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sashaice [31]
3 years ago
12

A curve has a radius of 50 meters and is banked 5 degrees. The road is covered with ice and is frictionless. What is the maximum

speed, in mph, a car can travel and negotiate the curve? A.) 15.65mph B.) 25.00mph C.)65.00mph D.) 6.55mph E.) None of the above
Physics
1 answer:
Salsk061 [2.6K]3 years ago
8 0

Answer:

E.) None of the above

Explanation:

Draw a free body diagram.  There are two forces acting on the car.  Normal force perpendicular to the road, and weight downward.

Sum of the forces perpendicular to the road:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces towards the center of the circle:

∑F = ma

N sin θ = mv²/r

mg cos θ sin θ = mv²/r

g cos θ sin θ = v²/r

v = √(gr cos θ sin θ)

Given r = 50 m and θ = 5°:

v = √(9.8 m/s² × 50 m × cos 5° × sin 5°)

v = 6.52 m/s

v = 14.6 mph

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Each plate of a parallel‑plate capacitor is a square of side 3.63 cm, and the plates are separated by 0.473 mm. The capacitor is
NARA [144]

Answer:

E= 55.53 x 10³ V/m

Explanation:

Given that

a=  3.63 cm

Area ,A= a²

distance ,d= 0.473 mm

Stored energy ,U = 8.49 nJ

Value of capacitor given as

C=\dfrac{\varepsilon _oA}{d}

By putting the values

C=\dfrac{8.85\times 10^{-12}\times 3.63^2\times 10^{-4}}{0.473\times 10^{-3}}

C=2.46 x 10⁻¹¹ F

U=\dfrac{1}{2}CV^2

V=Voltage difference

V=\sqrt{\dfrac{2U}{C}}

V=\sqrt{\dfrac{2\times 8.49\times 10^{-9}}{2.46\times 10^{-11}}}

V=26.27 V

V= E d

E=Electric filed

26.27 = E x 0.473 x 10⁻³

E= 55.53 x 10³ V/m

7 0
3 years ago
The most common form of angina is _______ angina. A. microvascular B. variant C. stable D. unstable
Paraphin [41]

The most common form of angina is stable angina.

C. stable

8 0
2 years ago
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A 544 g ball strikes a wall at 14.3 m/s and rebounds at 14.4 m/s. The ball is in contact with the wall for 0.042 s. What is the
Nezavi [6.7K]

Answer:

F = 371.738\,N

Explanation:

Let assume that ball strikes a vertical wall in horizontal direction. The situation can be modelled by the appropriate use of the definition of Moment and Impulse Theorem, that is:

(0.544\,kg)\cdot (14.3\,\frac{m}{s})-F\cdot \Delta t = -(0.544\,kg)\cdot (14.4\,\frac{m}{s} )

F\cdot \Delta t = 15.613\,\frac{kg}{m\cdot s}

The average force acting on the ball during the collision is:

F = \frac{15.613\,\frac{kg}{m\cdot s} }{0.042\,s}

F = 371.738\,N

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A 10.0 Ω lightbulb is connected to a 12.0 V battery. (a) What current flows through the bulb? (b) What is the power of the bulb?
andreev551 [17]

The current flowing through the bulb as well the power of the bulb are 1.2A and 14.4 Watts respectively.

<h3>What current flows through the bulb as well as the power of the bulb?</h3>

From ohm's law; V = I × R

Where V is the voltage, I is the current and R is the resistance.

Also, Power is expressed as; P = V × I

Where V is voltage and I is current.

Given that;

  • Resistance R = 10.0 ohms
  • Voltage V = 12.0V
  • Current I = ?
  • Power P = ?

First, we determine the current flow through the bulb.

V = I × R

12.0V = I × 10.0 ohms

I = 12.0 ÷ 10.0

I = 1.2A

Next, we determine the power of the bulb.

P = V × I

P = 12.0V × 1.2A

P = 14.4 Watts

Therefore, the current flowing through the bulb as well the power of the bulb are 1.2A and 14.4 Watts respectively.

Learn more about Ohm's law here: brainly.com/question/12948166

#SPJ1

6 0
1 year ago
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