<span>Assuming pulley is frictionless. Let the tension be ‘T’. See equation below.</span>
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Great question the answer is -25x.
Answer:
Fn: magnitude of the net force.
Fn=30.11N , oriented 75.3 ° clockwise from the -x axis
Explanation:
Components on the x-y axes of the 17 N force(F₁)
F₁x=17*cos48°= 11.38N
F₁y=17*sin48° = 12.63 N
Components on the x-y axes of the the second force(F₂)
F₂x= −19.0 N
F₂y= 16.5 N
Components on the x-y axes of the net force (Fn)
Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N
Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N
Magnitude of the net force.



Direction of the net force (β)

β=75.3°
Magnitude and direction of the net force
Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis
In the attached graph we can observe the magnitude and direction of the net force
<span>vf^2 = vi^2 + 2*a*d
---
vf = velocity final
vi = velocity initial
a = acceleration
d = distance
---
since the airplane is decelerating to zero, vf = 0
---
0 = 55*55 + 2*(-2.5)*d
d = (-55*55)/(2*(-2.5))
d = 605 meters
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Answer:
a)
b)S= 46.4 cm
Explanation:
Given that
Velocity = 16 Km/s
V= 16,000 m/s
E= 27 mV/m
E=0.027 V/m
d= 22.5 cm
d= 0.225 m
a)
lets time taken by electron is t
d = V x t
0.225 = 16,000 t

b)
We know that
F = m a = E q ------------1
Mass of electron ,m

Charge on electron

So now by putting the values in equation 1




Here initial velocity u= 0 m/s

S=0.464 m
S= 46.4 cm
S is the deflection of electron.