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sukhopar [10]
3 years ago
12

A small mailbag is released from a helicopter that is descending steadily at 2.52 m/s. (a) After 4.00 s, what is the speed of th

e mailbag
Physics
1 answer:
gogolik [260]3 years ago
6 0

Answer: 41.72m/s

Explanation: v= u + gt

V = 2.52 + 9.8(4.00)

V = 41.72m/s

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A is a hot object.  B is a cold object.  the two are brought into contact with each other so that heat is developed
Umnica [9.8K]

Answer:

A) Cold object will start getting hot

B) Heat exchange will stop as the two object acquire the same temperature.

Explanation:

A) When one hot object and one cold object are kept in contact then the heat is transferred from the hot object to the cold object via different modes of heat transmission. Hence, the cold object starts getting hot

B) The transmission of heat from the hot object to the cold object will stop as the temperature of the two object becomes equal to each other.

3 0
3 years ago
A skier is pulled by a towrope up a frictionless ski slope that makes an angle of 12 degrees with the horizontal. The rope moves
MArishka [77]

Answer:

Explanation:

Given,

  • Work done by the rope 900 m/s.
  • Angle of inclination of the slope = \theta\ =\ 12^o
  • Initial speed of the skier = v = 1.0 m/s
  • Length of the inclined surface = d = 8.0 m

part (a)

The rope is doing the work against the gravity on the skier to uplift up to the inclined surface. Therefore the work done by the rope is equal to the work done on the skier due to the gravity

\therefore W_r\ =\ W_g\ =\ 900\ J

In both cases the height attained by the skier is equal. and the work done by gravity does not depend upon the speed of the skier.

part (b)

  • Initial speed of the skier = v = 1.0 m/s.

Rate of the work done by the rope is power of the rope.

Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 1.0}{8.0}\\\Rightarrow P\ =\ 112.5\ Watt

Part (c)

  • Initial speed of the skier = v = 2.0 m/s.

Rate of the work done by the rope is power of the rope.

Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 2.0}{8.0}\\\Rightarrow P\ =\ 225\ Watt

4 0
3 years ago
Tech A says parallel circuits are like links in a chain. Tech B says total current in a parallel circuit equals the sum of the c
rosijanka [135]

Answer: Only Tech B is correct.

Explanation:

First, tech A is wrong.

The circuits that can be compared with links in a chain are the series circuit, and it can be related to the links in a chain because if one of the elements breaks, the current can not flow furthermore (because the elements in the circuit are connected in series) while in a parallel circuit if one of the branches breaks, the current still can flow by other branches.

Also in a parallel circuit, the sum of the currents of each path is equal to the current that comes from the source, so Tech B is correct, the total current is equal to the sum of the currents flowing in each branch of the circuit.

4 0
4 years ago
Which of the following tend to react by gaining 1 electron
Andrei [34K]
The correct answer should be C
8 0
3 years ago
Read 2 more answers
A projectile is launched vertically at 100 m/s. If air resistance can be ignored, at what speed will it return to its initial le
eduard
<h2>Speed with which it return to its initial level is 100 m/s</h2>

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 100 m/s  

Acceleration, a = -9.81 m/s²  

Final velocity, v = ?

Displacement, s = 0 m  

Substituting  

v² = u² + 2as

v² = 100² + 2 x -9.81 x 0

v² = 100²

v = ±100 m/s

+100 m/s is initial velocity and -100 m/s is final velocity.

Speed with which it return to its initial level is 100 m/s

7 0
3 years ago
Read 2 more answers
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