Answer:
Explanation:
As the circuit is parallel, then there is no effect of other branches as the potential difference across each arm is same.
Answer:
yes
Explanation:
its not a good thing for the rest of your life but you have a PS4
In contrast, extratropical cyclones have their strongest winds near the tropopause, which is about 8 miles above the surface. These differences are due to the tropical cyclone being “warm-core” in the troposphere, whereas extra-tropical cyclones are “warm-core” in the stratosphere and “cold-core” in the troposphere.
Answer:
a) 145.6kgm^2
b) 158.4kg-m^2/s
c) 0.76rads/s
Explanation:
Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation
(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and
(c) the angular speed of the merry-go-round and child after the child has jumped on.
a) From I = MK^2
I = (160Kg)(0.91m)^2
I = 145.6kgm^2
b) The magnitude of the angular momentum is given by:
L= r × p The raduis and momentum are perpendicular.
L = r × mc
L = (1.20m)(44.0kg)(3.0m/s)
L = 158.4kg-m^2/s
c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:
L = Iw
158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]
w = 158.6/208.96
w = 0.76rad/s
Answer:
304.89m
Explanation:
Given
acceleration a = 2.52m/s²
final speed v = 39.2m/s
initial speed = 0m/s (car accelerates from rest)
Using the equation of motion below to get the distance of Doc brown from Marty;
v² = u²+2as
substitute the given parameters
39.2² = 0²+2(2.52)s
1536.64 = 0+5.04s
divide both sides by 5.04
1536.64/5.04 = 5.04s/5.04
rearrange the equation
5.04s/5.04 = 1536.64/5.04
s = 304.89m
Hence He and Marty must stand at 304.89m to allow the car to accelerate from rest to a speed of 39.2 m/s?
