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user100 [1]
3 years ago
7

Which of the following can be seen as a result of the earth's tilt? 1 the seasons change throughout the year. II the tides chang

e throughout the day. III daytime is longer than nighttime during the summer than during the winter. A. II and III only B. I, II, and III C. III only D. I and III only
Engineering
2 answers:
Nina [5.8K]3 years ago
5 0
Seasons change through the year!!
kompoz [17]3 years ago
3 0

Answer:

d is the correct

Explanation:

because the tides arent involved with the sun if so than only a little bit

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At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho
victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0
4 years ago
In a food processing facility, a spherical container of inner radius r1 = 40 cm, outer radius r2 = 41 cm, and thermal conductivi
Rashid [163]

Answer:

attached below

Explanation:

5 0
3 years ago
Which of the following statements about resistance is TRUE?
valentina_108 [34]

Answer:

I think it's the no 3rd

Explanation:

hope this helps

3 0
2 years ago
A 60-cm-high, 40-cm-diameter cylindrical water tank is being transported on a level road. The highest acceleration anticipated i
dlinn [17]

Answer:

h_{max} = 51.8 cm

Explanation:

given data:

height of tank = 60cm

diameter of tank =40cm

accelration = 4 m/s2

suppose x- axis - direction of motion

z -axis - vertical direction

\theta = water surface angle with horizontal surface

a_x =accelration in x direction

a_z =accelration in z direction

slope in xz plane is

tan\theta = \frac{a_x}{g +a_z}

tan\theta = \frac{4}{9.81+0}

tan\theta =0.4077

the maximum height of water surface at mid of inclination is

\Delta h = \frac{d}{2} tan\theta

            =\frac{0.4}{2}0.4077

\Delta h  0.082 cm

the maximu height of wwater to avoid spilling is

h_{max} = h_{tank} -\Delta h

            = 60 - 8.2

h_{max} = 51.8 cm

the height requird if no spill water is h_{max} = 51.8 cm

3 0
3 years ago
Technician A says that ultraviolet light is bright, and very hot. Technician B says that ear protection does not need to be used
-Dominant- [34]

Answer:

a. technician a only

Explanation:

I took that worksheet

3 0
3 years ago
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