All categories

2 years ago

An orange light (f = 5.2 * 10'4Hz) is

Physics

1 answer:

KonstantinChe [14]2 years ago

7 0

Answer:

2.145×10^-10 V or 0.2145nV

Explanation:

From hf=eV

h= Plank's constant = 6.6×10^-34JS

f= frequency of the electromagnetic wave = 5.2×10^4 Hz

e= electronic charge= 1.6×10^-19 C

V= voltage

V= hf/e

V= 6.6×10^-34JS × 5.2×10^4 Hz/ 1.6×10^-19 C

V= 2.145×10^-10 V or 0.2145nV

Therefore the voltage created is 2.145×10^-10 V or 0.2145nV

You might be interested in

Calculate the acceleration of a 1000 kg car if the motor provides a small thrust of 1000 N and the static and dynamic friction c

grin007 [14]

Explanation :

It is given that,

Mass of the car, m = 1000 kg

Force applied by the motor, $F_A=1000\ N$

The static and dynamic friction coefficient is, $\mu=0.5$

Let a is the acceleration of the car. Since, the car is in motion, the coefficient of sliding friction can be used. At equilibrium,

$F_A-\mu mg=ma$

$\dfrac{F_A-\mu mg}{m}=a$

$a=\dfrac{1000-0.5(1000)(9.81)}{1000}$

$a=-3.905\ m/s^2$

So, the acceleration of the car is $-3.905\ m/s^2$ . Hence, this is the required solution.

6 0

2 years ago

Can the velocity of a body revese the direction when acceleration is constant?

TEA [102]

Answer:

Yes, the velocity of the object can reverse direction when its acceleration is constant. For example consider that the velocity of any object at any time t is given as: ... At At t = 0 sec, the magnitude of velocity is 2m/s and is moving in the forward direction i.e.v (t) = -2.

7 0

2 years ago

A boat travels north across a river at a velocity of 22 meters/second with respect to the water. The river's velocity is 2.2 met

MArishka [77]

The magnitude of the resultant is

√ (22² + 2.2²) = √ (484 + 4.84) = √488.84 = 22.11 m/s .

The direction of the resultant is

tan⁻¹(22N / 2.2E) = tan⁻¹(10) = 5.71° east of north .

7 0

3 years ago

A forklift raises a 1,020 N crate 3.50 m up to a shelf. How much work is done by the forklift on the crate?

Anarel [89]

Hope this will help u..:)

8 0

2 years ago

Read 2 more answers

a projectile is launched at an angle of 30 degrees and lands later at the same level. if it's initial speed is 50 m/s, solve for

Mrrafil [7]

$using \: the \: formula \\ t = \frac{2u \sin( \alpha ) }{g} where \: u = initial \: speed \: \\ \alpha = angle \: of \: projection \\ g = acceleration \: due \: to \: gravity \\ \frac{2 \times 50 \times \sin(30) }{10} \\ \frac{100 \times 0.5}{10} = \frac{50}{10} = 5seconds$

Maximum height
= (Usinα)^2/2g
(50*0.5)^2/20
25^2/20
625/20
=31.25metres
horizontal distance = Range= [U^2 * sin2α]/g
[50^2 * sin60]/10
2500 * 0.8660/10
2165/10=216.5metres

3 0

2 years ago