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3 years ago

Difference between angle of twist and angle of shear

1 answer:

3 0

The angle between 2 o'clock and 12 o'clock is referred to as the angle of twist. The angle between the planes of maximum shear which is bisected by the axis of greatest compression is angle of shear.

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Suppose you took a trip to the moon. Write a paragraph describing how and why your weight would change. Would your mass change t

noname [10]

Your weight would change but not your mass, the moon has less gravity so therefore you are going to be lighter :-)

5 0

3 years ago

In one cycle, a heat engine takes in 1000 J of heat from a high-temperature reservoir, releases 600 J of heat to a lower-tempera

Talja [164]

Answer:

η = 40 %

Explanation:

Given that

Qa ,Heat addition= 1000 J

Qr,Heat rejection= 600 J

Work done ,W= 400 J

We know that ,efficiency of a engine given as

$\eta=\dfrac{W(net)}{Q(heat\ addition)}$

Now by putting the values in the above equation ,then we get

$\eta=\dfrac{400}{1000}$

η = 0.4

The efficiency in percentage is given as

η = 0.4 x 100 %

η = 40 %

Therefore the answer will be 40%.

4 0

3 years ago

If 1 meter = 3.28 feet, what is the height of the Washington Monument in meters?

jasenka [17]

Since you didn't provide how tall the Monument was, I took the liberty to find it and it is 555 feet tall. So to convert to meters we must divide 555 by 3.28 or multiply it by 0.3048 (this is the method I used).
555 x 0.3048 = 169.164 meters

5 0

3 years ago

Read 2 more answers

14. A rocket is shot up into the air and then comes back down and hits the ground 9.2 second later.

sineoko [7]

Answer:

105.8 m

46 m/s

Explanation:

From the time the rocket is launched to the time it reaches its maximum height:

v = 0 m/s

a = -10 m/s²

t = 9.2 s / 2 = 4.6 s

Find: Δy and v₀

Δy = vt − ½ at²

Δy = (0 m/s) (4.6 s) − ½ (-10 m/s²) (4.6 s)²

Δy = 105.8 m

v = at + v₀

0 m/s = (-10 m/s²) (4.6 s) + v₀

v₀ = 46 m/s

3 0

4 years ago

94. The diagram shows the orbit of a satellite

Valentin [98]

Answer:

7.65x10^3 m/s

Explanation:

The computation of the satellite's orbital speed is shown below:

Given that

Earth mass, M_e = 5.97 × 10^24 kg

Gravitational constant, G = 6.67 × 10^-11 N·m^2/kg

Orbital radius, r = 6.80 × 10^6m

Based on the above information

the satellite's orbital speed is

V_o = √GM_e ÷ √r

= √6.67 × 10^-11 × 5.97 × 10^24 ÷ √6.80 × 10^6

= 7.65x10^3 m/s

4 0

3 years ago