How to solve for h using the formula eg=mgh?

Definiti: a)Marimile Fizice Vectoriale <br> b)Fortele Concurente

strojnjashka [21]

No lo se hsjdhdhfhjdhshdhdhhd

5 0

3 years ago

Star A and Star B have measured stellar parallax of 1.0 arc second and 0.75 arc second, respectively. Which star is closer? How

zhuklara [117]

Answer:

Star A is closer than Star B

Explanation:

As we know that in parallax method of distance measurement the angle subtended by the star when it covers a distance of one Parsec arc length, it is known as parallax angle

Here we can say

$angle = \frac{1 Parsec}{distance}$

so we have

$distance = \frac{1 Parsec}{angle}$

so here we have

angle subtended by Star A = 1 arc sec

angle subtended by star B = 0.75 arc sec

now we have

distance for star A is given as

$d_a = \frac{1 Parsec}{1} = 1 Parsec$

distance of star B is given as

$d_b = \frac{1 Parsec}{0.75} = 1.33 Parsec$

So star A is closer than star B

7 0

3 years ago

For a certain ideal Carnot engine, the hot reservoir is 35°C higher than the cold reservoir. If this engine is to have an effici

lord [1]

Answer:

Temperature of the hot reservoir is 1540K

Explanation:

$E= 1- \frac{T_{c}}[tex]{T_h}=308+{T_c}\\Efficiency of a carnot engine is given by the aboveTc=temperature of the cold reservoirTh= temperature of the hot reservoirK=273+ 35 (convert 35°C to kelvin)K=308k{T_h}={T_c}+308-----------------------(equation 1)20%=1-{T_c}/{T_h}$

0.2=({T_c}+308-{T_c})/{T_c}+308

.2({T_c}+61.6=308

0.2{T_c}=246.4

{T_c}=1232

recall from equation 1

{T_h}=308+1232

{T_h}=1540K

5 0

3 years ago

Calculate the entropy change that occurs when 1.0kg of water at 20.00 C is mixed with 2.0kg of water at 80.00 C

SOVA2 [1]

Answer:

The change in entropy ΔS = 0.0011 kJ/(kg·K)

Explanation:

The given information are;

The mass of water at 20.0°C = 1.0 kg

The mass of water at 80.0°C = 2.0 kg

The heat content per kg of each of the mass of water is given as follows;

The heat content of the mass of water at 20.0°C = h₁ = 83.92 KJ/kg

The heat content of the mass of water at 80.0°C = h₂ = 334.949 KJ/kg

Therefore, the total heat of the the two bodies = 83.92 + 2*334.949 = 753.818 kJ/kg

The heat energy of the mixture =

1 × 4200 × (T - 20) = 2 × 4200 × (80 - T)

∴ T = 60°C

The heat content, of the water at 60° = 251.154 kJ/kg

Therefore, the heat content of water in the 3 kg of the mixture = 3 × 251.154 = 753.462

The change in entropy ΔS = ΔH/T = (753.818 - 753.462)/(60 + 273.15) = 0.0011 kJ/(kg·K).

8 0

3 years ago

Suppose you wanted to use a non-reflecting layer for radar waves to make an aircraft invisible. What would the thickness of the

Xelga [282]

Answer:

the thickness of the film for destructive interference is 1 cm

Explanation:

We can assume that the radar wave penetrates the layer and is reflected in the inner part of it, giving rise to an interference phenomenon of the two reflected rays, we must be careful that the ray has a phase change when

* the wave passes from the air to the film with a higher refractive index

* the wavelength inside the film changes by the refractive index

λ = λ₀ / n

so the ratio for destructive interference is

2 n t = m λ

t = m λ / 2n

indicate that the wavelength λ = 2 cm, suppose that the interference occurs for m = 1, therefore it is thickness

t = 1 2/2 n

t = 1 / n

where n is the index of refraction of the anti-reflective layer. As they tell us not to take into account the change in wavelength when penetrating the film n = 1

t = 1 cm

So the thickness of the film for destructive interference is 1 cm

8 0

3 years ago