C. Acceleration is the rate of change of velocity. So at the top of the path, while the velocity is zero, the CONSTANT GRAVITATIONAL ACCELERATION is about 10 m/s^2 (9.8)
Answer:
Explanation:
From the given information:
radius = 15 m
Time T = 23 s
a) Speed (v) = 

v = 4.10 m/s
b) The magnitude of the acceleration is:

a = 1.12 m/s²
c) True weight = mg
Apparent weight = normal force
From the top;
the normal force = upward direction,
weight is downward as well as the acceleration.
true weight - normal force = ma
apparent weight =mg - ma


= 0.886 m/s²
d)
From the bottom;
acceleration is upward, so:
apparent weight - true weight = ma
apparent weight = true weight + ma



= 1.114 m/s²
Answer:
Yes you can. The charging current will be the lowest of the two current ratings of device and charger. The charger and the phone have complex internal circuitry that enable this behaviour.
But using a weaker charger for your phone will only lengthen charging time. And using a stronger charger than that the phone will accept doesn't affect charging at all and only wastes money.
Answer:
option (b)
Explanation:
for a telescope, the object is placed very far from the objective lens so the focal length of objective is large and the eye piece is of short focal length.
So, long focal length objective and short focal length eyepiece.
Option (b) is correct.
Answer:
The answer is t = 23.81 s
Explanation:
A car in this question means those partition that make up the train
Let denote the the of each car as Z
The equation of motion that has to do with distance is given as
S = ut + (1/2)at²
Where u = initial velocity = 0
t = time = 9 s
Substituting these values we have
Z = (1/2) a(9)²
= 40.5 a
making acceleration the subject we have that a = Z / 40.5
Let us denote the length of the 7th car as =7Z
Considering the distance at the 7th car we have
S₇ = ut₇ + (1/2)at²₇
Where u = initial velocity =0
Substituting values we have
7Z = (1/2)(Z/40.5)(t₇)²
7 = {1/2 ×40.5}(t₇)²
t₇² = 7 × 2 × 40.5
t₇² = 567
t₇ = 23.81 s