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Bess [88]
4 years ago
7

A hand pushes two blocks, A and B, along a frictionless table for a distance d. When the hand starts to push, the blocks are mov

ing with a speed of 2 m/s. Suppose that the work done on block A by the hand during a given displacement is 10 J. Determine the final energy of each block. Mass of A is 4kg, mass of B is 1kg.
Physics
1 answer:
Kitty [74]4 years ago
5 0
  <span>net work = change in kinetic energy 

for Block B, we just have the force from block A acting on it 
F(ab)d= .5(1)vf² - .5(1)(2²) 
F(ab)d= .5vf² - 2 

Block A, we have the force from the hand going in one direction and the force of block B on A going the opposite direction 

10-F(ba)d = .5(4)vf² - .5(4)(2²) 
10-F(ba)d = 2vf² - 8 
F(ba)d = 18 - 2vf² 

now we have two equations: 
F(ba)d = 18 - 2vf² 
F(ab)d= .5vf² - 2 

since the magnitude of F(ba) and F(ab) is the same, substitute and find vf (I already took into account the direction when solving for F(ab) 

10-.5vf² + 2 = 2vf² - 8 
12 - .5vf² = 2vf² - 8 
20 = 2.5vf² 
vf² = 8 

they both will have the same velocity 
KE of block A= .5(4)(2.828²) = 16 J 
KE of block B=.5(1)(2.828²) = 4 J</span>
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A container of gas is at a pressure of 1.3x10^5 Pa and a volume of 6 m^3. How much work is done by the gas if it expands at a co
Ivenika [448]
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
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Hope this answers the question. Have a nice day.
5 0
3 years ago
North America experienced all of the following during the last glacial period EXCEPT alpine glaciers covered the Rocky and Casca
Juli2301 [7.4K]

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.

5 0
3 years ago
Explain<br> (0) what is meant by regeneration,<br> (ii) why an analogue signal cannot be regenerated
Kamila [148]

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.

6 0
3 years ago
g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p
kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × 10^{-5} 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 = \frac{mv}{qB}

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 × 10^{-5}T, v = 121 m/s, Θ = 90^{0} (since it enters perpendicularly to the field), q = e  = 1.6 × 10^{-19}C and m = 9.11 × 10^{-31}Kg.

Thus,

         r = \frac{mv}{qB} ÷ sinΘ

But,  sinΘ =  sin 90^{0} = 1.

So that;

          r = \frac{mv}{qB}

            = (9.11 × 10^{-31} × 121) ÷ (1.6 × 10^{-19}  × 1.63 × 10^{-5})

            = 1.10231 × 10^{-28}   ÷ 2.608 × 10^{-24}

            = 4.2266 × 10^{-5}

            = 4.23 × 10^{-5} m

The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

B. The frequency 'f' of the motion is called cyclotron frequency;

           f = \frac{qB}{2\pi m}

             =  (1.6 × 10^{-19}  × 1.63 × 10^{-5}) ÷ (2 ×\frac{22}{7} × 9.11 × 10^{-31})

             =  2.608 × 10^{-24} ÷  5.7263 × 10^{-30}

             = 455442.4323

          f  = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.

3 0
4 years ago
Read 2 more answers
Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g
Ne4ueva [31]

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

7 0
3 years ago
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