10 s

now we can use the power P Eq.

P=W/Δt => ΔK/Δt

Where: the work done W change the kinetic energy K of the electron ,

ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2

P=1/2*m*vf^2-1/2*m*vi^2/Δt

P=3.42×10^-6 J/s

6 0

3 years ago

What is perfect machine?Why is this machine not possible in the real life?

andrezito [222]

Explanation:

The machine whose efficiency is 100% is known as perfect machine .This machine is not possible in real life because every machine is affected by the overcoming friction due to which is efficiency become less than hundred percent .

5 0

2 years ago

Science and math people where you at

PilotLPTM [1.2K]

For the second question you’re solving for resistance. resistance= voltage/ current. 120/0.5= 240. the answer is 240 ohms
for the third question you would do 2*4 since it’s asking for voltage, the answer is 8 volts :)

8 0

3 years ago

A racecar driver steps on the gas, and his racecar travels 20 meters in 2 seconds starting from rest. The acceleration of the ra

givi [52]

Answer:

2.5m/s^2

Explanation:

Step one:

given

distance = 20meters

time = 2 seconds

initial velocity u= 0m/s

let us solve for the final velocity

velocity = distance/time

velocity= 20/2

velocity= 10m/s

$v^2=u^2+2as\\\\10^2=0^2+2*a*20\\\\100=40a$

divide both sides by 40

$a= 100/40\\\\a=10/4\\\\a= 5/2\\\\a= 2.5m/s^2$

5 0

3 years ago

a projectile is shot horizontally from the edge of a cliff, 230m above the ground. the projectile lands 300m from base of the cl

bekas [8.4K]

Answer:

The time taken by the projectile to hit the ground is 6.85 sec.

Explanation:

Given that,

Vertical height of cliff = 230 m

Distance = 300 m

Suppose, determine the time taken by the projectile to hit the ground.

We need to calculate the time

Using second equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Where, s = vertical height of cliff

u = initial vertical velocity

g = acceleration due to gravity

Put the value in the equation

$230=0+\dfrac{1}{2}\times9.8\times t^2$

$t=\sqrt{\dfrac{230\times2}{9.8}}$

$t=6.85 sec$

Hence, The time taken by the projectile to hit the ground is 6.85 sec.

7 0

3 years ago