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salantis [7]
3 years ago
13

If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the

curve (a real problem on icy mountain roads). (a) Calculate the ideal speed in (m/s) to take a 100 m radius curve banked at 15°. m/s (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 15.0 km/h?
Physics
1 answer:
PolarNik [594]3 years ago
3 0

Answer:

a. v₁ = 16.2 m/s

b. μ = 0.251

Explanation:

Given:

θ = 15 ° , r = 100 m , v₂ = 15.0 km / h

a.

To determine v₁ to take a 100 m radius curve banked at 15 °

tan θ  = v₁² / r * g

v₁ = √ r * g * tan θ

v₁ = √ 100 m * 9.8 m/s² * tan 15° = 16.2 m/s

b.

To determine μ friction needed for a frightened

v₂ = 15.0 km / h * 1000 m / 1 km * 1h / 60 minute * 1 minute / 60 seg

v₂ = 4.2 m/s

fk = μ * m * g

a₁ = v₁² / r = 16.2 ² / 100 m = 2.63 m/s²

a₂ = v₂² / r = 4.2 ² / 100 m = 0.18 m/s²

F₁ = m * a₁  ,  F₂ = m * a₂

fk = F₁ - F₂   ⇒  μ * m * g = m * ( a₁ - a₂)

μ * g = a₁ - a₂   ⇒  μ = a₁ - a₂ / g

μ = [ 2.63 m/s² - 0.18 m/s² ] / (9.8 m/s²)

μ = 0.251

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Answer:

Explanation:

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Given

velocity = 139m/s

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Does electric and magnetic objects both have north and south poles
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Explanation:

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2 years ago
In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel
insens350 [35]

Answer:

59.26°

Explanation:

Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.

Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.

Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration =  acosθ.

So, y = ut + 1/2a't²

y = 0 × t + 1/2(acosθ)t²

y = 0 + 1/2(acosθ)t²

y = 1/2(acosθ)t²   (1)

Also, both particles must move the same horizontal distance to collide in time, t.

Let x be the horizontal distance,

x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision

Also,  using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration =  asinθ.

So, x = ut + 1/2a"t²

x = 0 × t + 1/2(ainsθ)t²

x = 0 + 1/2(asinθ)t²

x = 1/2(asinθ)t²  (3)

Equating (2) and (3), we have

vt = 1/2(asinθ)t²   (4)

From (1) t = √[2y/(acosθ)]

Substituting t into (4), we have

v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²  

v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)  

v√[2y/(acosθ)] = ytanθ

√[2y/(acosθ)] = ytanθ/v

squaring both sides, we have

(√[2y/(acosθ)])² = (ytanθ/v)²

2y/acosθ = (ytanθ/v)²

2y/acosθ = y²tan²θ/v²

2/acosθ = ytan²θ/v²

1/cosθ = aytan²θ/2v²

Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1

secθ = ay(sec²θ - 1)/2v²

2v²secθ = aysec²θ - ay

aysec²θ - 2v²secθ - ay = 0

Let secθ = p

ayp² - 2v²p - ay = 0

Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have

ayp² - 2v²p - ay = 0

0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0

10.85p² - 15.68p - 10.85 = 0

dividing through by 10.85, we have

p² - 1.445p - 1 = 0

Using the quadratic formula to find p,

p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125

Since p = secθ

secθ = 1.95625 or secθ = -0.51125

cosθ = 1/1.95625 or cosθ = 1/-0.51125

cosθ = 0.5112 or cosθ = -1.9956

Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.

So, cosθ = 0.5112

θ = cos⁻¹(0.5112)

θ = 59.26°

So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.

5 0
3 years ago
An unknown material, m1 = 0.45 kg, at a temperature of T1 = 91 degrees C is added to a Dewer (an insulated container) which cont
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Answer:

Explanation:

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heat lost = heat gained

m₂ cw (T2 - T ) = m₁ s (T 1 - T )

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45711.12 = 26.82 s

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3 years ago
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