"60 kg" is not a weight. It's a mass, and it's always the same
no matter where the object goes.
The weight of the object is
(mass) x (gravity in the place where the object is) .
On the surface of the Earth,
Weight = (60 kg) x (9.8 m/s²)
= 588 Newtons.
Now, the force of gravity varies as the inverse of the square of the distance from the center of the Earth.
On the surface, the distance from the center of the Earth is 1R.
So if you move out to 5R from the center, the gravity out there is
(1R/5R)² = (1/5)² = 1/25 = 0.04 of its value on the surface.
The object's weight would also be 0.04 of its weight on the surface.
(0.04) x (588 Newtons) = 23.52 Newtons.
Again, the object's mass is still 60 kg out there.
___________________________________________
If you have a textbook, or handout material, or a lesson DVD,
or a teacher, or an on-line unit, that says the object "weighs"
60 kilograms, then you should be raising a holy stink.
You are being planted with sloppy, inaccurate, misleading
information, and it's going to be YOUR problem to UN-learn it later.
They owe you better material.
Answer:
μ = 0.6
Explanation:
given,
speed of car = 29.7 m/s
Radius of curve = 50 m
θ = 30.0°
minimum static friction = ?
now,
writing all the forces acting along y-direction
N cos θ - f sinθ = mg
N cos θ -μN sinθ = mg

now, writing the forces acting along x- direction
N sin θ + f cos θ = F_{net}
N cos θ + μN sinθ = F_{net}

taking cos θ from nominator and denominator




now, inserting all the given values

μ = 0.6
Answer:
A) 
B) 
C) 
D) mosquitoes speed in part B is very much larger than that of part C.
Explanation:
Given:
- Distance form the sound source,

- sound intensity level at the given location,

- diameter of the eardrum membrane in humans,

- We have the minimum detectable intensity to the human ears,

(A)
<u>Now the intensity of the sound at the given location is related mathematically as:</u>
..........................................(1)



<em>As we know :</em>


is the energy transferred to the eardrums per second.
(B)
mass of mosquito, 
<u>Now the velocity of mosquito for the same kinetic energy:</u>



(C)
Given:
- Sound intensity,

<u>Using eq. (1)</u>



Now, power:



Hence:




(D)
mosquitoes speed in part B is very much larger than that of part C.
<span>-Mass (kg)
-Velocity (m/s)</span>