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sesenic [268]
2 years ago
6

A civil engineer wishes to redesign the curved roadway in the example What is the Maximum Speed of the Car? in such a way that a

car will not have to rely on friction to round the curve without skidding. In other words, a car of mass m moving at the designated speed can negotiate the curve even when the road is covered with ice. Such a road is usually banked, which means that the roadway is tilted toward the inside of the curve. Suppose the designated speed for the road is to be v = 12.8 m/s (28.6 mi/h) and the radius of the curve is r = 37.0 m. At what angle should the curve be banked?
Physics
1 answer:
vlabodo [156]2 years ago
3 0

Answer:

24.3 degrees

Explanation:

A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:

a_c = \frac{v^2}{r} = \frac{12.8^2}{37} = 4.43 m/s^2

Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.

Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.

So gsin\alpha = 4.43cos\alpha

\frac{sin\alpha}{cos\alpha} = \frac{4.43}{g} = \frac{4.43}{9.81} = 0.451

tan\alpha = 0.451

\alpha = tan^{-1}0.451 = 0.424 rad = 0.424*180/\pi \approx 24.3^0

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Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly
loris [4]

Answer: 36.86\°

Explanation:

According to the described situation we have the following data:

Horizontal distance between lily pads: d=2.4 m

Ferdinand's initial velocity: V_{o}=5 m/s

Time it takes a jump: t=0.6 s

We need to find the angle \theta at which Ferdinand jumps.

In order to do this, we first have to find the <u>horizontal component (or x-component)</u> of this initial velocity. Since we are dealing with parabolic movement, where velocity has x-component and y-component, and in this case we will choose the x-component to find the angle:

V_{x}=\frac{d}{t} (1)

V_{x}=\frac{2.4 m}{0.6 s} (2)

V_{x}=4 m/s (3)

On the other hand, the x-component of the velocity is expressed as:

V_{x}=V_{o}cos\theta (4)

Substituting (3) in (4):

4 m/s=5 m/s cos\theta (5)

Clearing \theta:

\theta=cos^{-1} (\frac{4 m/s}{5 m/s})

\theta=36.86\° This is the angle at which Ferdinand the frog jumps between lily pads

4 0
2 years ago
You’ve been given the challenge of balancing a uniform, rigid meter-stick with mass M = 95 g on a pivot. Stacked on the 0-cm end
Mariulka [41]

Answer: d = 4750n/3.1+95n

Explanation:

Using the principle of moment to solve the question.

Sum of clockwise moments = sum of anti clockwise moments

Since there are n identical coins with mass 3.1g placed at point 0cm, 1 coin will have mass of 3.1/n grams

Taking moment about the pivot,

Mass 3.1/n grams will move anti-clockwisely while the mass 95g will move in the clockwise direction.

Since its a meter rule (100cm) the distance from the center mass(95g) to the pivot will be 50-d (check attachment for diagram).

To get 'd'

We have 3.1/n × d = 95 × (50-d)

3.1d/n = 4750-95d

3.1d = 4750n-95dn

3.1d+95dn=4750n

d(3.1+95n) = 4750n

d = 4750n/3.1+95n

6 0
2 years ago
If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.
Anna35 [415]

This question is incomplete

Complete Question

Three equal point charges are held in place as shown in the figure below

If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.

A) F1=2F2

B) F1=3F2

C) F1=4F2

D) F1=9F2

Answer:

D) F1=9F2

Explanation:

We are told in the question that there are three equal point charges.

q, Q1, Q2 ,

q = Q1 = Q2

From the diagram we see the distance between the points d

q to Q1 = d

Q1 to Q2 = nd

Assuming n = 2

= 2 × d = 2d

Sum of the two distances = d + 2d = 3d

F1 is the force on q due to Q1 and

F2 is the force on q due to Q2,

Since we have 3 equal point charges and a total sum of distance which is 3d

Hence,

F1 = 9F2

6 0
2 years ago
A certain rigid aluminum container contains a liquid at a gauge pressure of P0 = 2.02 × 105 Pa at sea level where the atmospheri
MaRussiya [10]

Answer:

dz=19217687.07\ m

Explanation:

Given:

  • initial gauge pressure in the container, P_0=2.02\times 10^{5}\ Pa
  • atmospheric pressure at sea level, P_a=1.01\times 10^5\ Pa
  • initial volume, V_0=4.4\times 10^{-4}\ m^3
  • maximum pressure difference bearable by the container, dP_{max}=2.26\times 10^{5}\ Pa
  • density of the air, \rho_a=1.2\ kg.m^{-3}
  • density of sea water, \rho_s=1.2\ kg.m^{-3}

<u>The relation between the change in pressure with height is given as:</u>

\frac{dP_{max}}{dz} =\rho_a.g_n

where:

dz = height in the atmosphere

g_n= standard value of gravity

<em>Now putting the respective values:</em>

\frac{2.26\times 10^{5}}{dz} =1.2\times 9.8

dz=19217.687\ km

dz=19217687.07\ m

Is the maximum height above the ground that the container can be lifted before bursting. (<em>Since the density of air and the density of sea water are assumed to be constant.</em>)

7 0
1 year ago
In an isolated system, the total heat given off by warmer substances equals the total heat energy gained by cooler substances. N
galina1969 [7]

Answer:

The temperature of the cooler substance was close to the room temperature. Therefore, the system experienced less change

Explanation:

Generally, in a closed system containing two bodies at different temperatures, there is a flow of heat energy from the body at a higher temperature to the body at a lower temperature. The effect is more significant when there is a large temperature difference between the bodies. However, if the temperature difference is small or insignificant, the change will be less.

3 0
1 year ago
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