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topjm [15]
2 years ago
9

A car that is standing still accelerates up a hill with a slope of 6.4% at an average acceleration of 2.93 ft/s^2. The hill is 1

2.42 miles long, and the final velocity of the car is 203 mph. How long does it take for the car to climb to the top of the hill?
Physics
1 answer:
alexgriva [62]2 years ago
5 0

Answer:

213 s

Explanation:

Slope is the ratio of change in vertical distance to change in horizontal distance.

Slope = vertical height / horizontal height

Therefore:

6.4% = vertical height / 12.42

vertical height = 6.4% * 12.42

vertical height = 0.8 miles

The distance travelled by the car (s) is:

s² = 0.8² + 12.42²

s² = 154.9

s = 12.45 miles

Acceleration (a) =  2.93 ft/s^2 = 0.00055 mile/s²

initial velocity (u) = 0, final velocity = 203 mph

Using:

s = ut + 0.5at²

12.45 = 0.5(0.00055)t²

t =213 s

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Given:
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projector lens focal length, f
distance from the transparency to the projector lens, do

thin lens equation: 1/f = 1/di + 1/do
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convert feet to inches, for uniformity.
1 foot = 12 inches
8 feet * 12 inches/ft = 96 inches
 
1/f = 1/96 inches + 1/4 inches

Adding fractions, denominator must be the same.

1/f = (1/96 * 1/1) + (1/4 * 24/24)
1/f = 1/96 + 24/96
1/f = 25/96

to find the value of f, do cross multiplication
1*96 = f * 25
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The focal length of the project lens is 3.84 inches 

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2 years ago
How long did it take our planet to produce the fossil fuels we're using today?
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It takes millions sometimes hundreds of millions Explanation:

3 0
2 years ago
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A 27-g steel-jacketed bullet is fired with a velocity of 640 m/s toward a steel plate and ricochets along path CD with a velocit
Dmitry [639]

Answer:

F = - 3.56*10⁵ N

Explanation:

To attempt this question, we use the formula for the relationship between momentum and the amount of movement.

I = F t = Δp

Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say

v = d / t

t = d / v

Given that

m = 26 g = 26 10⁻³ kg

d = 50 mm = 50 10⁻³ m

t = d/v

t = 50 10⁻³ / 600

t = 8.33 10⁻⁵ s

F t = m v - m v₀

This is so, because the bullet bounces the speed sign after the crash is negative

F = m (v-vo) / t

F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵

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2 years ago
Be sure to answer all parts. By what factor does the fraction of collisions with energy equal to or greater than activation ener
qwelly [4]

Answer:

Factor = 8.77

Explanation:

Fraction of collision with energy greater than or equal to the activation energy  can be given by the formula:

f = \exp(\frac{-E_{a} }{RT} )

E_{a}  = Activation Energy\\E_{a}  = 100 kJ /mol\\E_{a}  = 10^{5} J /mol\\

R = 8.314 J/mol.K

When Temperature = 34⁰C = 34 + 273 = 307 K

f_{1}  = \exp(\frac{-10^{5}  }{8.314 * 307} )\\f_{1} = \exp(\frac{-10^{5}  }{2552.398} )\\f_{1}  = \exp(-39.18)\\f_{1} = 964.59 * 10^{-20}

When Temperature =  52⁰C = 52 + 273 = 325 K

f_{2}  = \exp(\frac{-10^{5}  }{8.314 * 325} )\\f_{2}= \exp(\frac{-10^{5}  }{2702.05} )\\f_{2}= \exp(-37.009)\\f_{2} = 8456.6 * 10^{-20}

\frac{f_{2}}{f_{1}}= \frac{8456.6 * 10^{-20} }{964.59 * 10^{-20} }

Factor = 8.77

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3 years ago
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