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aalyn [17]
3 years ago
11

A 1500 kg car is moving on a flat, horizontal flat road. If the radius of the curve is 35 m and

Physics
1 answer:
Troyanec [42]3 years ago
4 0

The net force on the car is the friction that keeps it on the road, which points toward the center of the circle of the curve. Then by Newton's second law, we have

• net vertical force:

∑ <em>F</em> = <em>N</em> - <em>W</em> = 0

• net horizontal force:

∑ <em>F</em> = <em>Fs</em> = <em>m a</em>

where

<em>N</em> = magnitude of normal force

<em>W</em> = car's weight

<em>Fs</em> = mag. of static friction

<em>m</em> = car's mass

<em>a</em> = <em>v</em> ²/<em>R</em> = mag. of the centripetal acceleration

<em>v</em> = car's speed

<em>R</em> = radius of curve

Now,

• compute the car's weight:

<em>W</em> = <em>m g</em> = (1500 kg) (9.8 m/s²) = 14,700 N

• solve for the mag. of the normal force:

<em>N</em> = 14,700 N

• solve for the mag. of the friction force, using the given friction coefficient:

<em>Fs</em> = 0.5 <em>N</em> = 7350 N

• solve for the (maximum) acceleration:

7350 <em>N</em> = (1500 kg) <em>a</em>   →   <em>a</em> = 4.9 m/s²

• solve for the (maximum) speed:

4.9 m/s² = <em>v</em> ²/ (35 m)   →   <em>v</em> ≈ 13 m/s

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Alexeev081 [22]

Answer: a=-2.4525 m/s^2

d=s=190.3 m

Explanation:The only force that is stopping the car and causing deceleration is the frictional force Fr

Fr = 25% of weight

W=mg

W=1750*9.81

W=17167.5

Hence

Fr=\frac{25}{100} * -17167.5\\\\Fr=-4291.875 N

Frictional force is negative as it acts in opposite direction

According to newton second law of motion

F=ma

hence

a=Fr/m

a=-4291.875/1750\\a=-2.4525

given

u= 110 km/h

u=110*1000/3600

u=30.55 m/s

to get t we know that final velocity v=0

v^2=u^2+2as\\0=30.55^2-2*2.4525*s\\s=190.34m

3 0
2 years ago
The law of reflection says that the angle of incidence is
nekit [7.7K]
The law of reflection states that the angle of incidence is equal to the angle of reflection. Furthermore, the law of reflection states that the incident ray, the reflected ray and the normal all lie in the same plane.

hope this helps :)

4 0
3 years ago
Help pleasee
agasfer [191]

Answer:

i/f = i/o + i/i       f = focal, o = object, i = image

1 / i = 1 / f - 1 / o  =    (o - f) / o f

i = o * f / ( o - f)      image distance

i = 12.5 * 22 / (12.5 - 22) = -28.9 cm

Image is real

Image is 28.9 cm to left of lens

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8 0
2 years ago
Temperature Dependence of the pH of pure Water
Mamont248 [21]

Answer:

At 100°C, the pH of pure water is 6.14. That is the neutral point on the pH scale at this higher temperature. A solution with a pH of 7 at this temperature is slightly alkaline because its pH is a bit higher than the neutral value of 6.14.

Explanation:

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8 0
3 years ago
An interference pattern is produced by light with a wavelength 520 nm from a distant source incident on two identical parallel s
Goshia [24]

Answer:

1) θ = 0.00118 rad, 2)  θ = 0.00236 rad , 3) I / I₀ = 0.1738, 4)  I / Io = 0.216

Explanation:

In the double-slit interference phenomenon it is explained for constructive interference by the equation

          d sin θ = m λ

1) the first order maximum occurs for m = 1

           sin θ = λ  / d

           θ = sin⁻¹ λ  / d

let's reduce the magnitudes to the SI system

           λ  = 520 nm = 520 10⁻⁹  θ = 0.00118 radm

           d = 0.440 mm = 0.440 10⁻³ m ³

let's calculate

           θ = sin⁻¹ (520 10⁻⁹ / 0.44 10⁻³)

            θ = sin⁻¹ (1.18 10⁻³)

            θ = 0.00118 rad

2) the second order maximum occurs for m = 2

            θ = sin⁻¹ (m λ  / d)

            θ = sin⁻¹ (2 5¹20 10⁻⁹ / 0.44 10⁻³)

            θ = 0.00236 rad

3) To calculate the intensity of the interference spectrum, the diffraction phenomenon must be included, so the equation remains

          I = I₀ cos² (π d sin θ /λ ) sinc² (pi b sin θ /λ )

where the function sinc = sin x / x

and b is the width of the slits

we caption the values

             x = π 0.310 10⁻³ sin 0.00118 / 520 10⁻⁹)

             x = 2.21

            I / I₀ = cos² (π 0.44 10⁻³ sin 0.00118 / 520 10⁻⁹) (sin (2.21) /2.21)²

remember angles are in radians

            I / I₀ = cos² (3.0945) [0.363] 2

            I / I₀ = 0.9978 0.1318

            I / I₀ = 0.1738

4) the maximum second intensity is

            I / I₀ = cos² (π d sinθ / λ) sinc² (πb sin θ /λ)

            x =π 0.310 10⁻³ sin 0.00236 / 520 10⁻⁹)

            x = 4.41

            I / Io = cos² (π 0.44 10⁻³ sin 0.00236 / 520 10⁻⁹) (sin 4.41 / 4.41)²

            I / Io = cos² 6.273    0.216

            I / Io = 0.216

.

7 0
2 years ago
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