How is using a digital signal like turning on and off a regular light<br> switch?

A submarine is 2.84 102 m horizontally from shore and 1.00 102 m beneath the surface of the water. A laser beam is sent from the

xxMikexx [17]

Answer:

468 m

Explanation:

So the building and the point where the laser hit the water surface make a right triangle. Let's call this triangle ABC where A is at the base of the building, B is at the top of the building, and C is where the laser hits the water surface. Similarly, the submarine, the projected submarine on the surface and the point where the laser hit the surface makes a another right triangle CDE. Let D be the submarine and E is the other point.

The length CE is length AE - length AC = 284 - 234 = 50 m

We can calculate the angle ECD:

$tan(\hat{ECD}) = \frac{ED}{EC} = \frac{100}{50} = 2$

$\hat{ECD} = tan^{-1} 2 = 63.43^o$

This is also the angle ACB, so we can find the length AB:

$tan(\hat{ACB}) = \frac{AB}{AC} = \frac{AB}{234}$

$2 = \frac{AB}{234}$

$AB = 2*234 = 468 m$

So the height of the building is 468m

5 0

3 years ago

An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm,

insens350 [35]

Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

$Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}$

Where

$t_{hf}$ = Temperature at the inside of the pipe = 300 °C

$t_{f}$ = Temperature at the outside of the pipe = 20 °C

r₁ =internal radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

$k_A$ = 15 W/m·K

$k_B$ = 0.038 W/m·K

$h_{hf}$ = 75 W/m²·K

$h_{cf}$ = 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

Q = 131.32 W

From which we have, for the film of air at the pipe outer boundary layer

$Q = \frac{t_A-t_B}{R_T}$ Where $R_T$ for the air film on the pipe outer surface is given by

$R_T= \frac{1}{\alpha A}$

where A =area of the outside of the pipe

= $\frac{1}{10*2\pi*0.07*1 }$ = 0.227 K/W

Therefore

131.32 W = $\frac{t_A-20}{0.227}$ which gives

$t_A$ = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

8 0

3 years ago

A car has a mass of 1000 km kilometers. The weight of the car is blank newtons use G equals 9.8 in KG for gravity? Help mehhhhh

Schach [20]

The weight of the car in the picture of the computer screen is 9,800 Newton's.

5 0

3 years ago

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

DiKsa [7]

Answer:0.061

Explanation:

Given

$T_C=300 k$

Temperature of soup $T_H=340 K$

heat capacity of soup $c_v=33 J/K$

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any instant

efficiency is given by

$\eta =\frac{dW}{Q}=1-\frac{T_C}{T}$

$dW=Q(1-\frac{T_C}{T})$

$dW=c_v(1-\frac{T_C}{T})dT$

integrating From $T_H$ to $T_C$

$\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT$

$W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT$

$W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}$

$W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]$

Now heat lost by soup is given by

$Q=c_v(T_C-T_H)$

Fraction of the total heat that is lost by the soup can be turned is given by

$=\frac{W}{Q}$

$=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}$

$=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}$

$=\frac{300-340-300\ln (\frac{300}{340})}{300-340}$

$=\frac{-40+37.548}{-40}$

$=0.061$

4 0

3 years ago

What is the net force exerted on a 28.8 kg shopping cart that accelerates at a rate of 2.88 m/s^2?

Mama L [17]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 28.8 × 2.88 = 82.944

We have the final answer as

Hope this helps you

6 0

3 years ago

How is using a digital signal like turning on and off a regular light switch?