An elevator has 750,000 J when it is lived 500 meters in the air
Answer: Option D
<u>Explanation:</u>
The expression to find the potential energy completely on the basis of forces that on two objects. It can be written as

Where,
m – mass of object
Acceleration due to gravity = 
h = 500 meters
Force = 1500 Newton
As we know, according to Newton second law of motion, the force can be expressed in terms of mass and gravity (F= ma). So, the potential energy can be calculated as

Hence, an elevator has 750,000 J when it is lived 500 meters in the air.
Answer:
4.763 × 10⁶ N/C
Explanation:
Let E₁ be the electric field due to the 4.0 μC charge and E₂ be the electric field due to the -6.0 μC charge. At the third corner, E₁ points in the negative x direction and E₂ acts at an angle of 60 to the negative x - direction.
Resolving E₂ into horizontal and vertical components, we have
E₂cos60 as horizontal component and E₂sin60 as vertical component. E₁ has only horizontal component.
Summing the horizontal components we have
E₃ = -E₁ + (-E₂cos60) = -kq₁/r²- kq₂cos60/r²
= -k/r²(q₁ + q₂cos60)
= -k/r²(4 μC + (-6.0 μC)(1/2))
= -k/r²(4 μC - 3.0 μC)
= -k/r²(1 μC)
= -9 × 10⁹ Nm²/C²(1.0 × 10⁻⁶)/(0.10 m)²
= -9 × 10⁵ N/C
Summing the vertical components, we have
E₄ = 0 + (-E₂sin60)
= -E₂sin60
= -kq₂sin60/r²
= -k(-6.0 μC)(0.8660)/(0.10 m)²
= -9 × 10⁹ Nm²/C²(-6.0 × 10⁻⁶)(0.8660)/(0.10 m)²
= 46.77 × 10⁵ N/C
The magnitude of the resultant electric field, E is thus
E = √(E₃² + E₄²) = √[(-9 × 10⁵ N/C)² + (46.77 10⁵ N/C)²) = (√226843.29) × 10⁴
= 476.28 × 10⁴ N/C
= 4.7628 × 10⁶ N/C
≅ 4.763 × 10⁶ N/C
Answer:
<h3>The answer is 2.51 s</h3>
Explanation:
The time taken can be found by using the formula

d is the distance
v is the velocity
From the question we have

We have the final answer as
<h3>2.51 s</h3>
Hope this helps you
Answer:
time will elapse before it return to its staring point is 23.6 ns
Explanation:
given data
speed u = 2.45 ×
m/s
uniform electric field E = 1.18 ×
N/C
to find out
How much time will elapse before it returns to its starting point
solution
we find acceleration first by electrostatic force that is
F = Eq
here
F = ma by newton law
so
ma = Eq
here m is mass , a is acceleration and E is uniform electric field and q is charge of electron
so
put here all value
9.11 ×
kg ×a = 1.18 ×
× 1.602 ×
a = 20.75 ×
m/s²
so acceleration is 20.75 ×
m/s²
and
time required by electron before come rest is
use equation of motion
v = u + at
here v is zero and u is speed given and t is time so put all value
2.45 ×
= 0 + 20.75 ×
(t)
t = 11.80 ×
s
so time will elapse before it return to its staring point is
time = 2t
time = 2 ×11.80 ×
time is 23.6 ×
s
time will elapse before it return to its staring point is 23.6 ns