I believe this is it
The centripetal force is given by
F = mv^2 / r
When v' = v/2,
F' = mv'^2/r = m(v/2)^2/r = mv^2/4r = F/4.
So the centripetal force is divided by 4.
Answer:
the middle
Explanation:
the left one bulb gets power from the outher bulb
the one on right has more bulbs
Answer:
1.9 MPa
Explanation:
Mass of person = 81 kg
Mass of chair = 3.8 kg
Diameter of contact point = 1.2 cm = D
Radius of contact point = 1.2/2 = 0.6 cm
Total mass of chair and person = 81 + 3.8 = 84.8 kg = M
Acceleration due to gravity = 9.81 m/s²
Force acting on the floor
<em>F = Mg</em>
<em>⇒F = 84.8×9.81</em>
<em>⇒F = 831.888 N</em>
Area of the contact point
<em>A = πR²</em>
<em>⇒A = π0.006²</em>
<em>⇒A = π0.000036 m²</em>
Area of the four points is
<em>4A = 0.000144π m²</em>
Pressure

Pressure exerted on the floor by each leg of the chair is 1.9 MPa
Answer
Pressure, P = 1 atm
air density, ρ = 1.3 kg/m³
a) height of the atmosphere when the density is constant
Pressure at sea level = 1 atm = 101300 Pa
we know
P = ρ g h


h = 7951.33 m
height of the atmosphere will be equal to 7951.33 m
b) when air density decreased linearly to zero.
at x = 0 air density = 0
at x= h ρ_l = ρ_sl
assuming density is zero at x - distance

now, Pressure at depth x


integrating both side


now,


h = 15902.67 m
height of the atmosphere is equal to 15902.67 m.
Explanation:
Given that,
The slope of the ramp, 
Mass of the box, m = 60 kg
(a) Distance covered by the truck up the slope, d = 300 m
Initially the truck moves with a constant velocity. We know that the net work done on the box is equal to 0 as per work energy theorem as :

u and v are the initial and the final velocity of the truck
(b) The work done on the box by the force of gravity is given by :

Here, 


W = -24550.13 J
(c) What is the work done on the box by the normal force is equal to 0 as the angle between the force and the displacement is 90 degrees.
(d) The work done by friction is given by :


Hence, this is the required solution.