Use Newton's method to determine the angle θ, between 0 and π/2 accurate to six decimal places. for which sin(θ) = 0.1. Show your work until you start computing x1, etc. Then just write down what your calculator gives you.

Answers

Answer 1
Answer:

Answer:

x3=0.100167

Explanation:

Let's find the answer.

Because we are going to find the solution for sin(Ф)=0.1 then:

f(x)=sin(Ф)-0.1 and:

f'(x)=cos(Ф)

Because 0<Ф<π/2 let's start with an initial guess of 0.001 (x0), so:

x1=x0-f(x0)/f'(0)

x1=0.001-(sin(0.001)-0.1)/cos(0.001)

x1= 0.100000

x2=0.100000-(sin(0.100000)-0.1)/cos(0.100000)

x2=0.100167

x3=0.100167


Related Questions

4. Use voltage divider concepts to find the voltages indicated in the following circuits. You may want to use some of your results from problem 1. You may need to use the voltage divider equation more than once.
WHEN A CAR WITH BRIGHT HEADLIGHTS COMES TOWARD YOU AT NIGHT, YOU SHOULD:A. Move toward the right edge of your lane B. Look above the oncoming headlights C. Look below the oncoming headlights D. Look toward the right edge of your lane Help
A insulated vessel s has two compartments separated by a membreane. On one side is 1kg of steam at 400 degC and 200 bar. The other side is evacuated . The membrane ruptures, filling the entire volume. The finial pressure is 100bar. Determine the final temperature of the steam and the volume of the vessel.
What is 29.95 inHg in kPa?
Air enters the first compressor stage of a cold-air standard Brayton cycle with regeneration and intercooling at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The overall compressor pressure ratio is 10, and the pressure ratios are the same across each compressor stage. The temperature at the inlet to the second compressor stage is 300 K. The turbine inlet temperature is 1400 K. The compressor stages and turbine each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k = 1.4, calculate: a. the thermal efficiency of the cycle b. the back work ratio c. the net power developed, in kW d. the rates of exergy destruction in each compressor stage and the turbine stage as well as the regenerator, in kW, for T 0 = 300 K.

What is the difference between absolute and gage pressure?

Answers

Explanation:

Step1

Absolute pressure is the pressure above zero level of the pressure. Absolute pressure is considering atmospheric pressure in it. Absolute pressure is always positive. There is no negative absolute pressure.

The expression for absolute pressure is given as follows:

P_(ab)=P_(g)+P_(atm)

Here, P_(ab) is absolute pressure, P_(g) is gauge pressure andP_(atm) is atmospheric pressure.

Step2

Gauge pressure is the pressure that measure above atmospheric pressure. It is not considering atmospheric pressure. It can be negative called vacuum or negative gauge pressure. Gauge pressure used to simplify the pressure equation for fluid analysis.  

After the load impedance has been transformed through the ideal transformer, its impedance is: + . Enter the real part in the first blank and the imaginary part in the second blank. If a value is negative, include the negative sign. Provide up to four digits of precision. If the exact value can be provided with fewer digits, merely provide the exact value. These instructions pertain to the following blanks as well. What is the total impedance seen by the source? + . What is the current phasor Ig (expressed in rectangular form)?

Answers

Answer:

Ig =7.2 +j9.599

Explanation: Check the attachment

The reservoir pressure of a supersonic wind tunnel is 5 atm. A static pressure probe is moved along the centerline of the nozzle, taking measurements at various stations. For these probe measurements, calculate the local Mach number and area ratio: a. 4 atm; b. 2.64 atm; c. 0.5 atm.

Answers

Answer

Given,

Reservoir pressure of a supersonic wind tunnel = 5 atm

Local Mach number = ?

Area ration = ?

a) 4 atm.

  Pressure ratio =(4)/(5)

                         = 0.8

From Isentropic Flow Tables

  M = 0.58   A/A* = 1.213

b) 2.64 atm

  Pressure ratio =(2.64)/(5)

                         = 0.528

From Isentropic Flow Tables

  M = 1  A/A* = 1

c)  0.5 atm

  Pressure ratio =(0.5)/(5)

                         = 0.1

From Isentropic Flow Tables

  M =2.10   A/A* = 1.8369

If you are pouring a large concrete slab and wish to avoid random cracks caused by shrinkage, you would likely provideA. expansion joints
B. isolation joints
C. control joints
D. construction joints

Answers

If you are pouring a large concrete slab and wish to avoid random cracks caused by shrinkage, you would likely provide Control joints. The correct answer would be C.

Control joints are used to prevent random cracks from forming in large concrete slabs caused by shrinkage. These joints are placed at strategic locations in the slab to allow for the concrete to expand and contract without cracking. Expansion joints, on the other hand, are used to separate concrete from other structures or materials, and isolation joints are used to separate different sections of concrete.

Construction joints are used to connect two different pours of concrete. Therefore, the best option for preventing random cracks caused by shrinkage would be to use control joints.

Learn more about Control joints:

brainly.com/question/1992029

#SPJ11

What is a SAFETY CHECK

Answers

Answer:

Safety check is defined as rounding to make sure that the patients and the milieu (patients living quarters) is secured and free of harmful items that can be used to hurt someone.

A site is compacted in the field, and the dry unit weight of the compacted soil (in the field) is determined to be 18 kN/m3. Determine the relative compaction if the maximum dry unit weight was measured to be 17 kN/m3. Express your answer as a percentage (but do not write the percentage sign in the answer box).

Answers

Answer:

the relative compaction is 105.88 %

Explanation:

Given;

dry unit weight of field compaction, W_d_((field)) = 18 kN/m³

maximum dry unit weight measured, W_d_((max)) = 17 kN/m³

Relative compaction (RC) of the site is given as the ratio of dry unit weight of field compaction and maximum dry unit weight measured

Relative compaction (RC) = dry unit weight of field compaction / maximum dry unit weight measured

RC = (W_d_((field)))/(W_d_((max)))

substitute the given values;

RC = (18)/(17) = 1.0588

RC (%) = 105.88 %

Therefore, the relative compaction is 105.88 %