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Romashka-Z-Leto [24]
4 years ago
7

A small piece of Styrofoam packing material is dropped from a height of 2.60 m above the ground. Until it reaches terminal speed

, the magnitude of its acceleration is given by a = g − Bv. After falling 0.600 m, the Styrofoam effectively reaches terminal speed, and then takes 4.50 s more to reach the ground.
(a) What is the value of the constant B? s-1.
(b) What is the acceleration at t = 0? [m/s2 (down)].
(c) What is the acceleration when the speed is 0.150 m/s? [m/s2 (down)].
Physics
1 answer:
Vadim26 [7]4 years ago
7 0

Answer:

Explanation:

a ) After the attainment of terminal speed , object takes 4.5 s to cover a distance of 2 m

So terminal speed V = 2 / 4.5

= .444 m /s

When it attains terminal speed , acceleration becomes zero

0 = g - B x .444

B = 22.25 s⁻¹

b ) At t = 0 , v = 0

a = g - B v

a = g at t = 0

c ) When v = .15

a = g - 22.25 x .15

= 9.8 - 3.31

= 6.5  m /s²

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Which type of planets have the most moons? Where did these moons likely originate?
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for example ,

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Imagine an alternative universe where the characteristic decay time of neutrons is 3 min instead of 15 min. All other properties
FrozenT [24]

Answer:

(a) [Y_{p} ]_{max} = \frac{2f}{1+f}

(b) f_{new} = 0.013; [Y_{p} ]_{max} = 0.026

Explanation:

Since the neutron-to-proton ratio at the time of nucleosynthesis is given:

f = \frac{n_{n} }{n_{p} }

Therefore:

n_{n} = f*n_{p}

Then, to determine the maximum ⁴He fraction if all the available n_{n} neutrons bind to all the protons. Since, there are 2 protons and 2 neutrons in a ⁴He nucleus, it shows that there would be n_{n}/2 nuclei of ⁴He.

In addition, a ⁴He nucleus has a mass of 4m_{p}, where m_{p} is the mass of one proton. Thus, n_{n}/2 nuclei of such nuclei will have a mass of n_{n}/2*4m_{p}.

Assuming that m_{p}=m_{n}, there would be a total of (n_{n}+n_{p}) protons and neutrons with a total mass of (n_{n}+n_{p})*m_{p}.

Thus:[Y_{p} ]_{max} = \frac{2f}{1+f}

(b) Given:

t_{nuc} = 200 s;   τ_{n} = 3*60s = 180 s

f_{new} = \frac{n_{nf} }{n_{pf} } = \frac{exp (-200/180)}{5 +[1- exp(-200/180)]} =\frac{0.077}{5.923} = 0.013

[Y_{p} ]_{max} = \frac{2f}{1+f} = (2*0.013)/(1+0.013) = 0.026

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