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kondor19780726 [428]
3 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]3 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]3 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:

   d = 19.92 m

Explanation:

As in this exercise there is friction we must use the relationship between work and energy

          W = ΔEm

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Initial. Fully compressed spring

          Em₀ = K_{e} = ½ k x²

Final. When the block stopped

      Em_{f} = 0

Let's look for the work of the rubbing force

       W = fr d cos θ

Since rubbing is always contrary to movement, θ = 180

      W = - fr d

Let's use Newton's second Law, to find the force of friction

Y Axis

          N- w = 0

          N = mg

The equation for the force of friction is

         fr = μ N

         fr = μ mg

We substitute in the work equation

          W = - μ m g d

We write the relationship of work and energy

     -μ m g d = 0 - ½ k x²

     d = ½ k x² / μ m g

Let's calculate

     d = ½ 131 2.1 2 / (0.74 2 9.8)

     d = 19.92 m

7 0
4 years ago
A projectile is fired with a velocity of 320 ms at an angle of 30 degree to a horizontal.1 Find the time to reach the maximum he
dedylja [7]

(a) The time taken for the projectile to reach the maximum height is 32.65 s.

(b) The horizontal range of the projectile is 9,049.1 m.

The given parameters:

  • Initial velocity of the projectile, u = 320 m/s
  • Angle of projection, = 30 degrees

The time taken for the projectile to reach the maximum height is calculated as follows;

v_f = u- gt\\\\0 = 320 - 9.8t\\\\9.8t = 320\\\\t = \frac{320}{9.8} \\\\t = 32.65 \ s

The horizontal range of the projectile is calculated as follows;

R = \frac{u^2 sin(2\theta)}{g} \\\\R = \frac{320^2 \times sin(2\times 30)}{9.8} \\\\R = 9,049.1 \ m

Learn more about horizontal range here: brainly.com/question/12870645

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A horizontal force of magnitude 34.0 n pushes a block of mass 3.93 kg across a floor where the coefficient of kinetic friction i
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A Cessna 150 aircraft has a lift-off speed of approximately 125 km/h. What minimum constant acceleration does this require if th
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Answer:

Part a)

a = 3.65 m/s^2

Part b)

t = 9.5 s

Part c)

v_f = 55.1 m/s

Explanation:

Part a)

As we know that it starts from rest and moves on runway by total distance 165 m

so we will have

v_f^2 - v_i^2 = 2ad

v_f^2 - 0 = 2(a)(165)

v_f = 125 km/h = 34.7 m/s

now we have

a = 3.65 m/s^2

Part b)

Now for take off time we will have

v_f - v_i = at

34.7 - 0 = 3.65 t

t = 9.5 s

Part c)

v_f = v_i + at

v_f = 0 + (3.65)(15.1)

v_f = 55.1 m/s

7 0
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