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8090 [49]
2 years ago
13

3. You are walking in Paris alongside the Eiffel Tower and suddenly a croissant smacks you on the head and knocks you to the gro

und. From your handy dandy tourist guidebook you find that the height of the Eiffel Tower is 300.5 m. If you neglect air resistance, calculate how many seconds the croissant dropped before it tagged you on the head.
Physics
1 answer:
Elza [17]2 years ago
5 0

As absurd as the concept is, we must assume that a croissant
can fall 300.5 meters through the moisture-laden, perfumed and
polluted Parisian air with no air resistance whatsoever.

Acceleration due to gravity on Earth:  9.8 m/s²

Distance in clean,
unimpeded free-fall         = (1/2) (acceleration) x (time²) 

                                            300.5 m  =  (1/2)  (9.8 m/s²) (T²)
Divide each side
by (4.9 m/s²):                  (300.5 m) / (4.9 m/s²)  =  T²

Take the square root
of each side:                     T = √(300.5/4.9) (s²)

                                                               =  7.831 seconds .

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two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i
NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

h = H - ut - \frac{1}{2}gt^2

0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 - 14.7t - 4.9t^2

4.9t^2 + 14.7t - 19.6 = 0

t^2 + 3t - 4 = 0

(t + 4)(t - 1) = 0

(t - 1) = 0

\boxed {t = 1 ~ second}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 + 14.7t - 4.9t^2

4.9t^2 - 14.7t - 19.6 = 0

t^2 - 3t - 4 = 0

(t - 4)(t + 1) = 0

(t - 4) = 0

\boxed {t = 4 ~ seconds}

The difference in the two ball's time in the air is:

\Delta t = 4 ~ seconds - 1 ~ second

\large {\boxed {\Delta t = 3 ~ seconds} }

<h2>Question B:</h2><h3>First Ball</h3>

v^2 = u^2 - 2gH

v^2 = (-14.7)^2 + 2(-9.8)(-19.6)

v^2 = 600.25

v = \sqrt {600.25}

\boxed {v = 24.5 ~ m/s}

<h3>Second Ball</h3>

v^2 = u^2 - 2gH

v^2 = (14.7)^2 + 2(-9.8)(-19.6)

v^2 = 600.25

v = \sqrt {600.25}

\boxed {v = 24.5 ~ m/s}

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

h = H - ut - \frac{1}{2}gt^2

h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2

\boxed {h = 11.025 ~ m}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2

\boxed {h = 25.725 ~ m}

The difference in the two ball's height after 0.500 s is:

\Delta h = 25.725 ~ m - 11.025 ~ m

\large {\boxed {\Delta h = 14.7 ~ m} }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0
2 years ago
A person's height will increase from birth until about age 25, and it may decrease starting at about age 70. This is an example
dsp73
LOL idk sorry maybe it b! Oh wait it is !
5 0
2 years ago
Read 2 more answers
A helicopter pulls upward by means of a rope on a 250 kg crate to lift it UNIFORMLY. What is the net force on the crate?
Cloud [144]

Answer:

The net force = 0

Explanation:

The given information includes;

The mass of the crate = 250 kg

The way the helicopter lifts the crate = Uniformly (constant rate (speed), no acceleration)

In order to pull the crate upwards, the helicopter has to provide a force equivalent to the weight of the crate keeping the helicopter on the ground.

The weight of the crate = The mass of the crate × The acceleration due gravity acting on the crate

The weight of the crate, F_w↓ = 250 kg × 9.81 m/s² = 2,452.5 N

The force the helicopter should provide to just lift the crate, F_{(helicopter)}↑ = The weight of the crate = 2,452.5 N

The net force, F_{(net)} = F_{(helicopter)}↑ - F_w↓ = 2,452.5 N - 2,452.5 N = 0

The net force = 0.

3 0
2 years ago
a cannonball with a mass of 1.0 kilogram is fired horizontally from a 500.- kilogram cannon, initially at rest, on a horizontal,
nirvana33 [79]

(a) According to Newton’s third law, for every action there is an equal yet opposite reaction. In this case, the cannonball is affected by 8.0x10^3 newtons in one direction. Now the cannon must also be affected by the same amount of net force on the opposite direction.

- 8.0x10^3 Newtons (the negative symbol only shows that it is opposite direction)

 

(b) We solve this using the formula:

F = m a

8.0x10^3 N = (1.0 kg) a

a = 8,000 m/s^2

8 0
2 years ago
An 8.0 m, 240 N uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground
sweet [91]

Answer:

5.7 m

Explanation:

AD = length of the ladder = L = 8 m

AB = distance of the center of mass of the ladder = (0.5) L = (0.5) 8 = 4 m

AC = distance of person on the ladder from the bottom end = x

W = weight of the ladder = 240 N

F_{g} = weight of the person = 710 N

F = force by the wall on the ladder

N = normal force by ground on the ladder = ?

Using equilibrium of force along the vertical direction

N = F_{g} + W

N = 710 + 240

N = 950 N

μ = Coefficient of static friction = 0.55

f =static frictional force on the ladder

Static frictional force is given as

f = μ N

f = (0.55) (950)

f = 522.5 N

Force equation along the horizontal direction is given as

F = f

F = 522.5 N

using equilibrium of torque about point A

F Sin50 (AD) = W Cos50 (AB) + (F_{g} Cos50 (AC))

(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)

x = 5.7 m

7 0
2 years ago
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