Calculate the intensity of current flowing through a computer that consumes 180W and operates at 120 V.

A photograph is taken by letting light fall on a light-sensitive medium, which then records the image onto that medium. True or

Jet001 [13]

Answer;

The above statement is true.

-A photograph is taken by letting light fall on a light-sensitive medium, which then records the image onto that medium.

Explanation;

-A photograph is created when light is allowed to fall on a light-sensitive medium. The pattern of light creates an image that is recorded by the photographic device. How light or dark a photograph is depends on how much light was allowed to fall on the light-sensitive medium.

-A camera is a light-tight box that contains a light-sensitive material or device and a way of letting in a desired amount of light at particular times to create an image on the light-sensitive material.

6 0

3 years ago

A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

3 0

3 years ago

A slit of width 2.0 μm is used in a single slit experiment with light of wavelength 650 nm. If the intensity at the central maxi

Valentin [98]

Answer:

The intensity at 10° from the center is 3.06 × 10⁻⁴I₀

Explanation:

The intensity of light I = I₀(sinα/α)² where α = πasinθ/λ

I₀ = maximum intensity of light

a = slit width = 2.0 μm = 2.0 × 10⁻⁶ m

θ = angle at intensity point = 10°

λ = wavelength of light = 650 nm = 650 × 10⁻⁹ m

α = πasinθ/λ

= π(2.0 × 10⁻⁶ m)sin10°/650 × 10⁻⁹ m

= 1.0911/650 × 10³

= 0.001679 × 10³

= 1.679

Now, the intensity I is

I = I₀(sinα/α)²

= I₀(sin1.679/1.679)²

= I₀(0.0293/1.679)²

= 0.0175²I₀

= 0.0003063I₀

= 3.06 × 10⁻⁴I₀

So, the intensity at 10° from the center is 3.06 × 10⁻⁴I₀

5 0

3 years ago

4.2 mol of monatomic gas A interacts with 3.2 mol of monatomic gas B. Gas A initially has 9500 J of thermal energy, but in the p

Komok [63]

Answer:

14657.32 J

Explanation:

Given Parameters ;

Number of moles mono atomic gas A , n 1 = 4 .2 mol

Number of moles mono atomic gas B , n 2 = 3.2mol

Initial energy of gas A , K A = 9500 J

Thermal energy given by gas A to gas B , Δ K = 600 J

Gas constant R = 8.314 J / molK

Let K B be the initial energy of gas B.

Let T be the equilibrium temperature of the gas after mixing.

Then we can write the energy of gas A after mixing as

(3/2)n1RT = KA - ΔK

⟹ (3/2) x 4.2 x 8.314 x T = 9500 - 600

T = (8900 x 3 )/(2x4.2x8.314) = 382.32 K

Energy of the gas B after mixing can be written as

(3/2)n2RT = KB + ΔK

⟹ (3/2) x 3.2 x 8.314 x 382.32 = KB + 600

⟹ KB = [(3/2) x 3.2 x 8.314 x 382.32] - 600

⟹ KB = 14657.32 J

6 0

3 years ago

What is the temperature of a sample of gas when the average translational kinetic energy of a molecule in the sample is 8.37 × 1

-BARSIC- [3]

Answer:

404K

Explanation:

Data given, Kinetic Energy.K.E=8.37*10^-21J

Note: as the temperature of a is increase, the rate of random movement will increase, hence leading to more collision per unit time. Hence we can say that the relationship between the kinetic energy and the temperature is a direct variation.

This relationship can be expressed as

$K.E=\frac{3}{2}KT$

where K is a constant of value 1.38*10^-23

Hence if we substitute the values, we arrive at

$T=\frac{2/3(8.37*10^{-21})}{1.38*10^-23}\\ T=404K$

converting to degree we have $131^{0}C$

4 0

3 years ago