Young's interference experiment demonstrated the particle nature of light T/F?

In a particular experiment to study the photoelectric effect, the frequency of the incident light and the temperature of the met

qwelly [4]

Answer:

B. The number of electrons emitted from the metal per second increases.

Explanation:

Light consists of photons . Energy of each photon depends upon frequency of light . The increase in intensity increases the number of photons . It does not increase energy of photons .

So if a high intensity light falls on a photosensitive plate , each photon ejects one electron . So number of electrons increases if we increase intensity of photon. It does not increase kinetic energy of ejected electrons . Work function depends upon the nature of plate.

6 0

3 years ago

The position of a car at time t is given by the function p(t)=t2 2t−4. What is the velocity when p(t)=11? assume t≥0

AysviL [449]

The velocity when function p(t)=11 is 8 .

According to the question

The position of a car at time t represented by function :

$p(t)=t^{2} +2t-4$

Now,

When function p(t) = 11 , t will be

$p(t)=t^{2} +2t-4$

11 = t²+2t-4

0 = t² + 2t - 15

or

t² +2t-15 = 0

t² +(5-3)t-15 = 0

t² +5t-3t-15 = 0

t(t+5)-3(t+5) = 0

(t-3)(t+5) = 0

t = 3 , -5

as t cannot be -ve as given ( t≥0)

so,

t = 3

Now,

the velocity when p(t)=11

As we know velocity = $\frac{position}{time}$

therefore to get the value of velocity from function p(t)

we have to differentiate the function with respect to time

$\frac{d(p(t))}{dt} =\frac{d}{dt} (t^{2} +2t-4)$

v(t) = 2t + 2

where v(t) = velocity at that time

as t = 3 for p(t)=11

so ,

v(t) = 2t + 2

v(t) = 2*3 + 2

v(t) = 8

Hence, the velocity when function p(t)=11 is 8 .

To know more about function here:

brainly.com/question/12431044

#SPJ4

4 0

2 years ago

What is the velocity of a wave with a wavelength of 1.5m and a frequency of 500 Hz

Schach [20]

Wave speed = (wavelength) x (frequency)

= (1.5 m) x (500 / sec)

= 750 m/s .

4 0

3 years ago

How far does light travel in the time it takes sound to travel 1 cm in air at 20°c?

kykrilka [37]

The speed of sound at $20^{\circ}C$ is approximately v=343 m/s. The distance covered by the sound wave is
$s=1 cm=0.01 m$
And the time it takes is
$t= \frac{S}{v}= \frac{0.01 m}{343 m/s}=2.9 \cdot 10^{-5} s$

Now we want to find how far the light travels during this time. Light travels at speed $c=3 \cdot 10^8 m/s$ , therefore the distance it covers during this time is
$S=ct = (3 \cdot 10^8 m/s)(2.9 \cdot 10^{-5} s)=8700 m$

8 0

3 years ago

Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting li

erica [24]

Answer:

(a) 10 s

(b) 236.5 m

(c) Kathy's speed = 47.3 m/s

Stan's speed = 42.9 m/s

Explanation:

<u>Given:</u>

$u_k$ = initial speed of Kathy = 0 m/s

$u_s$ = initial speed of Stan = 0 m/s
$a_k$ = acceleration of Kathy = $4.73\ m/s^2$

$a_s$ = acceleration of Stan = $3.9\ m/s^2$

<u>Assumptions:</u>

$v_k$ = final speed of Kathy when see catches Stan
$v_s$ = final speed of Stan when Kathy catches him
$s_k$ = distance traveled by Kathy to catch Stan
$s_s$ = distance traveled by Stan when Kathy catches him
$t_k$ = time taken by Kathy to catch Stan = $t$
$t_s$ = time interval in which Kathy catches Stan = $t+1$

Part (a):

Kathy will catch Stan only if the distances traveled by each of them are equal at the same instant.

$\therefore s_s=s_k\\\Rightarrow u_st_s+\dfrac{1}{2}a_st_s^2=u_kt_k+\dfrac{1}{2}a_kt_k^2\\ \Rightarrow (0)(t+1)+\dfrac{1}{2}(3.9)(t+1)^2=(0)(t)+\dfrac{1}{2}(4.73)t^2\\ \Rightarrow \dfrac{1}{2}(3.9)(t+1)^2=\dfrac{1}{2}(4.73)t^2\\\Rightarrow (3.9)(t+1)^2=(4.73)t^2\\\Rightarrow \dfrac{(t+1)^2}{t^2}=\dfrac{4.73}{3.9}\\\textrm{Taking square root in both sides}\\\dfrac{t+1}{t}= 1.1\\\Rightarrow t+1=1.1t\\\Rightarrow 0.1t = 1\\\Rightarrow t = 10\\$

Hence, Kathy catches Stan after 11 s from the Stan's starting times.

Part (b):

Distance traveled by Kathy to catch Stan will be distance the distance traveled by her in 10 s.

$s_s = u_kt_k+\dfrac{1}{2}a_kt_k^2\\\Rightarrow s_s= (0)(t)+\dfrac{1}{2}(4.73)t^2\\\Rightarrow s_s= \dfrac{1}{2}(4.73)(10)^2\\\Rightarrow s_s= 236.5$

Hence, Kathy traveled a distance of 236.5 m to overtake Stan.

Part (c):

$v_k = u_k+a_kt_k\\\Rightarrow v_k = 0+(4.73)(t)\\\Rightarrow v_k = (4.73)(10)\\\Rightarrow v_k =47.3$

The speed of Kathy at the instant she catches Stan is 47.3 m/s.

$v_s = u_s+a_st_s\\\Rightarrow v_s = 0+(3.9)(t+1)\\\Rightarrow v_s = (3.9)(10+1)\\\Rightarrow v_s =42.9$

The speed of Stan at the instant Kathy catches him is 42.9 m/s.

7 0

3 years ago