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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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If a metallic wire of cross sectional area 3.0 ´ 10-6 m2 carries a current of 6.0 A and has a mobile charge density of 4.24 ´ 10
elena-14-01-66 [18.8K]

Answer:

The drift velocity is v_d=2.9\times 10^{-4}\ m/s.

Explanation:

Given :

Area of metallic wire, A = 3\times 10^{-6}\ m^2.

Current through wire , I=6 \ A.

Mobile charge density , n=4.24\times 10^{28} \ carriers/m^3.

Charge value , e=1.6\times 10^{-19}\ C.

We need to find drift velocity , v_d.

Now, we know :

I=neAv_d

Therefore, v_d=\dfrac{I}{neA}

Putting all given values in above equation we get,

v_d=\dfrac{6}{4.24\times 10^{28}\times 1.6\times 10^{-19} \times 3 \times 10^{-6}}

v_d=2.9\times 10^{-4}\ m/s.

Hence, this is the required solution.

8 0
2 years ago
A one-dimensional conservative force is given by the function F(x) = (2.00 N/m) x +(1.00 N/m3) x3. What is the change in potenti
topjm [15]

Answer:

Explanation:

Given that

F(x) = (2.00 N/m) x +(1.00 N/m3) x3.

Which can be written as

F(x) = 2x +x³ N/m

Potential energy from x=1 to x=2

P.E is given as

P.E=∫F(x)dx

P.E=∫2x+x³ dx. From x=1 to x=2

P.E= 2x²/2 + x⁴/4 from x=1 to x=2

P.E= x² + x⁴/4 from x=1 to x=2

P.E=(2²+2⁴/4)-(1² +1⁴/4)

P.E=(4+4)-(1+1/4)

P.E, =8-5/4

P.E=27/4

P.E=6.75J

Since it is moving from lower position to upper position, then P.E should be negative

P.E = - 6.75J

8 0
2 years ago
Justine is ice-skating at the Lloyd Center what is her final velocity if she accelerates at a rate of 2.0 meters per second for
lilavasa [31]

2*3.5 = 7m/s

You multiply the acceleration per the time (they both are in seconds, otherwise, you should set them in the same units).

7 0
1 year ago
A refrigerator with a weight of 1,127 newtons needs to be moved into a house using a ramp. The length of the ramp is 2.1 meters,
lina2011 [118]
496/1127 = 0.44 = 44% 

<span>sin A = 0.85/2.1. </span>
<span>A = 23.9o. </span>

<span>Fp = 1127 sin23.9 = 457 N. = Force parallel to the ramp. </span>

<span>Fn = 1127 Cos23.9 = 1,030 N. = Force </span>
<span>perpendicular to the ramp = Normal force. </span>

<span>Eff. = Fp/Fap = 457/496 = 0.92 = 92%
Correct answer: 92%</span>
3 0
2 years ago
Read 2 more answers
The velocity of an object is given by the expression v(t) = 3.00 m/s + ( 4.00 m/s^3)t^2, where t is in seconds. Determine the po
bulgar [2K]

Answer:

Position of object is;

s(t) = 4t³/3 + 3t + 1

Explanation:

We are told that the velocity has an expression;

v(t) = 3.00 m/s + ( 4.00 m/s³)t²

Now, to get the expression for the position(s(t)) of the object, we have to integrate the velocity expression. Thus;

s(t) = ∫3 + 4t²

s(t) = 3t + 4t³/3 + c

Now, we were told that at x = 1.00 m, time t = 0.000 s

Thus, plugging the values in;

1 = 3(0) + 4(0³/3) + c

c = 1

Thus,the expression for the position of the object is;

s(t) = 4t³/3 + 3t + 1

6 0
2 years ago
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