Answer:
the net force is acting upon the object. The net force is the vector sum of all the forces that act upon an object. That is to say, the net force is the sum of all the forces, taking into account the fact that a force is a vector and two forces of equal magnitude. Hope this helps you.
Answer:
a) The trajectory will be a helical path.
b) θ = 2*π rad
Explanation:
a) Since the initial velocity of the particle has a component parallel (x-component) to the magnetic field B
, the trajectory will be a helical path.
b) Given
t = 2*π*m/(q*B)
We can use the equation
θ = ω*Δt
where
θ is the angular displacement
ω is the angular speed, which is obtained as follows:
ω = q*B/m
then we have
θ = (q*B/m)*2*π*m/(q*B)
⇒ θ = 2*π rad
22. a - (vf^2 - vi^2)/(2d)
a = (0 - 23^2)/(170)
a = -3.1 m/s^2
23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3
33 = 3t
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long.
24. The sprinter starts from rest. The average acceleration is found from:
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s
25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m
26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s
acceleration, a = -7.00m/s^2
displacement, s - 92m
Using v^2 = u^2 - 2as
0^2 - u^2 + 2 (-7.00) (92)
initial velocity, u = sqrt (1288) = 35.9 m/s
This is the speed pf the car just bore braking.
I hope this helps!!
Answer:
KE=800,000
Explanation:
The formula for kinetic energy is KE=1/2mv^2 or Kinetic Energy= 0.5*mass*velocity^2
so 1000 is the mass and 40 is the velocity
KE=0.5*1000*40^2
KE=0.5*1,000*1,600
KE=800,000 Joules
