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gladu [14]
3 years ago
14

At the same moment from the top of a building 3.0 × 10 2 m tall, one rock is dropped and one is thrown downward with an initial

velocity of 26 m/s. both of them experience negligible air resistance. how much earlier does the thrown rock strike the ground?
Physics
2 answers:
Bess [88]3 years ago
5 0
The equation that relates distance, velocities, acceleration, and time is,
                   d = V₀t + 0.5gt²
where d is distance,
V₀ is the initial velocity,
t is time, and 
g is the acceleration due to gravity (equal to 9.8 m/s²)

(1) Dropped rock,
                  (3 x 10² m ) = 0(t) + 0.5(9.8 m/s²)(t²)
The value of t from this equation is 24.73 s

(2) Thrown rock with V₀ = 26 m/s
                (3 x 10² m) = (26)(t) + 0.5(9.8 m/s²)(t²)
The value of t from the equation is 5.61 s

The difference between the tim,
        difference = 24.73 s - 5.61 s
          difference = 19.12 s

<em>ANSWER: 19.12 s</em>
erik [133]3 years ago
3 0

Answer:

t = 5.607 sec

Explanation:

s = ut + (1/2)at²

you can figure out the time that the dropped ball to hit the ground quite easily because u = 0 m/s, therefore

s = (1/2)at²

t² =2s/a

t = sqrt(2s/a)

assuming the acceleration from gravity is 9.81 m/s²

t = sqrt(2*300m/9.81 m/s²)

t = 7.8 s

You can use the quadratic equation to find t for the thrown rock and then subtract them.  Or you can calculate the final velocity of the thrown rock with v^2 = u^2 + 2as and then use v = u + at to find the time for the thrown rock.

v² = (26 m/s)² + 2 * (9.81 m/s^2 * 300 m)

v² =6562

v = 81 m/s

v = u + at

t = (v - u)/s

t = (81 m/s - 26 m/s)/9.81 m/s^2

t = 5.607 sec

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A stone that is dropped freely from rest traveled half the total height in the last second. with what velocity will it strike th
alexira [117]

Answer:

hellooooo :) ur ans is 33.5 m/s

At time t, the displacement is h/2:

Δy = v₀ t + ½ at²

h/2 = 0 + ½ gt²

h = gt²

At time t+1, the displacement is h.

Δy = v₀ t + ½ at²

h = 0 + ½ g (t + 1)²

h = ½ g (t + 1)²

Set equal and solve for t:

gt² = ½ g (t + 1)²

2t² = (t + 1)²

2t² = t² + 2t + 1

t² − 2t = 1

t² − 2t + 1 = 2

(t − 1)² = 2

t − 1 = ±√2

t = 1 ± √2

Since t > 0, t = 1 + √2.  So t+1 = 2 + √2.

At that time, the speed is:

v = at + v₀

v = g (2 + √2) + 0

v = g (2 + √2)

If g = 9.8 m/s², v = 33.5 m/s.

4 0
3 years ago
Write the chemical formula for the following diagrams.
seraphim [82]

Answer:

hydrogen chloride...........

4 0
3 years ago
What is the net force exerted on a 28.8 kg shopping cart that accelerates at a rate of 2.88 m/s^2?​
Mama L [17]

Answer:

<h2>82.94 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

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force = 28.8 × 2.88 = 82.944

We have the final answer as

<h3>82.94 N</h3>

Hope this helps you

6 0
3 years ago
Help me asap its due today
Margaret [11]

Answer:

3.54* 10^{22} N

Explanation:

Using the formula you gave:

F_g = \frac{6.67*10^{-11}*2.0*10^{30}*5.97^{24}  }{(1.5*10^{11})^2 }

3 0
3 years ago
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