When light strikes an object, what happens to some of the light?

9. If the frequency of a certain light is 3.8 x 1024 Hz, what is the energy of this light?

german

Answer:

E=hf

Were, h = Planck constant 6.67*10^11

E=3.8*10^24 * 6.67*10^11= 2.508*q0^36j

4 0

2 years ago

Sino tao ang gutom?<br>A.you<br>B.mama mo!<br>C.tamod

Fiesta28 [93]

The answer to that question is c. tamod

5 0

3 years ago

Please check my answers!

Alecsey [184]

A.) The higher the altitude, the colder the climate will be

B.) Areas near the equator have warmer climates than areas for form the equator.

D.) Winds that blow inland from oceans or large lakes contain a lot of water vapor that will cause precipitation.

C.) Monsoons.

8 0

3 years ago

Read 2 more answers

The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ

prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

$\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})$

= $(30)^{2} + 2(-0.25(412 - 0))$

= 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

y = $200 e^{\frac{x}{1000}}$

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

Now, second derivative of the train is calculated as follows.

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

$\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}$

Radius of curvature of the train is as follows.

$\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}$

= $\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}$

= 3808.96 m

Now, we will calculate the normal component of the train as follows.

$a_{n} = \frac{\nu^{2}_{B}}{\rho}$

= $\frac{(26.34)^{2}}{3808.96}$

= 0.1822 $m/s^{2}$

The magnitude of acceleration of train is calculated as follows.

a = $\sqrt{(a_{t})^{2} + (a_{n})^{2}}$

= $\sqrt{(-0.25)^{2} + (0.1822)^{2}}$

= $0.309 m/s^{2}$

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is $0.309 m/s^{2}$ .

6 0

3 years ago

A circular loop of flexible iron wire has an initial circumference of 165cm , but its circumference is decreasing at a constant

ArbitrLikvidat [17]

Answer:

emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise

Explanation:

Given data

initial circumference = 165 cm

rate = 12.0 cm/s

magnitude = 0.500 T

tome = 9 sec

to find out

emf induced and direction

solution

we know emf in loop is - d∅/dt ........1

here ∅ = ( BAcosθ)

so we say angle is zero degree and magnetic filed is uniform here so that

emf = - d ( BAcos0) /dt

emf = - B dA /dt ..............2

so area will be

dA/dt = d(πr²) / dt

dA/dt = 2πr dr/dt

we know 2πr = c,

r = c/2π = 165 / 2π

r = 26.27 cm

c is circumference so from equation 2

emf = - B 2πr dr/dt ................3

and

here we find rate of change of radius that is

dr/dt = 12/2π = 1.91 $10^{-2}$ cm/s

so when 9.0s have passed that radius of coil = 26.27 - 191 (9)

radius = 9.08 $10^{-2}$ cm

so now from equation 3 we find emf

emf = - (0.500 ) 2π(9.08 $10^{-2}$ ) 1.91 $10^{-2}$

emf = - 0.005445

and magnitude of emf = 0.005445 V

so

emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise

4 0

3 years ago