The force needed to accelerate an elevator upward at a rate of
is 2000 N or 2 kN.
<u>Explanation:
</u>
As per Newton's second law of motion, an object's acceleration is directly proportional to the external unbalanced force acting on it and inversely proportional to the mass of the object.
As the object given here is an elevator with mass 1000 kg and the acceleration is given as
, the force needed to accelerate it can be obtained by taking the product of mass and acceleration.


So 2000 N or 2 kN amount of force is needed to accelerate the elevator upward at a rate of
.
The masses of the objects and how much distance there is between them
Answer:
C
Explanation:
Im not sure but I did somthing simalier
Answer:
I = 18 x 10⁻⁹ A = 18 nA
Explanation:
The current is defined as the flow of charge per unit time. Therefore,
I = q/t
where,
I = Average Current passing through nerve cell
q = Total flow of charges through nerve cell
t = time period of flow of charges
Here, in our case:
I = ?
q = (9 pC)(1 x 10⁻¹² C/1 pC) = 9 x 10⁻¹² C
t = (0.5 ms)(1 x 10⁻³ s/1 ms) = 5 x 10⁻⁴ s
Therefore,
I = (9 x 10⁻¹² C)/(5 x 10⁻⁴ s)
<u>I = 18 x 10⁻⁹ A = 18 nA</u>
Answer:43.34 m
Explanation:
Given
acceleration(a)
Initial Velocity(u)=0 m/s
After 6 s fuel runs out
Velocity after 6 s
v=u+at

After this object will start moving under gravity
height reached in first 6 s


s=36 m
After fuel run out distance traveled in upward direction is

here v=0
u=12 m/s



