Answer:
300 J
Explanation:
The gravitational potential energy of the performer is given by:
where
(mg) is the weight of the performer, equal to the product between mass (m) and gravitational acceleration (g)
h is the height of the performer above the ground
In this problem, we have
(mg) = 60.0 N
h = 5.0 m
Substituting into the formula, we find
Answer:
The final angular speed of the carousel after the person climbs aboard is 1.70 rad/s
Explanation:
Conservation of angular moment states that if there are not external torques on a system its angular moment will be conserved i.e. final angular momentum is equal to initial angular momentum, mathematically:
(1)
With Li the initial angular momentum and Lf the final angular momentum, we are interested only on the magnitude of the angular momenta. Initially, the total angular momentum of the system carousel-person is only the angular momentum of the playground carousel because the person isn’t moving, so it is the product of its moment of inertia (I) and angular velocity ( :
(2)
But the final angular momentum of the system is the angular momentum of the carousel plus the angular momentum of the person moving on a circle, again the angular momentum of the carousel is the product of its moment of inertia and angular velocity, but the angular momentum of the person is the product of the radius (R) of the carousel, the person’s mass (m) and its tangential velocity :
(3)
The angular velocity of the person is the same of the carousel because the person is on the carousel, and its related with the tangential velocity by:
(4)
Replacing (4) on (3) and adding this to the angular moment of the carousel we obtain the final angular momentum of the system:
(5)
Equating this with the initial angular momentum an solving for we obtain:
Answer:
4 days
either multiply 128 by .5 until you get to 2 counting each time or use 2 formulas ln(n2/n1)=-k(t2-t1) to get k then input k into ln(2)=k*t1/2
n2 is final amount and n1 is beginning and t is either time elapsed as in the first formula or the actual half life that is t1/2
Explanation:
Answer: a) 7.71 m/s², b) - 6.67 m/s²
Explanation: first thing to note is that
1 mile = 1609.34
1 hour = 3600s
Hence, 100mph to m/s = (100 × 1609.34)/3600 = 44.71 m/s
Initial velocity (u) = 0, final velocity (v) = 44.71 m/s, t = 5.8s, a = acceleration =?
By using newton's laws of motion
v = u + at
44.71 = 0 + a(5.8)
44.71 = 5.8a
a = 44.71/5.8
a = 7.71 m/s²
Question b)
The car is completing a stop which implies that the car is coming to rest, and when a car is coming to rest, the final velocity (v) is zero.
Hence u = 34 m/s, v = 0, a =?, t = 5.1 s
v = u + at
0 = 34 + a(5.1)
a(5.1) = - 34
a = - 34/5.1
a = - 6.67 m/s².
The negative sign beside the acceleration shows that the body is decelerating