1 g = 1 ÷ 1000 kg
= 0.001 kg
1 cm³ = 1 ÷ 100 ÷ 100 ÷ 100 m³
= 0.000001 m³
1 g/cm³ = 1 g / 1 cm³
= 0.001 kg / 0.000001 m³
= 1000 kg/m³
The density is 1000 kg/m³.
See the attached picture:
Answer:
Explanation:
a )
change in the gravitational potential energy of the bear-Earth system during the slide = mgh
= 45 x 9.8 x 11
= 4851 J
b )
kinetic energy of bear just before hitting the ground
= 1/2 m v²
= .5 x 45 x 5.8²
= 756.9 J
c ) If the average frictional force that acts on the sliding bear be F
negative work done by friction
= F x 11 J
then ,
4851 J - F x 11 = 756.9 J
F x 11 = 4851 J - 756.9 J
= 4094.1 J
F = 4094.1 / 11
= 372.2 N
Answer:
The magnitude of the force required to bring the mass to rest is 15 N.
Explanation:
Given;
mass, m = 3 .00 kg
initial speed of the mass, u = 25 m/s
distance traveled by the mass, d = 62.5 m
The acceleration of the mass is given as;
v² = u² + 2ad
at the maximum distance of 62.5 m, the final velocity of the mass = 0
0 = u² + 2ad
-2ad = u²
-a = u²/2d
-a = (25)² / (2 x 62.5)
-a = 5
a = -5 m/s²
the magnitude of the acceleration = 5 m/s²
Apply Newton's second law of motion;
F = ma
F = 3 x 5
F = 15 N
Therefore, the magnitude of the force required to bring the mass to rest is 15 N.
Explanation:
Given that,
The initial velocity of a skater is, u = 5 m/s
She slows to a velocity of 2 m/s over a distance of 20 m.
We can find the acceleration of skater. It is equal to the rate of change of velocity. So, it can be calculated using third equation of motion as follows :
a = acceleration
So, her acceleration is 
and she is deaccelerating. Also, her initial velocity is given i.e. 5 m/s.
