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zubka84 [21]
2 years ago
14

To practice Problem-Solving Strategy 6.1: Circular motion A highway curve with radius R = 274 m is to be banked so that a car tr

aveling v = 25.0 m/s will not skid sideways even in the absence of friction. At what angle should the curve be banked?
Physics
1 answer:
Sedaia [141]2 years ago
4 0

Answer:

The curve should be banked at an angle of 13 degrees.

Explanation:

We have,

Radius of a highway curve is 274 m

Speed of car on this curve is 25 m/s

Let \theta is the banking angle. On a banked curve, the angle of safe diving is given by following expression.

\tan\theta=\dfrac{v^2}{Rg}

g = 10 m/s²

Plugging all the values in above formula,

\tan\theta=\dfrac{(25)^2}{274\times 9.8}\\\\\theta=\tan^{-1}\left(\dfrac{(25)^{2}}{274\times9.8}\right)\\\\\theta=13^{\circ}

So, the curve should be banked at an angle of 13 degrees.

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Alex goes cruising on his dirt bike. He rides 700m north, 300m east, 400 m north, 600m west, 1200m south, 300m east and finally
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Find Displacement and Distance

displacement ...

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south=1200m

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east is 300+300=600m

west is 600m

600-600=0

back at dtart. displ zero


distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m  = 3600m


3 0
2 years ago
You start with spring that's already been stretched an unknown amount from equilibrium. After stretching it an additional 2.0 cm
maxonik [38]

Answer: 35*10^3 N/m

Explanation: In order to explain this problem we know that the potential energy for spring is given by:

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We  also know that with additional streching of 2 cm of teh spring,  the potential energy is 18J. Then it applied another additional streching of 2 cm and the energy is 25J.

Then the difference of energy for both cases is 7 J so:

ΔUp= 1/2*k* (0.02)^2 then

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7 0
2 years ago
A 4.00-kg box sits atop a 10.0-kg box on a horizontal table. The coefficient of kinetic friction between the two boxes and betwe
natta225 [31]
First, we have to calculate the normal forces on different surfaces.The normal force on the 4.00 kg, N1 = (4)(9.8) = 39.2 N. The normal force on the 10.0 kg, N2 = (14)(9.8) = 137.2 N. Looking at the 10.0 kg block, the static forces that counteract the pulling force equals the sum of the friction from the two surfaces. Fc = N1 * 0.80 + N2 * 0.80 = 141.12 N. Since the counter force is less than the pulling force, the blocks start to move and hence, kinetic frictions are considered.


Therefore, f1 = uk * N1 = (0.60)(39.2) = 23.52 N.
4 0
2 years ago
A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 cans, 0.355 - l each,
Alona [7]
Flow rate = 220*0.355 l/m = 78.1 l/min = 1.3 l/s = 0.0013 m^3/s

Point 2:
A2= 8 cm^2 = 0.0008 m^2
V2 = Flow rate/A2 = 0.0013/0.0008 = 1.625 m/s
P1 = 152 kPa = 152000 Pa

Point 1:
A1 = 2 cm^2 = 0.0002 m^2
V1 = Flow rate/A1 = 0.0013/0.0002 = 6.5 m/s
P1 = ?
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Applying Bernoulli principle;
P2+1/2*V2^2/density = P1+1/2*V1^2/density +density*gravitational acceleration*height
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=> 153320.31 = P1 + 34368.5
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3 0
2 years ago
Read 2 more answers
A motor does a total of 480 joules of work in 5.0 seconds to lift a 12-kilogram block to the top of a ramp. The average power de
stepan [7]
Power may be defined as the rate of doing work or the rate of using energy. <span> It is the amount of energy consumed per unit time. It is calculated as follows:

P = E / t
P = 480 / 5
P = 96 W <-----OPTION 3

Hope this answers the question. Have a nice day.</span>
7 0
2 years ago
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