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Bingel [31]
3 years ago
7

A ball is released from a tower at a height of 100 meters toward the roof of another tower that is 25 meters high. The horizonta

l distance between the two towers is 20 meters. With what horizontal velocity should the ball be imparted so that it lands on the rooftop of the second building?
Physics
1 answer:
I am Lyosha [343]3 years ago
7 0
-- During the time the ball is flying from the high roof to the low roof,
it's going to fall (100-25) = 75 meters.
How long does it take an object dropped from rest to fall 75 meters ?

Distance = (1/2) · (gravity) · (time)²

75 m = (4.9 m/s²) · (time)²

Time² = (75 m) / (4.9 m/s²)
Time² = 15.31 sec²
Time = √(15.31 sec²)  =  3.91 seconds 

So the ball has to cover the horizontal distance of 20 meters 
in 3.91 seconds.

Distance = (speed) · (time)

20 m = (speed) · (3.91 sec)

Speed = (20 m) / (3.91 sec)

Speed  =  5.11 m/s
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How would local winds be affected if water and land release heat at the same time
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8 0
2 years ago
Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final
Serhud [2]

Complete question:

Consider the hypothetical reaction 4A + 2B → C + 3D

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?

Answer:

the final concentration of A is 0.992 M.

Explanation:

Given;

time of reaction, t = 4.0 s

rate of change of the concentration of B =  -0.0760 M/s

initial concentration of A = 1.600 M

⇒Determine the rate of change of the concentration of A.

From the given reaction: 4A + 2B → C + 3D

2 moles of B ---------------> 4 moles of A

-0.0760 M/s of B -----------> x

x = \frac{4(-0.076)}{2} \\\\x = -0.152 \ M/s

⇒Determine the change in concentration of A after 4s;

ΔA = -0.152 M/s  x 4s

ΔA = -0.608 M

⇒ Determine the final concentration of A  after 4s

A = A₀ + ΔA

A = 1.6 M + (-0.608 M)

A = 1.6 M - 0.608 M

A = 0.992 M

Therefore, the final concentration of A is 0.992 M.

5 0
2 years ago
A charge +Q is located at the origin and a second charge, +4Q, is at distance (d) on the x-axis.
tatiyna

Answer:

a)   x = ⅔ d , b) the charge must be negative, c) Q

Explanation:

a) In this exercise the force is electric between the charges, we are asked that the system of the three charges is in equilibrium, we use Newton's second law. Balance is on the third load that we are placing

         ∑ F = 0

        -F₁₂ + F₂₃ = 0

         F₁₂ = F₂₃

         

let's replace the values

        k Q Q / r₁₂² = k Q 4Q / r₂₃²

            Q² / r₁₂² = 4 Q² / r₂₃²

suppose charge 3 is placed at point x

        r₁₂ = x

        r₂₃ = d-x

             

we substitute

             1 / x² = 4 / (d-x) 2

             1 / x = 2 / (d-x)

             x = 2 (x-d)

             x = 2x -2d

            3x = 2d

              x = ⅔ d

b) The sign of the charge must be negative, to have an attractive charge on the two initial charges

c)  Q

5 0
3 years ago
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