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2 years ago

How is light energy transported?

Physics

1 answer:

REY [17]2 years ago

4 0

Answer:

By reflection

Explanation:

The light energy can be transported by the means of transmitting it by reflecting it to several surfaces . as we know , light doesn't need any medium to travel . Electromagnetic waves are used to transmit them .

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A block of density pb = 9.50 times 10^2 kg/m^3 floats face down in a fluid of density pt = 1.30 times 10^3 kg/m^3. The block has

Nookie1986 [14]

Answer:

The depth and acceleration are 0.1919291 ft and 3.61 m/s².

Explanation:

Given that,

Density of block $\rho_{b} =9.50\times10^2\ kg/m^3$

Density of fluid $\rho_{t} =1.30\times10^3\ kg/m^3$

We need to calculate the depth

Using balance equation

$mg=\rho g V$ ....(I)

We know that,

The density is

$\rho=\dfrac{m}{V}$

$m=\rh0\times V$

Put the value of m in equation (I)

$\rho_{b}\times V\times g=\rho_{t}\times g\times V$

$\rho_{b}\times A\times H\times g=\rho_{t}\times g\times A\times h$

$h=\dfrac{\rho_{b}H}{\rho_{t}}$

Put the value into the formula

$h=\dfrac{9.50\times10^2\times8.00\times10^{-2}}{1.30\times10^3}$

$h= 5.85\ cm$

$h=0.1919291\ ft$

We need to calculate the acceleration

Using formula of net force

$F_{net}=\rho_{t}\times g\times V- \rho_{b}\times g\times V$

$ma=\rho_{t}\times g\times V- \rho_{b}\times g\times V$

$\rho_{b}\times V\times a=\rho_{t}\times g\times V- \rho_{b}\times g\times V$

$a=(\dfrac{\rho_{t}}{\rho_{b}}-1)g$ ....(II)

Put the value in the equation (II)

$a=(\dfrac{1.30\times10^3}{9.50\times10^2}-1)\times9.8$

$a=3.61\ m/s^2$

Hence, The depth and acceleration are 0.1919291 ft and 3.61 m/s².

4 0

2 years ago

A bubble, located 0.200 m beneath the surface in a glass of beer, rises to the top. The air pressure at the top is 1.01 ???? 10

SIZIF [17.4K]

Answer:

the ratio of the bubble’s volume at the top to its volume at the bottom is 1.019

Explanation:

given information

h = 0.2 m

$P_{0}$ = 1.01 x $10^{5}$ Pa

$P_{1} V_{1} = P_{2} V_{2}$

$\frac{V_{2} }{V_{1}} = \frac{P_{1} }{P_{2}}$

$P_{1}$ = $P_{0}$ + ρgh, ρ = 1000 kg/ $m^{3}$

$P_{1}$ = 1.01 x $10^{5}$ Pa + (1000 x 9.8 x 0.2) = 1,0296 x $10^{5}$ Pa

$P_{2}$ = $P_{0}$ = $10^{5}$ Pa

thus,

$\frac{V_{2} }{V_{1}} = 1,0296 x [tex]10^{5}$ / $10^{5}$ = 1.019

8 0

3 years ago

To increase the current in a circuit, the BLANK<br> can be increased. HELP ASAP

sladkih [1.3K]

Voltage is your answer

3 0

3 years ago

Read 2 more answers

A parallel-plate capacitor has a plate area of 0.2m^2 and a plate separation of 0.1mm. To obtain an electric field of 2.0 × 10^6

Oduvanchick [21]

Answer:

3.536*10^-6 C

Explanation:

The magnitude of the charge is expresses as Q = CV

C is the capacitance of the capacitor

V is the voltage across the capacitor

Get the capacitance

C = ε0A/d

ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m

A is the area = 0.2m²

d is the plate separation = 0.1mm = 0.0001m

Substitute

C = 8.84 x 10-12 * 0.2/0.0001

C = 1.768 x 10-8 F

Get the potential difference V

Using the formula for Electric field intensity

E = V/d

2.0 × 10^6 = V/0.0001

V = 2.0 × 10^6 * 0.0001

V = 2.0 × 10^2V

Get the charge on each plate.

Q = CV

Q = 1.768 x 10-8 * 2.0 × 10^2

Q = 3.536*10^-6 C

Hence the magnitude of the charge on each plate should be 3.536*10^-6 C

5 0

2 years ago

_____ you remember to put the lid back on the jar of mayonnaise

Lilit [14]

Answer:

Did you remember to put the lid back on the jar of mayonnaise?

Explanation: Hope this helps :)

7 0

2 years ago