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erastovalidia [21]
4 years ago
10

The uniform crate has a mass of 50 kg and rests on the cart having an inclined surface. Determine the smallest acceleration that

will cause the crate either to tip or slip relative to the cart. What is the magnitude of this acceleration? The coefficient of static friction between the crate and cart is ms = 0.5.
Engineering
1 answer:
Lostsunrise [7]4 years ago
5 0

Answer: smallest will be 000,.111 84.

Explanation:

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The federal highway administration reports nearly
levacccp [35]

Explanation:

The federal highway administration reports nearly 800 work zone fatalities per year.

8 0
3 years ago
Which of the following types of valves have a valve stem that moves up and down in a valve guide? Select one: a. Poppet valves b
Alja [10]

Answer:c

Explanation:

thats wht it is

5 0
3 years ago
Read 2 more answers
A very long pipe of 0.05 m (r1) radius and 0.03 m thickness (r2 - r1) is buried at a depth of 2m (z) to transport liquid nitroge
Elenna [48]

Answer:

The surface temperature of the ground is = 296.946K

Explanation:

Solution

Given

r₁= 0.05m

r₂= 0.08m

Tn =Ti = 77K

Ki = 0.0035 Wm-1K-1

Kg =  1 Wm-1K-1

Z= 2m

Now,

The outer type temperature (Skin temperature pipe)

Q = T₀ -T₁/ln (r2/r1)/2πKi = 2πKi  T0 -T1/ln (r2/r1)

Thus,

10 w/m = 2π * 0.0035 = T0 -77/ln 0.08/0.05

⇒ T₀ -77 = 231.72

    T₀= 290.72K

The shape factor between the cylinder and he ground

S = 2πL/ln 4z/D

where L = length of pipe

D = outer layer of pipe

S = 2π * 1/4 *2/ 2 * 0.08 = 1.606m

The heat gained in the pipe is = S  * Kg * (Tg- T₀)

(10* 1) = 1.606 * 1* (Tg- 290.72)

Tg - 290.72 = 6.2266

Tg = 296.946K

Therefore the surface temperature to the ground is 296.946K

6 0
3 years ago
Hello, I want to introduce you to our hosting - VPSDOM
Solnce55 [7]

Answer:

pogchamp

Explanation:

sussy balls

4 0
3 years ago
While driving down the highway, your car must overcome the forces of friction and air resistance in order to keep the car moving
ElenaW [278]

Answer:

Efficiency of the engine equals 20%

Explanation:

We know that when the car moves it must do work against the resisting forces to keep moving and this work is spend as energy by the engine to keep the car moving.

we know that

Work=Force\times Displacement

Thus to keep the car moving for 100,000 meters the theoretical work that requires to be done equals  

Work=560N\times 100,000m=56\times 10^{6}Joules

Now the actual energy spend by the car equals the energy spend by burning 2.8 gallons of gasoline.

Thus the energy produced by burning 2.8 gallons of gasoline equals

2.8\times 100\times 10^{6}=280\times 10^{6}Joules

Thus the efficiency is calculated as

\eta =\frac{56\times 10^{6}}{280\times 10^{6}}\times 100\\\\\therefore \eta =20percent

5 0
3 years ago
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