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Vera_Pavlovna [14]
3 years ago
15

You pull the plunger back on a pinball machine. The spring is pulled back 25 cm from its rest position and has a spring constant

of 78 N/m. What speed will a 50.0 g ball in front of the spring reach when the spring has returned to its rest position?
Physics
1 answer:
NemiM [27]3 years ago
5 0

Answer:

9.9 m/s

Explanation:

from the question we are given the following

length (L) = 25cm = 25 / 100 = 0.25m

spring constant (K) = 78 N/m

mass of the ball (m) = 50g = 50 / 1000 = 0.05 kg

At maximum compression all energy is potential energy and at maximum velocity all energy is kinetic energy, and from the conservation of energy the two energy must be he same. Therefore

potential energy at max compression = kinetic energy at max velocity

(1/2) x K x L^2 = (1/2) x m x V^2

(1/2) x 78 x 0.25^2 = (1/2) x 0.05 x V^2

V^2 = 97.6

V =  9.9 m/s

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In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin
alexandr402 [8]

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a) v=0.999124c

b) E=7.566*10^{22}

c) E_a=760 times\ larger

Explanation:

From the question we are told that

Distance to Betelgeuse d_b=430ly

Mass of Rocket M_r=20000

Total Time in years traveled T_d=36years

Total energy used by the United States in the year 2000 E_{2000}=1.0*10^20

Generally the equation of speed of rocket v mathematically given by

v=\frac{2d}{\triangle t}

v=860ly/ \triangle t

where

\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}

\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}

\triangle t=\sqrt{(860)^2+(36)^2}

\triangle t=860.7532

Therefore

v=\frac{860ly}{ 860.7532}

v=0.999124c

b)

Generally the equation of the energy E required to attain prior speed mathematically given by

E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2

E=7.566*10^{22}

c)Generally the equation of the energy E_a required to accelerate the rocket mathematically given by

E_a=\frac{E}{E_{2000}}

E_a=\frac{7.566*10^{22}}{1.0*10^{20}}

E_a=760 times\ larger

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