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3 years ago

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Kelli weighs 425 N, and she is sitting on a playground swing that hangs 0.36 m above the ground. Her mom pulls the swing back an

d releases it when the seat is 1.00 m above the ground. Assume that air resistance is negligible. (a) How fast is Kelli moving when the swing passes through its lowest position?

1 answer:

kodGreya [7K]3 years ago

7 0

Answer:

V = 3.54 m/s

Explanation:

Using the conservation of energy:

$E_i = E_f$

so:

$wh = \frac{1}{2}mV^2$

where w is te weigh of kelly, h the distance that kelly decends, m is the mass of kelly and V the velocity in the lowest position.

So, the mass of kelly is:

m = 425N/9.8 = 43.36 Kg

and h is:

h = 1m-0.36m =0.64m

then, replacing values, we get:

$(425N)(0.64m) = \frac{1}{2}(43.36kg)v^2$

Solving for v:

V = 3.54 m/s

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A 24.7-g bullet is fired from a rifle. It takes 2.73 × 10-3 s for the bullet to travel the length of the barrel, and it exits th

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Answer:

$F_a_v_g=7093333.33N*s$

Explanation:

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Where:

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$v_2=final\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}at\hspace{3}the\hspace{3}end\hspace{3}of\hspace{3}the\hspace{3}time\hspace{3}interval$

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$t_2=final\hspace{3}time$

$t_1=initial\hspace{3}time$

Asumming v1=0 and t1=0:

$F_a_v_g=m* \frac{v_2}{t_2} =(24.7)*\frac{784}{2.73*10^{*3} } =7093333.333N*s$

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3 years ago

The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming a

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Answer:

The radius of the gold nucleus is 7.1x10⁻¹⁵m

Explanation:

The nearest distance is:

$r=\frac{kze^{2} }{mpV^{2} }$ (eq. 1)

Where

z = atomic number of gold = 79

e = electron charge = 1.6x10⁻¹⁹C

k = electrostatic constant = 9x10⁹Nm²C²

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At the potential energy is zero, all the energy will be kinetic energy:

$E_{k} =\frac{1}{2} m V^{2}$

Where

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$5.12x10^{-12} =\frac{1}{2} *4*m_{p}* V^{2} \\m_{p}* V^{2} = 2.56x10^{-12}$

Replacing in equation 1

$r=\frac{9x10^{9}*79*(1.6x10^{-19})^{2} }{2.56x10^{-12} } =7.1x10^{-15} m$

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2 years ago

An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b

lys-0071 [83]

Answer:

a)Distance traveled during the first second = 4.905 m.

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c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

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a) We have equation of motion

S = ut + 0.5at²

Here u = 0, and a = g

S = 0.5gt²

Distance traveled during the first second ( t =1 )

S = 0.5 x 9.81 x 1² = 4.905 m

Distance traveled during the first second = 4.905 m.

b) We have equation of motion

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Here u = 0, s= 75 m and a = g

v² = 0² + 2 x g x 75 = 150 x 9.81

v = 38.36 m/s

Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

75 = 0.5 x 9.81 x t²

t = 3.91 s

We need to find distance traveled last second

That is

S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

Distance traveled during the last second of motion before hitting the ground = 33.45 m

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3 years ago

A baseball has mass 0.147 kg. If the velocity of a pitched ball has a magnitude of 44.5 m/s and the batted ball's velocity is 55

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Explanation:

We have,

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Change in momentum,

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2 years ago

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Answer:

1) It expresses the rate (top speed) at which it can move with time.

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Explanation:

1) Power is the rate of transfer of energy.

⇒ Power = $\frac{Energy(or workdone)}{Time}$

i.e P = $\frac{E}{t}$

Thus a car's engine power is 44000W implies that the engine of the car can propel the car at this rate. This expresses the rate (top speed) at which it can move with time.

2) m = 400g = 0.4 kg

t = 20 s

h = 100m

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P = $\frac{mgh}{t}$

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= $\frac{400}{20}$

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2 years ago