Answer:
Given that
T= 0.43 s
Radius of the ball path's , r=2.1 m
a)
We know that
f= 1/T
Here f= frequency
T= Time period
Now by putting the values
f= 1/T
T= 0.43 s
f= 1/0.43
f=2.32 Hz
b)
We know that
V= ω r
ω = 2 π f
ω=Angular speed
V= Linear speed
ω = 2 π f=ω = 2 x π x 2.32 =14.60 rad/s
V= ω r= 14.60 x 2.1 = 30.66 m/s
c)
Acceleration ,a
a =ω ² r
a= 14.6 ² x 2.1 = 447.63 m/s²
We know that g = 10 m/s²
So
a= a/g= 447.63/10 = 44.7 g m/s²
a= 44.7 g m/s²
Answer:
Explanation:
1.
vf = vi + a(t)
vi = 0 m/s
a = 17/13
= 1.3 m/s^2
2.
F = M * a
= 15 * 1.3
= 19.62 N
3.
Normal force, Fn = m * g
= 15 * 9.8 m/s²
= 147 N.
Then find the maximum force of friction, knowing that μs = 0.76
Ff = Fn × μs
= 147 * 0.76
= 111.83 N
Maximum acceleration,
Ff = m × a
111.83 = 15 * a
a = 7.4556 m/s²
4.
In order to find the acceleration for the box, you need to know the net force of the box moving in the x direction and the frictional force, and you will end up with the Force of the vehicle minus the Frictional force (Ff) between the box and the vehicle, resulting your net force in the x direction.
F = m*a and Ff = μ * N
m*a = μk * N
m*a = μk * m * g
a = μk * g
a = 0.61 * 9.81
5.98 m/s^2 for the acceleration of the box on top of the vehicle.
5.
Assuming the box has re settled and is no longer sliding when braking begins and the surface remains horizontal, the maximum negative acceleration will again be
a = -7.4556 m/s².
Time - 0.792s
acceleration - 9.8m/s2
vertical height - ?
remember that height is in meters
so from the acceleration
9.8 = x
----
0.792*0.792
cross multiply
=6.147m approximately 6.2m
Answer:
(b) B
Explanation:
The direction of force on a current carrying wire in a magnetic field can be found using the right hand rule, which states that-"stretch the thumb in the direction of the current, and point the fingers in the direction of magnetic field. The direction of palm will then give the direction of force on the wire
On wire B the forces due to A and C act in the same direction and so strengthen each other. they get added up because the forces act in the same direction.
on wires A and C the forces (due to B and C and A and B
respectively) act in opposite directions and therefore tend to cancel out.