A +2e charge is at the point (-1,0) mm in the x,y plane. A –e charge is at the point (0,1) mm. What is the electric field at the
1 answer:
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Answer:
For the complete question provided in explanation, if the elevator moves upward, then the apparent weight will be 1035 N. While for downward motion the apparent weight will be 435 N.
Explanation:
The question is incomplete. The complete question contains a velocity graph provided in the attachment. This is the velocity graph for an elevator having a passenger of 75 kg.
From the slope of graph it is clear that acceleration at t = 1 sec is given as:
Acceleration = a = (4-0)m/s / (1-0)s = 4 m/s^2
Now, there are two cases:
1- Elevator moving up
2- Elevator moving down
For upward motion:
Apparent Weight = m(g + a)
Apparent Weight = (75 kg)(9.8 + 4)m/s^2
<u>Apparent Weight = 1035 N</u>
For downward motion:
Apparent Weight = m(g - a)
Apparent Weight = (75 kg)(9.8 - 4)m/s^2
<u>Apparent Weight = 435 N</u>
Answer:
2) a_y= -g 3) vₓ=constant v_y = v_{oy} - g t, 4) vₓ = v₀ₓ - ax t
5) changes the horizontal speed, should change range
7) changes the vertical speed change the maximum height
Explanation:
1) After reading your long writing, we are going to solve the exercise, in the attachment you can see the different vectors.
2) The acceleration vectors are vertical and directed downwards due to the attraction of the Earth (gravity force) this force is constant, on the x axis there is no acceleration
3) the velocity vectors on the x-axis are constant because there are no relationships and the y-axis changes value according to the expression
v_y = v_{oy} - gt
at the point of maximum height, vy = 0 is equal to the maximum height
4) For someone to change the horizontal acceleration we must assume a friction with the air, in this case they relate it would be in the opposite direction to the horizontal speed
In the graph it would be directed to the left, therefore the velocity would be
vₓ = v₀ₓ - ax t
5 and 6) If someone changes the horizontal speed, they should change the range of the shot for greater horizontal speed, the rock goes further.
the equations of motion are
x = v₀ₓ t
y = v_{oy} t - ½ g t²
7) If someone changes the vertical speed change the maximum height, but not the scope of the shot, for higher speed higher maximum height,
the equations of motion are the same.
