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Veronika [31]
3 years ago
13

When an object is falling and reaches a constant velocity, the net force on the object is ____ and the weight of the object is e

qual to____
A. zero; the mass of the object
B. 10; the speed at which the object is falling
C. 10; acceleration due to gravity
D. Zero; the force of air resistance against the object
Physics
1 answer:
Maslowich3 years ago
5 0

When an object is falling and reaches a constant velocity, the net force on the object is <em>zero</em> (it's not accelerating), and the weight of the object is equal to <em>the force of air resistance against the object</em>.  (choice-D)

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Calculate the power of the elevator if it can raise 290 kg of mass 30 flights of stairs (300 m) in 17 seconds
Advocard [28]

Answer:

Power =  50204 [watts]

Explanation:

We know that the power is defined by the following expression:

Power = Work/time

where:

Power [watts]

time [seconds]

The work done will be the following:

Work = Force * distance [Joules]

Force[Newtons]

distance[meters]

Force = mass* gravity

Force=290 [kg]*9.81[m/s^2] =2844.9[N]

Work = 2844.9[N]*300[m] = 853470[J]

Therefore

Power = 853470 / 17 = 50204 [watts]

5 0
4 years ago
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Nesterboy [21]

Answer:

what is Jose potential energy

7 0
3 years ago
The potential (relative to infinity) at the midpoint of a square is 3.0 V when a point charge of +Q is located at one of the cor
Sveta_85 [38]

Answer:

d) 12 V

Explanation:

Due to the symmetry of the problem, the potential (relative to infinity) at the midpoint of the square, is the same for all charges, provided they be of the same magnitude and sign, and be located at one of the corners of the square.

We can apply the superposition principle (as the potential is linear with the charge) and calculating the total potential due to the 4 charges, just adding the potential due to any of  them:

V = V(Q₁) + V(Q₂) +V(Q₃) + V(Q₄) = 4* 3.0 V = 12. 0 V

4 0
3 years ago
A 50 gram meterstick is placed on a fulcrum at its 50 cm mark. A 20 gram mass is attached at the 12 cm mark. Where should a 40 g
alekssr [168]

Answer:

The 40g mass will be attached at 69 cm

Explanation:

First, make a sketch of the meterstick with the masses placed on it;

--------------------------------------------------------------------------

               ↓                    Δ                      ↓

             20 g.................50 cm.................40g

                         38 cm                  y cm  

Apply principle of moment;

sum of clockwise moment = sum of anticlockwise moment

40y = 20 (38)

40y = 760

y = 760 / 40

y = 19 cm

Therefore, the 40g mass will be attached at 50cm + 19cm = 69 cm

              12cm             50 cm              69cm

--------------------------------------------------------------------------

               ↓                    Δ                      ↓

             20 g.................50 cm.................40g

                         38 cm                 19 cm                                              

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                   

5 0
3 years ago
Help I don’t understand this question
7nadin3 [17]
Metals
3/4 of the periodic table
Good conductors of heat and electricity
Malleable

Nonmetals
1/4 of the periodic table
Bad conductors of heat and electricity
Not bendable
5 0
3 years ago
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