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prisoha [69]
3 years ago
7

You are driving down the road at 30 mph (miles per hour) when you see the traffic light change to red. The red light is 300 feet

from you. How much time will it take you to stop? (please provide explanation)
Physics
1 answer:
RSB [31]3 years ago
8 0

Answer:

45 s

Explanation:

To find the time it takes to stop, we first find the deceleration, a of the car from

v² = u² + 2as and a = (v² - u²)/2s were v = final velocity of car = 0 mph = 0 m/s, u = initial velocity of car = 30 mph = 30 × 1609.34 ft ÷ 3600 s = 13.41 ft/s and s = distance = 300 ft. Substituting the values into a, we ave

a = (v² - u²)/2s = (0² - 13.41²)/2×300 = -0.3 ft/s²

We then find the time for this deceleration from v = u + at ⇒ t = (v - u)/a

t = (v - u)/a = (0 - 13.41 ft/s)/-0.3 ft/s² = - 13.41 ft/s/-0.3 ft/s² = 44.7 s ≅ 45 s

So it takes 45 seconds to stop.

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Samantha is refinishing her rusty wheelbarrow. She moves her sandpaper back and forth 45 times over a rusty area, each time movi
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Answer:

W = 12.96 J

Explanation:

The force acting in the direction of motion of the sand paper is the frictional force. So, we first calculate the frictional force:

F = μR

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F = Friction Force = ?

μ = 0.92

R = Normal Force = 2.6 N

Therefore,

F = (0.92)(2.6 N)

F = 2.4 N

Now, the displacement is given as:

d = (0.12 m)(45)

d = 5.4 m

So, the work done will be:

W = F d

W = (2.4 N)(5.4 m)

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A metal and a non-metal typically form covalent compounds.<br>True<br>False​
alexandr1967 [171]

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Explanation:

4 0
3 years ago
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Experimenting with free fall, Mariana observes that her baseball takes 1.5 s to travel the last 30m before hitting the ground. F
Art [367]

Answer:

37.8 m

Explanation:

At point 0, the ball is at height y₀.

At point 1, the ball is at height 30 m.

At point 2, the ball is at height 0 m.

Given:

y₁ = 30 m

y₂ = 0 m

v₀ = 0 m/s

a = -10 m/s²

t₂ − t₁ = 1.5 s

Find: y₀

Use constant acceleration equation.

y = y₀ + v₀ t + ½ at²

Evaluate at point 1.

y₁ = y₀ + v₀ t₁ + ½ at₁²

30 m = y₀ + (0 m/s) t₁ + ½ (-10 m/s²) t₁²

30 = y₀ − 5t₁²

Evaluate at point 2.

y₂ = y₀ + v₀ t₂ + ½ at₂²

0 m = y₀ + (0 m/s) t₂ + ½ (-10 m/s²) t₂²

0 = y₀ − 5t₂²

y₀ = 5t₂²

Substitute:

y₀ = 5 (1.5 + t₁)²

y₀ = 5 (2.25 + 3t₁ + t₁²)

y₀ = 11.25 + 15t₁ + 5t₁²

30 = 11.25 + 15t₁ + 5t₁² − 5t₁²

30 = 11.25 + 15t₁

t₁ = 1.25

30 = y₀ − 5t₁²

30 = y₀ − 5(1.25)²

y₀ ≈ 37.8

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3 years ago
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