Its total mechanical energy is <em>2,000 J</em>.
We don't have enough information to say anything about its heat energy, its chemical energy, or the energy due to any electrical charge it may be carrying or any magnetic field it may have.
What you need to know is that the force is
F=ma
The force is the product of mass and acceleration
this means that the acceleration is
a=F/m
a) The force is halved?
this means that f will be
now:
a=
So the accelaration will also he halved (it's the original acceleratation divided by 2)
b) The object's mass is halved?
a=
=a=
which is the original acceleration times two!! so it will double
c) The force and the object's mass are both halved?
now we have
a=
=a=
=a=
so they will cancel each other out and the acceleration will stay the same!
Answer:
The time is 1.8s
Explanation:
The ball droped, will freely fall under gravity.
Hence we use free fall formula to calculate the time by the ball to hit the ground
Where h is the height from which the ball is droped, g is the acceleration due to gravity that acted on the ball, and t is the time taken by the ball to hit the ground.
From the question,
h=16m
Also, let take
By substitution we obtain,
Diving through by 9.8
square root both sides, we obtain
Answer:
The final speed of the crate is 12.07 m/s.
Explanation:
For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:
Now, we can calculate the final speed of the crate at the end of 10.0 m:
For the next 10.5 meters we have frictional force:
So, the acceleration is:
The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:
Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.
I hope it helps you!
Answer:
a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions
Explanation:
a. Its angular speed in radians per second ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s
b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m
So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s
c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min
α = (ω₁ - ω)/t
= (1410 - 207)/(80.5/60)
= 60(1410 - 207)/80.5
= 60(1203)80.5
= 896.65 rev/min² ≅ 897 rev/min²
d. Using θ = ωt + 1/2αt²
where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and t = 80.5/60 min = 1.342 min
θ = ωt + 1/2αt²
= 207 × 1.342 + 1/2 × 896.65 × 1.342²
= 277.725 + 807.417
= 1085.14 revolutions ≅ 1085 revolutions