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DerKrebs [107]
2 years ago
10

A minivan starts from rest on the road whose constant radius of curvature is 40 meters and whose bank angel is 10 degrees. the m

otion occurred in a horizontal plane. If the constant forward acceleration of the minivan is 1.8 m/s², determine the magnitude a of it's total acceleration 5 second after starting.
Engineering
1 answer:
zimovet [89]2 years ago
8 0

Based on the calculations, the magnitude (a) of it's total acceleration is equal to 2.71 m/s².

<u>Given the following data:</u>

  • Angle of inclination = 10°.
  • Radius of curvature, r = 40 meters.
  • Acceleration of the minivan, A = 1.8 m/s².
  • Initial velocity, u = 0 m/s (since it's starting from rest).
  • Time, t = 5 seconds.

<h3>How to determine the magnitude (a) of it's total acceleration?</h3>

First of all, we would determine the final velocity of the minivan by applying the first equation of motion as follows:

V = u + at

V = 0 + 1.8 × 5

V = 9 m/s.

Next, we would calculate the centripetal acceleration of this minivan:

Ac = V²/r

Ac = 9²/40

Ac = 2.025 m/s².

Now, we can determine the magnitude (a) of it's total acceleration:

a = √(Ac² + A²)

a = √(2.025² + 1.8²)

a = 2.71 m/s².

Read more on acceleration here: brainly.com/question/24728358

#SPJ1

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Answer:

I would say false but I am not for sure

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A second inventor was driving down the highway in her Prius one day with her hand out the window. She happened to be driving thr
Eva8 [605]

Answer:

Explanation:

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6 0
3 years ago
1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil
kakasveta [241]

Answer:

a) n = 1000\,users, b)\Delta t_{min} = \frac{1}{30}\,h, \Delta t_{max} = \frac{\sqrt{2} }{30}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h, c) n = 10000000\,users, \Delta t_{min} = \frac{1}{3000}\,h, \Delta t_{max} = \frac{\sqrt{2} }{3000}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

Explanation:

a) The total number of users that can be accomodated in the system is:

n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )

n = 1000\,users

b) The length of the side of each cell is:

l = \sqrt{1\,km^{2}}

l = 1\,km

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{30}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{30}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h

c) The total number of users that can be accomodated in the system is:

n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )

n = 10000000\,users

The length of each side of the cell is:

l = \sqrt{100\,m^{2}}

l = 10\,m

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{3000}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

8 0
3 years ago
A person is planning a bungee jump from a 40 meter high bridge. Under the bridge is a river with crocodiles, so the person does
Nonamiya [84]

Answer:

a. l = 19.7m, b. 18.55m, c. Impact Force = 3889.84 N

Explanation:

The total energy of the system when the person is at top of the bridge is

Potential energy = mgh, Kinetic energy = 0

The total energy of the  system when the person reaches just above the surface

Potential energy = 0, Kinetic energy = 0, Spring energy = ½ K X2, where k is the spring constant and X is the deflection

Applying conservation of energy

mgh = 0 + 0 + ½ K X²

80 x 9.81 x 40 = ½ (3600/l) X²

31392 = ½ (3600/l) X²

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l+ X + 1.75 = 40

l + X = 38.25

a. <u>Substitute the value of x from above into the energy conversion expression</u>

31392 = ½ (3600/l)(38.25 - l)²

31392 x 2/3600 = (38.25 + l² – 2l(38.25))/l

17.44l = l2 – 76.5l + 38.25²

l² – 76.5l – 17.44l +1463.0625 = 0

Solving for l we get

L = 19.7

Hence, length of the rope is 19.7m

b. <u>The deflection is calculated by using the relation between l and X</u>

L + X = 38.25

X = 38.25 – 19.7 = 18.55m

c. <u>The impact force is calculated using the impact force formula which relates the impact force with the deflection</u>

F = KX

F = (3600/l) . X

F = (3600/19.7) . (18.55) = 3889.84 N

Thus, the impact force is 3889.84 N

3 0
3 years ago
What's the relationship between energy and time<br>​
boyakko [2]

Answer:

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Explanation:

8 0
3 years ago
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