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3 years ago

A ball is thrown at 60 degrees and lands in 18.5 seconds what is the velocity of the ball at the start

Physics

1 answer:

n200080 [17]3 years ago

3 0

Answer:

the initial velocity of the ball is 104.67 m/s.

Explanation:

Given;

angle of projection, θ = 60⁰

time of flight, T = 18.5 s

let the initial velocity of the ball, = u

The time of flight is given as;

$T = \frac{2u\times sin(\theta)}{g} \\\\2u\times sin(\theta) = Tg\\\\u = \frac{Tg}{2\times sin(\theta)} \\\\u = \frac{18.5 \times 9.8}{2\times sin(60^0)} \\\\u = 104.67 \ m/s$

Therefore, the initial velocity of the ball is 104.67 m/s.

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A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

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3 years ago

The part of a sound wave in which the particles are most spread out is called a(n)

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Seismic wave is the answer

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3 years ago

8. Il An 8.00 kg package in a mail-sorting room slides 2.00 m down a

Vitek1552 [10]

Answer:

See below

Explanation:

Normal force = m g cos 53 = 8 kg * 9.8 m/s^2 * cos 53 = 47.1823 N

no work is done by this force

Force friction = coeff friction * force normal = .4 * 47.1823 = 7.55 N

work of friction = 7.55 * 2 m = 15.1 j

Force Downplane = mg sin 53 = 62.61 N

work = 62.61 * 2 = 125.22 j

Net Force downplane = force downplane - force friction = 55.06 N

net Work = force * distance = 55.06 N * 2 M = 110.12 j

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2 years ago

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3 years ago

While eating lunch high up in a skyscraper, two construction workers calculate their gravitational potential

Maurinko [17]

Answer:

The mass of the other worker is 45 kg

Explanation:

The given parameters are;

The gravitational potential energy of one construction worker = The gravitational potential energy of the other construction worker

The mass of the lighter construction worker, m₁ = 90 kg

The height level of the lighter construction worker's location = h₁

The height level of the other construction worker's location = h₂ = 2·h₁

The gravitational potential energy, P.E., is given as follows;

P.E. = m·g·h

Where;

m = The mass of the object at height

g = The acceleration due to gravity

h = The height at which is located

Let P.E.₁ represent the gravitational potential energy of one construction worker and let P.E.₂ represent the gravitational potential energy of the other construction worker

We have;

P.E.₁ = P.E.₂

Therefore;

m₁·g·h₁ = m₂·g·h₂

h₂ = 2·h₁

We have;

m₁·g·h₁ = m₂·g·2·h₁

m₁ = 2·m₂

90 kg = 2 × m₂

m₂ = (90 kg)/2 = 45 kg

The mass of the other construction worker is 45 kg.

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