Answer:
s = 20 m
Explanation:
given,
mass of the roller blader = 60 Kg
length = 10 m
inclines at = 30°
coefficient of friction = 0.25
using conservation of energy
u = 9.89 m/s
Using second law of motion
ma =μ mg
a = μ g
a = 0.25 x 9.8
a = 2.45 m/s²
Using third equation of motion ,
v² - u² = 2 a s
0² - 9.89² = 2 x 2.45 x s
s = 20 m
the distance moved before stopping is 20 m
Complete question:
A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?
Answer:
the weight of the air is 76.44 lbs
Explanation:
Given;
dimension of the dormitory, = 14 ft by 13 ft by 6 ft
density of the air, = 0.07 lbs/ft³
The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft
= 1092 ft³
The weight of the air = density x volume
= 0.07 lbs/ft³ x 1092 ft³
= 76.44 lbs
Therefore, the weight of the air is 76.44 lbs
Answer: Things continue doing what they are doing unless a force is applied to it. Objects have a natural tendency to resist change. This is INERTIA. Heavier objects (objects with more mass) are more difficult to move and stop. Heavier objects (greater mass) resist change more than lighter objects, so true
Explanation:
Pushing a bicycle or a Cadillac, or stopping them once moving. The more massive the object (more inertia) the harder it is to start or stop. The Cadillac has more of a tendency to stay stationary (or continue moving), and resist a change in motion than a bicycle.
Answer:
$175
Explanation:
Insurance premium is expressed as a rate $1000
($3.50 per $1000)
Therefore;
Annual premium= $50000x$3.50/$1000
= $175
Answer:
0.012-m
Explanation:
∆L = α × Lo × (T-To)
α is the coefficient of linear expansion = 12 × 10-6 K-1
Lo = Initial length = 25-m
∆L = Change in length
(T-To) = 40 K
∆L = 12 × 10-6 × 25 × 40
∆L = 0.012-m
