since airplane is thrown towards west with speed 6 m/s
while air is blowing with speed 8 m/s towards north
so here the net speed of air plane will be the resultant of airplane speed and wind speed always
SO here we can say it would be a combination of vector along west with must be of length 6 m/s and other vector is towards north with is of length 8 m/s
so correct answer must be 1st option
Apply Newton's second law to the person's motion:
F = ma
F = net force, m = mass, a = acceleration
Given values:
m = 50.8kg, a = 3.50m/s²
Plug in and solve for F:
F = 50.8(3.50)
F = 178N
Answer:
"2Ω" is the net resistance in the circuit.
Explanation:
The given resistors are:
R1 = 3Ω
R2 = 6Ω
The net resistance will be:
⇒ 
On substituting the values, we get
⇒ 
On taking L.C.M, we get
⇒ 
⇒ 
⇒ 
On applying cross-multiplication, we get
⇒ 
The distance at which the man slips is 0.3 m
Newton's Second Law, F = ma, is used to calculate the braking distance. By dividing the mass of the car by the gravitational acceleration, one may determine its weight. The weight of the car multiplied by the coefficient of friction equals the brake force.
Given-
mass of man= 70 kg
frictional coefficient μ=0.02
mass of body thrown= m2 = 3kg
let s be the stopping distance
we know that frictional force = F= μN
=μMg= 0.02 x 70 x 10
=14 N
∴acceleration, a= 14/70 = 0.2 m/s²
now on applying conservation of linear momentum
pi=pf pi=0 (initially at rest)
0=m1v1-m2v2 (v1= velocity of man) (v2=velocity of body= 8m/s
v1= m2v2 /m1= 0.3 m/s
we know,
v²- u² = -2as
0- (0.3) ²= -2 x 0.2 x 5
s= 0.09/0.4 ≈ 0.3 m
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