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3 years ago

Difference between angle of shear and angle of twist.

2 answers:

7 0

Angle shear equals to shear stress divided by shear strain which is also equals to constant value known as the shear modulus and angle of twist is when a torque is applied to a section that is relative to another.

Natalka [10]3 years ago

6 0

Answer:

The angle of shear can be defined as the angle of the deformation that occurs on the sides when the object is exerted upon by shearing stress or deformation force.

The angle of twist can be defined as the angle through which the rotating machine element rotates or twists with respect to its free end. The angle of twist exists because of the torque.

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Question 1 Gary is on the space shuttle. It takes off and lifts him to a height of 300 km above Earth's surface. a. How has Gary

balandron [24]

A) The mass is an intrinsic property of an object: it means it depends only on the properties of the object, so it does not depend on the location of the object. Therefore, Gary's mass at 300 km above Earth's surface is equal to his mass at the Earth's surface.

b) The weight of an object is given by
$W=mg$
where
m is the mass
$g= \frac{GM}{r^2}$ is the gravitational acceleration at the location of the object, with G being the gravitational constant, M the mass of the planet and r the distance of the object from the center of the planet.

At the Earth's surface, $g=9.81 m/s^2$ , so Gary's weight is
$W=mg=9.81 m$ (1)
where m is Gary's mass.

Then, we must calculate the value of g at 300 km above Earth's surface. the Earth's radius is
$R=6370 km$
So the distance of Gary from the Earth's center is
$r=R+h=6370 km+300 km =6670 km = 6.67 \cdot 10^6 m$

The Earth's mass is $M=5.97 \cdot 10^{24} kg$ , so the gravitational acceleration is
$g'=G \frac{M}{r^2}= (6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )\frac{5.97 \cdot 10^{24} kg}{(6.67 \cdot 10^6 m)^2}=8.95 m/s^2$

Therefore, Gary's weight at 300 km above Earth's surface is
$W' = mg' = 8.95 m$ (2)

If we compare (1) and (2), we find that Gary's weight has changed by
$\frac{W'}{W}= \frac{8.95 m}{9.81 m}=0.91$
So, Gary's weight at 300 km above Earth's surface is 91% of his weight at the surface.

6 0

3 years ago

An ocean current moving from the equator toward a pole is

Viefleur [7K]

Answer:

A warm current

Explanation:

3 0

3 years ago

Read 2 more answers

Which of the following scenarios would be optimal for obtaining a date from radioactive decay using these isotopes: 87Rb, 147Sm,

REY [17]

Answer:

a) 238U, 40K and 87Rb, b) 235U and to a lesser extent 40K , c) he 235U,

d) possibility is 14C , e)this period would be ideal for 14C , f) 14C should be used since it is the one with the least average life time, even though the measurements must be very careful

Explanation:

One of the applications of radioactive decay is the dating of different systems.

To do this, the quantity of radioactive material in a meter is determined and with the average life time, the time of the sample is found.

Let's write the half-life times of the given materials

87Rb T ½ = 4.75 1010 years

147Sm T ½ = 1.06 1011 years

235U = 7,038 108 years

238U = 4.47 109 years

40K = 1,248 109 years

14C = 5,568 103 years

we already have the half-life of the different elements given

a) meteors. As these decomposed in the formation of the solar system, their life time is around 3 109 to 5 109 years, so it is necessary to look for elements that have a life time of this order, among the candidates we have 238U, 40K and 87Rb if these elements were at the moment of the formation of these meteors, there must still be rations in them, instead elements 14C already completely adequate

b) rock. The formation period is 4.20-108 years, therefore one of the most promising elements is 235U and to a lesser extent 40K since it is more abundant in rocks. The other elements with higher life times have not decayed and therefore will not give a true value and the 14C is completely decayed

c) volcanic ash. Formation time 6107 years, the only element that has the possibility of having a count is the 235U, the others have a life time so long that they have not decayed and the 14C is complete, unbent

d) scarp of an earthquake formation time 5 101 years, The only one that has any possibility is 14C even when it has declined very little, all the others, you have time to long that has not decayed

e) INCA excavation. The time of this civilization is about 10000 to 500 years (104 to 5 102 years), we see that this period would be ideal for 14C since it has some period of cementation, the others have not decayed

f) Tree in Blepharitis. 14C should be used since it is the one with the least average life time, even though the measurements must be very careful because of a period of disintegration. We have such a long time that they have not decayed

8 0

4 years ago

A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at

GaryK [48]

Answer:

50.93 m/s

199.5 kW

Explanation:

From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=

0.025m

We can calculate the The cross sectional area of the nozzle as

A= πr^2

A= π ×0.025^2

= 1.9635 ×10^- ³ m²

From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at

(0.1 / 1.9635 ×10^- ³ m²)

= 50.93 m/s

During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation

mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

From the question;

The total irreversible head loss of the system = 3 m,

the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

=(3 + 3 + 132.21m)

=138.21m

mass of water pumped in a seconds can be calculated since we know that mass is a product of volume and density

Volume= 0.1m³

Density of sea water=1030 kg/m

(0.1 m^3× 1030)

= 103kg

We can calculate the Potential enegry, which is = mgh

= (103 ×9.81 × 138.21)

= 139651.5 Watts

= 139.65kW

To determine required shaft power input to the pump and the water discharge velocity

Energy= efficiency × power

But we are given efficiency of 70 percent, then

139651.5 Watts = 0.7P

=199502.18 Watts

P=199.5 kW

Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

5 0

3 years ago

water can be thought of as a mass of moving particles. What evidence supports this idea? How does this idea help to explain boil

Illusion [34]

Answer:

May 29, 2020 — Using activities like POE (Predict-Observe-Explain) can help think about and then question their existing ideas.

Explanation:

YAY

7 0

3 years ago