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Verizon [17]
2 years ago
5

a ball is shot from the ground straight up into the air with initial velocity of 50 50 ft/sec. assuming that the air resistance

can be ignored, how high does it go?
Physics
1 answer:
Anna [14]2 years ago
3 0

The height attained by the ball is 11.86m

a ball is shot from the ground straight up into the air its initial and final velocity is

initial velocity, u = 50 ft/s = 50×0.305  = 15.25m/s

final velocity ,v = 0 m/s

gravity =-9.8 m/s²

( negative sign shows acceleration in opposite direction)

height =?

using the newton motion of equation

v² = u² + 2as

where

a= acceleration due to gravity(g)

s = height

v² = u² + 2gs

(0)² = (15.25)² + 2×(-9.8)×s

0  = (15.25)² -  19.6 × s

s= - (15.25)²/ 19.6

s = 11.86m

after ignoring the air resistance the maximum height of the ball is 11.86m

To learn more about motion under gravity -

brainly.com/question/27962354

#SPJ4

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3 years ago
Two particles with charges +6e and -6e are initially very far apart (effectively an infinite distance apart). They are then fixe
JulijaS [17]

Answer:

\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}\ J.

Explanation:

Given charges are:

\rm q_1 = +6e.\\q_2 = -6e.

The electric potential energy of a charge due to the electric field of another charge is given by

\rm EPE=\dfrac{kq_1q_2}{r}.

where,

  • k = Coulomb's constant, having value = \rm 9\times 10^9\ Nm^2/C^2.
  • r = distance between the charges.

When the charges are infinite distance apart, \rm r = \infty,

\rm EPE_{initial} = \dfrac{kq_1q_2}{r}=0\ J.

When the charges are \rm 5.61\times 10^{-12}\ m apart, \rm r=5.61\times 10^{-12}\ m,

\rm EPE_{final}=\dfrac{kq_1q_2}{r}\\=\dfrac{(9\times 10^9)\times (+6e)\times (-6e)}{5.61\times 10^{-12}}\\=-5.775\ e^2\times 10^{22}.

Here, e is the charge on one electron, such that, \rm e = -1.6\times 10^{-19}\ C.

Therefore,

\rm EPE_{final}=-5.775\times (-1.6\times 10^{-19})^2\times 10^{22} = -1.478\times 10^{-15}\ J.

Thus,

\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}-0=-1.478\times 10^{-15}\ J.

4 0
3 years ago
Water from a leaking tap begins to fall into
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Answer:

It will take 20,000 seconds to get a pressure of 200Pa at the bottom of the tank.

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The pressure at the bottom of the tank will be

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where M is the mass of the water, and A is the base area of the tank.

The base area of the tank is

A = 0.5m* 2m = 1m^2,

and if we want the pressure at the bottom to be 200pa, then it must be that

200Pa = \dfrac{M(10m/s^2)}{1m^2},

solving for M we get:

M = 20kg\\

which is the required mass of the water in the tank.

Now, the tank fills at a rate of 2 drops per second or

2 (0.05g)/s = 0.10g/s = 0.0001kg/s

since each drop weights 0.05g.

Therefore, the time t it takes to collect 20kg of water will be

t = 20kg \div \dfrac{0.0001kg}{s}

t = 2*10^5s

which is 55.56 hours or 2 days and 7.56 hours.

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Answer:

here is explanation

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(OR ABOUT 268,770 AU)

5 0
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