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Oksana_A [137]
3 years ago
7

(a) Calculate the buoyant force on a 2.00-L helium balloon.

Physics
1 answer:
iragen [17]3 years ago
6 0

Answer:

0.0239364 N

0.0057879 N

Explanation:

\rho = Density of the gas

g = Acceleration due to gravity = 9.81 m/s²

V = Volume

Mass of rubber = 1.5 g

Buoyant force is given by

F_b=\rho gV\\\Rightarrow F_b=1.22\times 9.81\times 2\times 10^{-3}\\\Rightarrow F_b=0.0239364\ N

The buoyant force is 0.0239364 N

Net vertical force is given by

F_n=F_b-W_{He}-W_{r}\\\Rightarrow F_n=0.0239364-0.175\times 2\times 10^{-3}\times 9.81-1.5\times 10^{-3}\times 9.81\\\Rightarrow F_n=0.0057879\ N

The net vertical force is 0.0057879 N

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What must the charge (sign and magnitude) of a particle of mass 1.49 g be for it to remain stationary when placed in a downward-
Kipish [7]

Answer;

= -2.18 × 10^-5 C

Explanation;

m = 1.49 × 10^-3 kg  

Take downward direction as positive.  

Fg = m g  

E = 670 N/C  

Fe = q E  

Fe + Fg = 0  

q E + m g = 0  

q = -m g/E

   = -1.49 × 10^-3 × 9.81/670

    = -2.18 × 10^-5 C  

=  -2.18 × 10^-5 C

6 0
3 years ago
Isaac drops a rubber ball drom height of 2.0m and it bounces to a height of 1.5m. a) What fraction of it's initial energy is los
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Answer:

a)  ΔE = 25 %

b) v = 8,85 m/s

c) The energy was used against air resistance

Explanation:

In any situation total energy of a body is equal to potential energy plus

kinetic energy, then, just at the moment when Isaac dop the ball the situation is:

Ei = Ep + Ek         where     Ep = m*g*h    and  Ek = 1/2*m*v²

As v = 0  (Isaac drops the ball)

Ei = Ep  =  m*g*h    = 2*m*g

At the end (when the ball bounced to 1,5 m

E₂ = Ep₂  + Ek₂         again at that point  v =0 and

E₂ = 1,5*m*g*

Ei =  E₂ + E(lost)

E(lost) = Ei - E₂

E(lost) = 2*m*g* - 1,5*m*g      and the fraction of energy lost is

E(lost)/Ei

ΔE = (2*m*g* - 1,5*m*g )/ 2*m*g

ΔE = 0,5*m*g / 2*m*g

ΔE = 0,5/2

ΔE = 0,25      or     ΔE = 25 %

b) The speed of the ball is

Potential energy is converted in kinetic energy just when the ball is touching the ground, then

m*g*h = 1/2*m*v²

2*h*g = 1/2 *v²

v² = 4*g*h

v² = 4*2*9,8

v² = 78,4

v = 8,85 m/s

If the impact is an elastic collision, then Ek before and after the impact is the same.

7 0
3 years ago
Round the following off to required number of significant figures.
Stells [14]

Answer:

140

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Explanation:

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