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Ad libitum [116K]

2 years ago

13

At what angle should the roadway on a curve with a 50m radius be banked to allow cars to negotiate the curve at 12 m/s even if t

he roadway is frictionless?

1 answer:

kari74 [83]2 years ago

8 0

Answer:

The road bank angle is 16.38⁰.

Explanation:

radius of curvature of the road, r = 50 m

allowable speed of car on the road, v = 12 m/s

The bank angle is calculated as;

$\theta = tan^{-1} (\frac{v^2}{gr} )$

where;

θ is the road bank angle

g is acceleration due to gravity = 9.8 m/s²

$\theta = tan^{-1} (\frac{v^2}{gr} )\\\\\theta = tan^{-1} (\frac{12^2}{9.8 \times 50} )\\\\\theta = tan^{-1} ( 0.2939)\\\\\theta = 16.38 ^0$

Therefore, the road bank angle is 16.38⁰.

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You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th

PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight = $m\times g$

m = mass of the ball = 10 kg

g = acceleration due to gravity = $9.81m/s^2$

l = length of wire = 2.6 m

A = area of cross section = $\pi r^2$

r = radius of the wire = $\frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m$ (Conversion factor: 1 m = 1000 mm)

$\Delta l$ = change in length = 1.99 mm = $1.99\times 10^{-3}m$

Putting values in above equation, we get:

$Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2$

Hence, the Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

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3 years ago

The cube with 2.00m wide and 2.00m long and 2.00m high has a weight of 960.00 N what pressure does it exert

andre [41]

2m X 2m = 4m ^2
960n / 4 = 240pa

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2 years ago

Define resistance and describe what would happen to a light bulb if the voltage increased but the resistance stayed the same. (h

PtichkaEL [24]

The light bulb would glow brighter.

<h3>What is Resistance?</h3>

a force that works against a body's direction of motion and seeks to stop or slow down motion, such as friction. a measure of how much a material prevents an electric current from flowing as a result of a voltage.

What is the law of resistance?

Resistance and Ohm's Law. According to Ohm's law, the resistance of the circuit and the current or energy travelling through the resistance are both exactly proportional to the voltage or potential difference between two places.

The current would grow since it is exactly proportionate to the voltage, increasing the light bulb's brilliance, or simply making it brighter.

to learn more about Resistance go to - brainly.com/question/15728236

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9 months ago

A person wants to make a metronome for music practice. He uses a 35-g object attached to a spring to serve as the time standard.

alukav5142 [94]

To develop this problem it will be necessary to apply the concepts related to the frequency of a spring mass system, for which it is necessary that its mathematical function is described as

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Here,

k = Spring constant

m = Mass

Our values are given as,

$m = 35g = 35*10^{-3}kg$

$f = 1 Hz$

Rearranging to find the spring constant we have that,

$k = (2\pi f \sqrt{m})^2$

$k = 4\pi^2 f^2 m$

$k = (4) (\pi)^2 (1) (35*10^{-3})$

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Therefore the spring constant is 1.38N/m

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3 years ago

An elevator is moving down with an acceleration of 3.36 m/s2.

sergeinik [125]

Answer : 413.44N

Here it is given that an elevator is moving down with an acceleration of 3.36 m/s² . And we are interested in finding out the apparent weight of a 64.2 kg man . For the diagram refer to the attachment .

From the elevator's frame ( non inertial frame of reference) , we would have to think of a pseudo force.
The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
When a elevator accelerates down , the weight recorded is less than the actual weight .

From the Free body diagram ,

$\sf\longrightarrow Weight = mg - ma \\$

$\sf\longrightarrow Weight = m ( g - a ) \\$

Mass of the man = 64.2 kg

$\sf\longrightarrow Weight = 64.2( 9.8 - 3.36) N\\$

$\sf\longrightarrow Weight = 64.2 * 6.44 N\\$

$\sf\longrightarrow \underline{\boxed{\bf Weight_{apparent}= 413.44 N }} \\$

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1 year ago