How does the length and direction of a shadow change from sunrise to noon in the northern hemisphere? A: The shadow gets shorter and points more east B: The shadow gets shorter and points more north C: The shadow gets longer and points more north D: The shadow gets shorter and points more west E: The shadow gets longer and points more south F: The shadow gets longer and points more east

Answers

Answer 1

Answer: the answer is A

Explanation: because the sun rises in the west and goes down in the east

Answer 2
A Is the answerrrrrrr

Related Questions

What is the tallest building ever made

Answers

Answer:

Burj Khalifa

Explanation:

The world's tallest artificial structure is the 829.8-metre-tall (2,722 ft) Burj Khalifa in Dubai (of the United Arab Emirates). The building gained the official title of "tallest building in the world" and the tallest self-supported structure at its opening on January 9, 2010.

a brainliest would be appriciated

Answer:

Burj khalifa

Explanation:

829.8 meters tall

Most methods of transportation rely on some sort of infrastructure to drive, steer, navigate, or direct at some point or another in a journey. Which category of transportation system is least reliant on infrastructure?(1 point)

Answers

Answer:

Most methods of transportation rely on some sort of infrastructure to drive, steer, navigate, or direct at some point or another in a journey. Which category of transportation system is least reliant on infrastructure?(1 point). road

Explanation:

The alternator must be operated with the battery disconnected or with the terminals at the back of the alternator
disconnected.
True or false

Answers

Answer:

true

Explanation:

True

I think so


Hope it helps

What causes the charging system warning lamp to go out when the engine starts up?
Select one:
a. It turns off when ground is supplied to the lamp.

b. It turns off because voltage is applied to both sides of the lamp.
c. It turns off automatically after about 5 seconds.
d. It turns off because voltage is applied to one side of the bulb and ground to the other side.

Answers

B is the answer I believe so

A continuous and aligned fiber-reinforced composite is to be produced consisting of 30 vol% aramid fibers and 70 vol% of a polycarbonate matrix; mechanical characteristics of these two materials are as follows:
Modulus of Elasticity [GPa] Tensile Strength [MPa] Aramid fiber 131 3600 Polycarbonate 2.4 65
Also, the stress on the polycarbonate matrix when the aramid fibers fail is 45 MPa. For this composite, compute the following:
(a) the longitudinal tensile strength, and
(b) the longitudinal modulus of elasticity

Answers

Answer:

1. 1111.5MPa

2. 56.1GPa

Explanation:

1. Longitudinal tensile stress can be obtained by obtaining the strength and volume of the fiber reinforcement. The derived formula is given by;

σcl = σm (1 - Vf) + σfVf

Substituting the figures, we will have;

45(1 - 0.30) + 3600(0.30)

45(0.70) + 1080

31.5 + 1080

= 1111.5MPa

2. Longitudinal modulus of elasticity or Young's modulus is the ability of an object to resist deformation. The derived formula is given by;

Ecl = EmVm + EfVf

Substituting the formula gives;

= 2.4 (1 - 0.30) + 131 (0.30)

= 2.4(0.70) + 39.3

= 16.8 + 39.3

= 56.1GPa

Using the appropriate relation, the longitudinal tensile stress and the longitudinal modulus are 1111.50 and 56.10 respectively.

Longitudinal tensile stress can be obtained using the relation :

σcl = σm (1 - Vf) + σfVf

Substituting the values into the relation:

45(1 - 0.30) + 3600(0.30)

45 × 0.70 + 1080

31.5 + 1080

= 1111.50 MPa

2.)

Longitudinal modulus of elasticity is obtained using the relation :

Ecl = EmVm + EfVf

Substituting the values thus :

2.4 (1 - 0.30) + 131 (0.30)

= 2.4 × 0.70 + 39.3

= 16.8 + 39.3

= 56.10 GPa

Hence, the longitudinal tensile stress and the longitudinal modulus are 1111.50 and 56.10 respectively.

Learn more : https://brainly.com/question/22664384

Hard steering can be caused by

Answers

Answer:

Lack of fluid oil – lack of fluid oil in your vehicle, or a fluid leakage, can lead to heavy steering. If there is a lack of fluid oil, or a leak, this can reduce the pressure in the system, meaning the steering wheel does not receive enough supply of fluid to perform freely.

The most common cause of a stiff steering wheel is the lack of enough power steering fluid in the system. This condition can arise if there's fluid leakage from the pressurized hose area. Refilling the liquid in the power steering tank will fix the problem for the time being, but the leak must be taken care of.

An interior beam supports the floor of a classroom in a school building. The beam spans 26 ft. and the tributary width is 16 ft. Dead load is 20 psf. Find:

a. Basic floor live load Lo in psf
b. Reduced floor live load L in psf
c. Uniformly distributed total load to the beam in lb/ft.
d. Compare the loading in part c with the alternate concentrated load requried by the Code. Which loading is more critical for bending, shear, and deflection.?

Answers

Answer:

a. [tex]L_o[/tex]  = 40 psf

b. L ≈ 30.80 psf

c. The uniformly distributed total load for the beam = 812.8 ft./lb

d. The alternate concentrated load is more critical to bending , shear and deflection

Explanation:

The given parameters of the beam the beam are;

The span of the beam = 26 ft.

The width of the tributary, b = 16 ft.

The dead load, D = 20 psf.

a. The basic floor live load is given as follows;

The uniform floor live load, = 40 psf

The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²

Therefore, the uniform live load, [tex]L_o[/tex]  = 40 psf

b. The reduced floor live load, L in psf. is given as follows;

[tex]L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)[/tex]

For the school, [tex]K_{LL}[/tex] = 2

Therefore, we have;

[tex]L = 40 \times \left ( 0.25 + \dfrac{15}{\sqrt{2 \times 416} } \right) = 30.80126 \ psf[/tex]

The reduced floor live load, L ≈ 30.80 psf

c. The uniformly distributed total load for the beam, [tex]W_d[/tex] = b × [tex]W_{D + L}[/tex] =

∴  [tex]W_d[/tex] =  = 16 × (20 + 30.80) ≈ 812.8 ft./lb

The uniformly distributed total load for the beam, [tex]W_d[/tex] = 812.8 ft./lb

d. For the uniformly distributed load, we have;

[tex]V_{max}[/tex] = 812.8 × 26/2 = 10566.4 lbs

[tex]M_{max}[/tex] =  812.8 × 26²/8 = 68,681.6 ft-lbs

[tex]v_{max}[/tex] = 5×812.8×26⁴/348/EI = 4,836,329.333/EI

For the alternate concentrated load, we have;

[tex]P_L[/tex] = 1000 lb

[tex]W_{D}[/tex] = 20 × 16 = 320 lb/ft.

[tex]V_{max}[/tex] = 1,000 + 320 × 26/2 = 5,160 lbs

[tex]M_{max}[/tex] =  1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs

[tex]v_{max}[/tex] = 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI

Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load

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