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nirvana33 [79]
2 years ago
13

There are two routes similar in terrain; however, one route has an incline of approximately 28 degrees, and you must decide on w

hich route to take. what is the maximum road grade that your fully loaded ecv can climb?
Physics
1 answer:
Mnenie [13.5K]2 years ago
4 0

We should chose the second route has inclination less than 28 because the car move easily on this route.

<h3>On which route you will move?</h3>

I will chose the second route that has less inclination because on less incline route, the car move easily as compared to the route that have high inclination. The heavier ECVs when fully loaded, they are capable of climbing a grade of 40 percent which is about 22 degrees.

So we can conclude that we should chose the second route has inclination less than 28 because the car move easily on this route.

Learn more about incline here:  brainly.com/question/8837820

#SPJ1

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Komok [63]

Answer:

A

Explanation:

6 0
3 years ago
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Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du
Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time t is

\theta=\dfrac\alpha2t^2

It rotates 44.5 rad in this time, so we have

44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}

b. Since acceleration is constant, the average angular velocity is

\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2

where \omega_f is the angular velocity achieved after 6.00 s. The velocity of the disk at time t is

\omega=\alpha t

so we have

\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}

making the average velocity

\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}

Another way to find the average velocity is to compute it directly via

\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}

c. We already found this using the first method in part (b),

\omega=14.8\dfrac{\rm rad}{\rm s}

d. We already know

\theta=\dfrac\alpha2t^2

so this is just a matter of plugging in t=12.0\,\mathrm s. We get

\theta=179\,\mathrm{rad}

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2

Then for t=6.00\,\rm s we would get the same \theta=179\,\rm rad.

7 0
4 years ago
Could anyone help with number 9?
Alisiya [41]
The answer would be A
3 0
3 years ago
A sample of carbon dioxide (C°p,m = 37.11 J K−1 mol−1) of mass 2.80 g at 27°C is allowed to expand reversibly and adiabatically
My name is Ann [436]

Answer:

182 to 3 s.f

Explanation:

Workdone for an adiabatic process is given as

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

where γ = ratio of specific heats. For carbon dioxide, γ = 1.28

For an adiabatic process

P₁V₁ʸ = P₂V₂ʸ = K

K = P₁V₁ʸ

We need to calculate the P₁ using ideal gas equation

P₁V₁ = mRT₁

P₁ = (mRT₁/V₁)

m = 2.80 g = 0.0028 kg

R = 188.92 J/kg.K

T₁ = 27°C = 300 K

V₁ = 500 cm³ = 0.0005 m³

P₁ = (0.0028)(188.92)(300)/0.0005

P₁ = 317385.6 Pa

K = P₁V₁¹•²⁸ = (317385.6)(0.0005¹•²⁸) = 18.89

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

V₁ = 0.0005 m³

V₂ = 2.10 dm³ = 0.002 m³

1 - γ = 1 - 1.28 = - 0.28

W =

18.89 [(0.002)⁻⁰•²⁸ - (0.0005)⁻⁰•²⁸]/(-0.28)

W = -67.47 (5.698 - 8.4)

W = 182.3 = 182 to 3 s.f

7 0
3 years ago
Help plss i just need it plss​
Travka [436]

Answer:

ooh thanks but don't give me advise

4 0
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