All categories

3 years ago

Question 10 of 10

Physics

2 answers:

Alex787 [66]3 years ago

5 0

Answer:

It allows only a reduced number of electrons to flow through it

Explanation:

No B is the best answer

Alinara [238K]3 years ago

4 0

it allows only a reduced number of electrons to flow through it.

You might be interested in

From a vantage point very close to the track at a stock car race, you hear the sound emitted by a moving car. You detect a frequ

V125BC [204]

Answer:

55.8 m/s

Explanation:

$f$ = Actual frequency of sound emitted by car

$f'$ = frequency observed when the car moves = $(0.86)f$

$V$ = Speed of sound = 343 m/s

$v$ = Speed of car

Frequency observed is given as

$f' = \frac{Vf}{V+v}$

$(0.86) = \frac{(343)}{343 + v}\\(0.86) = \frac{(343)}{343 + v}\\294.98 + (0.86) v = 343 \\v = 55.8 ms^{-1}$

6 0

3 years ago

Pls help I need this done soon 50 points

lidiya [134]

Answer:

1,3,5

Explanation:

i think maybe dont come at me

6 0

3 years ago

Read 2 more answers

The velocity of a particle moving along the x-axis varies with time according to v(t) = A + Bt−1 , where A = 2 m/s, B = 0.25 m,

kondaur [170]

Answer:

$a= -2\ m/s^2$

$a=-12.5\ m/s^2$

x=2.17 m

x=8.4 m

Explanation:

Given that

$v=A+Bt^{-1}$

$v=2+0.25t^{-1}$

To find acceleration :

we know that

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

$\dfrac{dv}{dt}=0-0.5t^{-2}$

$a=-0.5t^{-2}$

Acceleration at t= 2 s

$a=-0.5\times 2^{-2}$

$a= -2\ m/s^2$

Acceleration at t= 5 s

$a=-0.5\times 5^{-2}$

$a=-12.5\ m/s^2$

We know that

$v=\dfrac{dx}{dt}$

$dx=\left(2+\dfrac{1}{4t}\right)dt$

Position at t= 2 s:

$\int_{0}^{x}dx=\int_{1}^{2} \left(2+0.25\dfrac{1}{t}\right)dt$

$x=\left [2t+0.25\ lnt \right ]_{1}^{2}$

x=2+0.25 ln2

x=2.17 m

Position at t= 5 s:

$\int_{0}^{x}dx=\int_{1}^{5} \left(2+0.25\dfrac{1}{t}\right)dt$

$x=\left [2t+0.25\ lnt \right ]_{1}^{5}$

x=8+0.25 ln5

x=8.4 m

4 0

3 years ago

Interference occurs with not only light waves but also all frequencies of electromagnetic waves and all other types of waves, su

telo118 [61]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.

4 0

3 years ago

Can anybody help me solve this problem? Thank you so much!

ser-zykov [4K]

Please ignore my comment -- mass is not needed, here is how to solve it. pls do the math

at bottom box has only kinetic energy
ke = (1/2)mv^2
v = initial velocity
moving up until rest work done = Fs
F = kinetic fiction force = uN = umg x cos(a)
s = distance travel = h/sin(a)
h = height at top
a = slope angle
u = kinetic fiction
work = Fs = umgh x cot(a)
ke = work (use all ke to do work)
(1/2)mv^2 = umgh x cot(a)
u = (1/2)v^2 x tan (a) / gh

4 0

2 years ago