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Alex73 [517]
3 years ago
11

suppose a ball is thrown vertically upward. Eight seconds later it returns to its point of release. What is the initial velocity

of the ball?
Physics
1 answer:
valentinak56 [21]3 years ago
6 0
The ball took half of the total time ... 4 seconds ... to reach its highest
point, where it began to fall back down to the point of release.

At its highest point, its velocity changed from upward to downward. 
At that instant, its velocity was zero.

The acceleration of gravity is 9.8 m/s².  That means that an object that's
acted on only by gravity gains 9.8 m/s of downward speed every second. 

-- If the object is falling downward, it moves 9.8 m/s faster every second.

-- If the object is tossed upward, it moves 9.8 m/s slower every second.

The ball took 4 seconds to lose all of its upward speed.  So it must have
been thrown upward at  (4 x 9.8 m/s)  =  39.2 m/s .

(That's about  87.7 mph straight up.  Somebody had an amazing pitching arm.)
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Answer:

The correct answer is "The median year built is greater than the mean year built., The median year built is between 1980 and 2000."

Explanation:

Wake country is located in the north Carolina, it is the most fastest growing country among the united nations. The town of Cary and also city of Raleigh is the 8th and the 15th fastest growing cities. Hence, the city was growing along with infrastructures as well as with the population. For proper shelter more number of the homes were built. The literacy rates were also good. A quite good number of schools and colleges are there in the city. With better medical facilities.

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When work is done on a spring to stretch it, elastic potential energy is stored in the spring.
aniked [119]

Explanation:

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3 years ago
An object initially at rest experiences an acceleration of 0.281 m/s2 to the South for a time of 5.44 seconds. It then increases
andre [41]

Answer:

12.0 meters

Explanation:

Given:

v₀ = 0 m/s

a₁ = 0.281 m/s²

t₁ = 5.44 s

a₂ = 1.43 m/s²

t₂ = 2.42 s

Find: x

First, find the velocity reached at the end of the first acceleration.

v = at + v₀

v = (0.281 m/s²) (5.44 s) + 0 m/s

v = 1.53 m/s

Next, find the position reached at the end of the first acceleration.

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²

x = 4.16 m

Finally, find the position reached at the end of the second acceleration.

x = x₀ + v₀ t + ½ at²

x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²

x = 12.0 m

5 0
3 years ago
Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R.From the express
AlladinOne [14]

Answer:

Vb = k Q / r        r <R

Vb = k q / R³ (R² - r²)    r >R

Explanation:

The electic potential is defined by

             ΔV = - ∫ E .ds

We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product

             VB - VA = - ∫ E dr

Let's substitute every equation they give us and we find out

r> R

           Va = - ∫ (k Q / r²) dr

           -Va = - k Q (- 1 / r)

We evaluate with it Va = 0 for r = infinity

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We perform the calculation of the power with the expression of the electric field that they give us

           Vb = - int (kQ / R3 r) dr

  We integrate and evaluate from the starting point r = R to the final point r <R

         Vb = ∫kq / R³ r dr

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This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity

8 0
3 years ago
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