the end of the tracks, 8.8 m lower vertically, is a horizontally situated spring with constant 5 × 105 N/m. The acceleration of
1 answer:
Answer
Assuming the mass of the car, m = 43000 kg
initial speed u = 0
vertical distance moved, h = 8.8 m
spring constant k = 5 x 10⁵ N / m
acceleration of gravity = 9.8 m/s²
From law of conservation of energy ,
Gravitational potential energy at starting position =potential energy of the spring at maximum compression
x = 14.83 m
If the mass of the car is equal to 43000 Kg the spring is compressed to 14.83 m
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Answer:
distance r from the uranium atom is 18.27 nm
Explanation:
given data
uranium and iron atom distance R = 44.10 nm
uranium atom = singly ionized
iron atom = doubly ionized
to find out
distance r from the uranium atom
solution
we consider here that uranium electron at distance = r
and electron between uranium and iron so here
so we can say electron and iron distance = ( 44.10 - r ) nm
and we know single ionized uranium charge q2= 1.602 × C
and charge on iron will be q3 = 2 × 1.602 × C
so charge on electron is q1 = - 1.602 × C
and we know F =
so now by equilibrium
Fu = Fi
=
put here k = and find r
=
r = 18.27 nm
distance r from the uranium atom is 18.27 nm
Answer:
Angular acceleration,
Explanation:
It is given that,
Mass of the solid sphere, m = 245 g = 0.245 kg
Diameter of the sphere, d = 4.3 cm = 0.043 m
Radius, r = 0.0215 m
Force acting at a point, F = 0.02 N
Let is its angular acceleration. The relation between the angular acceleration and the torque is given by :
I is the moment of inertia of the solid sphere
For a solid sphere,
So, its angular acceleration is . Hence, this is the required solution.