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Fittoniya [83]
2 years ago
12

A civil engineer plans to design a curved ramp such that a car may not have to rely on friction to round the curve without skidd

ing. The roadway is tilted toward the inside of the curve. Suppose the designated speed is v and the radius of the curve is R. At what angle should the curve be tilted?

Physics
1 answer:
Delicious77 [7]2 years ago
8 0

Answer:\theta =\tan ^{-1}(\frac{v^2}{gR})

Explanation:

let \theta be the inclination at which curve is tilted

v is the speed of car and R is the radius of curve

m is the mass of car

Suppose R is the reaction offered by road to car

Resolving R in x and y direction we get

R\cos \theta will balance weight and R\sin \theta will provide the necessary centripetal Force

thus R\sin \theta =\frac{mv^2}{R}  ------------1

R\cos \theta =m g  ----------------2

Divide 1 & 2 we get

\frac{R\sin \theta }{R\cos \theta }=\frac{mv^2}{mgR}

\tan \theta =\frac{v^2}{gR}

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

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The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5
KiRa [710]

Answer:

0.68 s

Explanation:

We are given that

Initial velocity of box=u=13m/s

Final velocity of box=v=11.5 m/s

Distance=d=8.5 m

We have to find the time taken by box to slow by this amount.

We know that

v^2-u^2=2as

Substitute the values

(11.5)^2-(13)^2=2a(8.5)

132.25-169=17a

-36.75=17a

a=\frac{-36.75}{17}=-2.2m/s^2

We know that

Acceleration=a=\frac{v-u}{t}

Substitute the values

-2.2=\frac{11.5-13}{t}

-2.2=\frac{-1.5}{t}

t=\frac{1.5}{2.2}=0.68 s

Hence, the time taken by box to slow by this amount=0.68 s

8 0
3 years ago
A light bulb dissipates 100 Watts of power when it is supplied a voltage of 220 volts.
ycow [4]

Given Information:

Power = P = 100 Watts

Voltage = V = 220 Volts

Required Information:

a) Current = I = ?

b) Resistance = R = ?

Answer:

a) Current = I = 0.4545 A

b) Resistance = R = 484 Ω

Explanation:

According to the Ohm’s law, the power dissipated in the light bulb is given by

P = VI

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.

Re-arranging the above equation for current I yields,

I = \frac{P}{V}  \\\\I = \frac{100}{220} \\\\I = 0.4545 \: A \\\\

Therefore, 0.4545 A current is flowing through the light bulb.

According to the Ohm’s law, the voltage across the light bulb is given by

V = IR

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.

Re-arranging the above equation for resistance R yields,

R = \frac{V}{I} \\\\R = \frac{220}{0.4545} \\\\R = 484 \: \Omega

Therefore, the resistance of the bulb is 484 Ω

3 0
3 years ago
Read 2 more answers
North Dakota Electric Company estimates its demand trend line​ (in millions of kilowatt​ hours) to​ be: D​ = 75.0 ​+ 0.45​Q, whe
Alborosie

Answer:

The demand forecast for winter is 96.36 millions KWH

The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

Explanation:

Given that,

The demand trend line​ is

D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

We need to calculate the demand forecast for winter

Using given formula

D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

Put the value into the formula

D=(75.0+0.45\times101)\times0.80

D=96.36\ millions\ KWH

We need to calculate the demand forecast for spring

Using given formula

D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

Put the value into the formula

D=(75.0+0.45\times102)\times1.20

D=145.08\ millions\ KWH

We need to calculate the demand forecast for summer

Using given formula

D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

Put the value into the formula

D=(75.0+0.45\times103)\times1.40

D=169.89\ millions KWH

We need to calculate the demand forecast for fall

Using given formula

D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

Put the value into the formula

D=(75.0+0.45\times104)\times0.60

D=73.08\ millions KWH

Hence, The demand forecast for winter is 96.36 millions KWH

The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

3 0
3 years ago
How many lenses with different focal lengths can be obtained by combining two surfaces whose radii of curvature are 4.00 cm and
Dovator [93]

Answer:

The lenses with different focal length are four.

Explanation:

Given that,

Radius of curvature R₁= 4

Radius of curvature R₂ = 8

We know ,

Refractive index of glass = 1.6

When, R₁= 4, R₂ = 8

We need to calculate the focal length of the lens

Using formula of focal length

\dfrac{1}{f}=(n-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}+\dfrac{1}{8})

\dfrac{1}{f}=\dfrac{9}{40}

f=4.44\ cm

When , R₁= -4, R₂ = 8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}+\dfrac{1}{8})

\dfrac{1}{f}=-\dfrac{3}{40}

f=-13.33\ cm

When , R₁= 4, R₂ = -8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}-\dfrac{1}{8})

\dfrac{1}{f}=\dfrac{3}{40}

f=13.33\ cm

When , R₁= -4, R₂ = -8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}-\dfrac{1}{8})

\dfrac{1}{f}=-\dfrac{9}{40}

f=-4.44\ cm

Hence, The lenses with different focal length are four.

8 0
2 years ago
How is energy transferred through a generator that produces electric current? A.from armature to commutator to brushes to turbin
vaieri [72.5K]
Armature to commutator to brushes to turbine
7 0
2 years ago
Read 2 more answers
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