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sesenic [268]
2 years ago
14

What is the magnitude of the speed of a

Engineering
1 answer:
Mars2501 [29]2 years ago
6 0

The equatorial diameter of the earth is 12,756 km.

Therefore the equatorial radius is

r = 12756/2 = 6378 km.

The angular velocity of the earth is  

ω = (2π radians)/(24 hours) = 0.2618 rad/h

The tangential velocity of a person at the equator is

v = rω

  = (6378 km)*( 0.2618  rad/h)

  =  1670 km/h

Answer: 1670 km/h

I hope this helps you.

Brainly, thanks you, rating are all appreciated greatly!

Smile and have an amazing day ; )

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I have to find the critical points of this function of two variables <img src="https://tex.z-dn.net/?f=%5C%5Cf%28x%2Cy%29%3Dx%5E
liraira [26]

Answer:

no i dont think there is

Explanation:

because theres not

4 0
3 years ago
Which kind of fracture (ductile or brittle) is associated with each of the two crack propagation mechanisms?
Nina [5.8K]

dutile is the correct answer

6 0
2 years ago
Which of the following is NOT true about hydraulic systems?
Dmitry [639]

Answer: The answer is D

D.In hydraulic systems, the operating temperatures must be kept between 170�F and 180°F 

Explanation:

The operating temperature for hydraulic systems is 140°F and below. Anything above this temperature is too high and will reduce the useful life of hydraulic fluid.

Most often problems associated with hydraulic systems are caused by fluid contaminated with particulate matter.

7 0
3 years ago
The modulus of elasticity for a ceramic material having 6.0 vol% porosity is 303 GPa. (a) Calculate the modulus of elasticity (i
Phantasy [73]

Answer:

modulus of elasticity for the nonporous material is 340.74 GPa

Explanation:

given data

porosity = 303 GPa

modulus of elasticity = 6.0

solution

we get here  modulus of elasticity for the nonporous material Eo that is

E = Eo (1 - 1.9P + 0.9P²)    ...............1

put here value and we get Eo

303 = Eo ( 1 - 1.9(0.06) + 0.9(0.06)² )  

solve it we get

Eo = 340.74 GPa

8 0
2 years ago
The driveshaft of an automobile is being designed to transmit 134 hp at 2900 rpm. Determine the minimum diameter d required for
Misha Larkins [42]

Answer:

The minimum diameter is 1.344 in

Explanation:

The angular speed of the driveshaft is equal to:

w=\frac{2\pi N}{60}

Where

N = rotational speed of the driveshaft = 2900 rpm

w=\frac{2\pi *2900}{60} =303.69rad/s

The torque in the driveshaft is equal to:

\tau=\frac{P}{w}

Where

P = power transmitted by the driveshaft = 134 hp = 73700 lb*ft/s

\tau=\frac{73700}{303.69}  =242.68lb*ft

The minimum diameter is equal to:

d_{min} =(\frac{16T}{\pi *\tau } )^{1/3}

Where

T = shear stress = 6100 psi

τ = 242.68 lb*ft = 2912.16 lb*in

d_{min} =(\frac{16*2912.16}{\pi *6100} )^{1/3} =1.344in

4 0
2 years ago
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