Answer:
r2 = 2.401557 cm
distance = 0.10 cm
Explanation:
given data
radius = 2.50 cm
density = 15.0 nC/m
voltmeter read = 175
solution
we know here potential difference that is express as
ΔV = 
...........1
so here
as here 
is linear charge density
r2 = r1 × 
r2 = 2.40 × 
r2 = 2.401557 cm
and
here distance above surface will be
distance = r1 - r2
distance = 2.50 - 2.40
distance = 0.10 cm
Answer:
time is 0.42 sec
Explanation:
Given data
radius = 23 m
angular acceleration = 5.7 rad/s²
to find out
time
solution
we know that radius is constant so that
tangential acceleration At = angular acceleration × radius ............. 1
tangential acceleration = 5.7 × 23 = 131.1 m/s²
and
radial acceleration Ar = (angular velocity)² × radius ........................2
we consider angular velocity = ω
this is acting toward center
so
compare 1 and 2
At = Ar
5.7 r =ω³ r
ω = √5.7 = 2.38746 rad/s
so
ω = 5.7 t
2.387 = 5.7 t
t = 2.387 / 5.7
t = 0.4187
time is 0.42 sec
Answer:
the the right because opposites attract to each other
Answer:
the 8 feet long plank has the most mechanical advantage
Explanation:
Answer:
Explanation:
I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air. Before , the ball gets separated from my hand , both me the ball will be moving with velocity of 50 m /s in horizontal direction .
As soon as ball is separated from the hand , it acquires addition velocity in upward direction and acceleration in downward direction . This will give relative velocity to the ball with respect to me . So I will see the ball going in upward direction under gravitational acceleration . It appears as if I am sitting at rest and ball is going in upward direction under deceleration . My motion at 50 m/s will have no effect on the motion of ball in upward direction , according to first law of Newton . It is so because ball too will be moving in forward direction with the same speed which will not be visible to me because I too am moving with the same speed.
If I am sitting at rest at home and I threw a ball straight up into the air , I will have the same experience of seeing ball going in similar way as described above.
