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kondor19780726 [428]
2 years ago
12

A cartoon shows two friends watching an unoccupied car in free fall after it has rolled off a cliff. one friend says to the othe

r "It goes from zero to sixty miles per hour in about three seconds." is this statement correct?
Physics
2 answers:
Alona [7]2 years ago
6 0
The acceleration of gravity is 32.2 ft/sec^2.
After 3 seconds of free fall, the car is falling at 96.6 ft/sec.

60 mph is 88 ft/sec.

Let's look at it in two ways:

==> 96.6 is 9.8% greater than 88 .

==> It takes the car (88/32.2)=2.73 seconds to reach 60 mph. 3 is 9.8% greater than 2.73 .

Either way you look at it, the rough number is within 10% of the real number. That's plenty close enough to say that they're "about" the same.

It really depends on what the friends mean when they say "about". As an engineer, my personal feeling is that anything less than 10% off is a beautiful match.

enyata [817]2 years ago
3 0
1 mile = 1609 meters.60 miles = 96 540 meters.1 hour = 3 600 seconds.
Then:The velocity V = 96 540 / 3600 ≈ 26.8 m/s
Acceleration of free falla = V/t = 26.8/3 ≈ 8.9 m/c².
But it should be (excluding air resistance of 9.8 m/s².
Answer: the statement is wrong!




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Answer:

Acid mine drainage is dissolved toxic materials wash from mines into nearby lakes and streams.

Explanation:

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8 0
3 years ago
Which of these is an example of acceleration ?
anygoal [31]

Answer:

no picture or anything so cant anwser

Explanation:

8 0
3 years ago
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Gravity does not actually "pull" objects at
Radda [10]
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5 0
3 years ago
A pendulum is hanging from a tower. The pendulum almost touches the ground and has a period of 18.0 s. What is the height of the
nadya68 [22]
We apply the following equation

T = 2π * sqrt (L/g)

Where g is the gravity = 9.8 m/s^2

L is the longitude of the pendulum (Height of the tower)

T is the period. (T = 18s)

We find L.............> (T /2π)^2 = L/g

L = g*(T /2π)^2...........> L = 80.428 meters
5 0
3 years ago
car 1 is traveling south at 18 m/s and has a full load, giving it a total mass of 14,650 kg. Car 2 is traveling north at 11 m/s
kvasek [131]

Answer:

v_{4}= 80.92[m/s] (Heading south)

Explanation:

In order to calculate this problem, we must use the linear moment conservation principle, which tells us that the linear moment is conserved before and after the collision. In this way, we can propose an equation for the solution of the unknown.

ΣPbefore = ΣPafter

where:

P = linear momentum [kg*m/s]

Let's take the southward movement as negative and the northward movement as positive.

-(m_{1}*v_{1})+(m_{2}*v_{2})=-(m_{1}*v_{3})+(m_{2}*v_{4})

where:

m₁ = mass of car 1 = 14650 [kg]

v₁ = velocity of car 1 = 18 [m/s]

m₂ = mass of car 2 = 3825 [kg]

v₂ = velocity of car 2 = 11 [m/s]

v₃ = velocity of car 1 after the collison = 6 [m/s]

v₄ = velocity of car 2 after the collision [m/s]

-(14650*18)+(3825*11)=(14650*6)-(3825*v_{4})\\v_{4}=80.92[m/s]

4 0
2 years ago
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