In thermodynamics, work of a system at constant pressure conditions is equal to the product of the pressure and the change in volume. It is expressed as follows:
W = P(V2 - V1)
W = 1.3x10^5 (2x6 - 6 )
<span>W = 780000 J
</span>
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Answer:
the Hudson Bay was covered with alpine glaciers
Explanation:
During the last glacial period, large portions of North America were covered with ice. The majority of the ice was from the ice sheets that were covering Canada and the northern part of the United States, and the alpine glaciers on the mountain ranges. Hudson Bay was all frozen at this point of time. It was not covered with alpine glaciers though, instead it was covered with the ice of the extended ice sheets, with the ice cover reaching up to 2 km in thickness.
Explanation:
<h3>1.) Regeneration is the natural process of replacing or restoring damaged or missing cells, tissues, organs, and even entire body parts to full function in plants and animals.</h3>
2.) When noise is added to analogue signals, it usually sounds like background hiss. Such noise can not be removed so the original clean signal can not be re-created or re-generated.
Answer:
i. The radius 'r' of the electron's path is 4.23 ×
m.
ii. The frequency 'f' of the motion is 455.44 KHz.
Explanation:
The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.
r = 
Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.
From the question, B = 1.63 ×
T, v = 121 m/s, Θ =
(since it enters perpendicularly to the field), q = e = 1.6 ×
C and m = 9.11 ×
Kg.
Thus,
r =
÷ sinΘ
But, sinΘ = sin
= 1.
So that;
r = 
= (9.11 ×
× 121) ÷ (1.6 ×
× 1.63 ×
)
= 1.10231 ×
÷ 2.608 × 
= 4.2266 ×
= 4.23 ×
m
The radius 'r' of the electron's path is 4.23 ×
m.
B. The frequency 'f' of the motion is called cyclotron frequency;
f = 
= (1.6 ×
× 1.63 ×
) ÷ (2 ×
× 9.11 ×
)
= 2.608 ×
÷ 5.7263 × 
= 455442.4323
f = 455.44 KHz
The frequency 'f' of the motion is 455.44 KHz.
Answer:
C
Explanation:
- Let acceleration due to gravity @ massive planet be a = 30 m/s^2
- Let acceleration due to gravity @ earth be g = 30 m/s^2
Solution:
- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:
t = v / a
t = v / 30
- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:
t = v / g
t = v / 9.81
- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C