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ikadub [295]
3 years ago
7

A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 1.90 s. You may ignore air

resistance, so the brick is in free fall.(a) How tall, in meters, is the building?(b) What is the magnitude of the brick’s velocity just before it reaches the ground?(c) Sketch ay−t,vy−t, and y-t graphs for the motion of the brick.

Physics
1 answer:
kozerog [31]3 years ago
5 0

Answer:

h=17m

v=18.6 m/s

Explanation: The question can be solved by applying kinematic equations of motion

Data

u=0

a=g

t=1.9 secs

firstly to calculate the height

s=ut+0.5at^2\\h=ut+0.5at^2\\h=0*1.9+0.5*9.81*1.9^2\\h=17.707 m

to find the final velocity

v=u+at\\v=0+9.81*1.9\\v=18.639

The acceleration graph is straight line of equation y=9.8 as acceleration is constant:

Velocity graph is given by y=9.8x ( y as velocity and x as time):

Displacement graph is given by y=4.9x^2 ( x as time, y as displacement):

These graphs are only applicable from x=0 to x=1.9 ... ignore the other graph sections

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never [62]

Answer:

    Em₀ = 245 J

Explanation:

We can solve this problem with the concepts of energy conservation, we assume that there is no friction with the air.

Initial energy the highest point

        Em₀ = U

        Em₀ = m g h

The height can be found with trigonometry

The length of the pendulum is L and the length for the angle of 60 ° is L ’, therefore the height from the lowest point is

         h = L - L’

         cos θ = L ’/ L

         L ’= L cos θ

          h = L (1 - cos θ)

We replace

         Em₀ = m g L (1- cos θ)

Let's calculate

         Em₀ = 10 9.8 5.0 (1 - cos 60)

         Em₀ = 245 J

3 0
3 years ago
An arrow is shot at a target 20 m away. The arrow is shot with a horizontal velocity of 80 m/s.
Verdich [7]

(1) The time of motion of the arrow is 0.25 s.

(2) The vertical height dropped by the arrow as it approaches the target is 0.31 m.

The given parameters:

  • <em>Horizontal distance of the arrow, X = 20 m</em>
  • <em>Horizontal speed of the arrow, v = 80 m/s</em>

<em />

The time of motion of the arrow is calculated as follows;

t = \frac{X}{v} \\\\t = \frac{20 }{80} \\\\t  = 0.25 \ s

The vertical height dropped by the arrow as it approaches the target is calculated as follows;

h = v_0_y t + \frac{1}{2} gt^2\\\\h = 0 \ + \ \frac{1}{2} \times 9.8 \times 0.25^2\\\\h =0.31 \ m

Learn more about time of motion of projectile here:  brainly.com/question/1912408

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2 years ago
A woman climbs up a ladder in 1.37 s at 2.20 m/s. How tall is the ladder?
ArbitrLikvidat [17]

Answer:

The ladder is 3.014 m tall.

Explanation:

To solve this problem, we must use the following formula:

v = x/t

where v represents the woman’s velocity, x represents the distance she climbed (the height of the ladder), and t represents the time it took her to move this distance

If we plug in the values we are given for the problem, we get:

v = x/t

2.20 = x/1.37

To solve this equation for x (the height of the ladder), we must multiply both sides by 1.37. If we do this, we get:

x = (2.20 * 1.37)

x = 3.014 m

Therefore, the ladder is 3.014 m tall.

Hope this helps!

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4 years ago
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