Answer:
C) Mass
Explanation:
The mass of an object divided by its volume equals the density.
Hope this helps!
Answer:
4.24m/s
Explanation:
Potential energy at the top= kinetic energy at the button
But kinetic energy= sum of linear and rotational kinetic energy of the hoop
PE= mgh
KE= 1/2 mv^2
RE= 1/2 I ω^2
Where
m= mass of the hoop
v= linear velocity
g= acceleration due to gravity
h= height
I= moment of inertia
ω= angular velocity of the hoop.
But
I = m r^2 for hoop and ω = v/r
giving
m g h = 1/2 m v^2 + 1/2 (m r^2) (v^2/r^2) = 1/2 m v^2 + 1/2 m v^2 = m v^2
and m's cancel
g h = v^2
Hence
v= √gh
v= √10×1.8
v= 4.24m/s
A - the objects are too small
GRAVITATIONAL FORCE IS EXPERIENCED BY ALL OBJECTS IN THE UNIVERSE ALL THE TIME. BUT THE ORDINARY OBJECTS YOU SEE EVERY DAY HAVE MASSES SO SMALL THAT THEIR ATTRACTION TOWARD EACH OTHER IS HARD TO DETECT. -https://www.ftsd.org/cms/lib6/MT01001165/Centricity/ModuleInstance/630/CHAPTER_2_NOTES_FOR_EIGHTH_GRADE_PHYSICAL_SCIENCE.pdf
Answer:
The necessary information is if the forces acting on the block are in equilibrium
The coefficient of friction is 0.577
Explanation:
Where the forces acting on the object are in equilibrium, we have;
At constant velocity, the net force acting on the particle = 0
However, the frictional force is then given as
F = mg sinθ
Where:
m = Mass of the block
g = Acceleration due to gravity and
θ = Angle of inclination of the slope
F = 5×9.81×sin 30 = 24.525 N
Therefore, the coefficient of friction is given as
24.525 N = μ×m×g × cos θ = μ × 5 × 9.81 × cos 30 = μ × 42.479
μ × 42.479 N= 24.525 N
∴ μ = 24.525 N ÷ 42.479 N = 0.577
Those two units can be compared to a 'mile per hour' and a 'mile per hour - hour'.
One is a rate. The other is a quantity, after maintaining a rate for some time.
-- 'Joule' is a unit of energy. It's the amount of work (energy) you do
when you push with a force of 1 newton though a distance of 1 meter.
Lifting 10 pound of beans 3 feet off the floor takes about 40.7 joules of energy.
-- 'Watt' is a <u><em>rate</em></u> of using energy . . . 1 joule per second.
If you lift 10 pounds 3 feet off the floor in 1 second, your <em>power</em> is 40.7 watts.
-- 'Watt-second' is the amount of energy used in one second,
at the rate of 1 joule per second . . . 1 joule.
-- 'Watt-hour' is the amount of energy used in one hour,
at the rate of 1 joule per second . . . 3,600 joules.
-- 'Kilowatt' is a bigger <em>rate</em> of using energy . . . 1,000 joules per second.
-- 'Kilowatt - second' is the amount of energy used in one second,
at the rate of 1,000 joules per second . . . 1,000 joules .
-- 'Kilowatt - hour' is the amount of energy used in one hour,
at the rate of 1,000 joules per second . . . 3,600,000 joules .
Depending on where you live, 3,600,000 joules of energy bought
from the electric company costs something between 5¢ and 25¢.
