Initial speed(u)=0m/s
Final speed(v)= 27m/s
Time(t)=7.6s
Use the equation of motion: v = u + at
27 = 0 + a(7.6)
27/7.6 = a
a = 3.55 m/s^2 (3 s.f)
Answer:
The magnitude of the static frictional force is 1200 N
Explanation:
given information :
radius, r = 0.380 m
applied-torque, τ1 = 456 N
The car has a constant velocity, thus the acceleration is zero
α = 0
Στ = I α
τ1 - τ2 = I α
τ2 = counter-torque
τ1 - τ2 = 0
τ1 = τ2
r x 
= τ1
= the static frictional force (N)
= τ1 /r
= 456 N/0.380 m
= 1200 N
By definition,
q = 1.22y/D
Where,
q = min. angle
y = wavelength
D = Aperture diameter = diameter of the antenna
At distance "x" from the antenna,
L =xq = 1.22xy/D
Where, L = Min. distance
But, y =c/f = (3*10^8)/(16*10^9) = 0.01875 m
Substituting;
L = 1.22*5*10^3*0.01875/2.1 = 54.46 m
Answer:
1.2cm
Explanation:
V=(2ev/m)^1/2
=(2*1.6*10^19 x2500/ 1.67*10^27)^1/2
=6.2x10^5m/s
Radius of resulting path= MV/qB
= 1.67*10^-27x6.92*10^6/1.6*10^-16 x0.6
=0.012m
=1.2cm
Answer:
a) v, v
b) 2mv^2
c) Elastic collion
Explanation:
(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction
Here x, y represent direction.They are not variable. 1 and 2 represent before and after.
2vm=v1xm+v2xm, we find v1x=v.
From momentum conservation in y-direction
0 =v1ym+v2ym, we findv1y=v.
(b) By energy conservation principle
Before: K=1/2m(2v)^2=2mv^2.
After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2
(c) The collision is elastic
