A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the ca
r from sliding down the incline.
1 answer:
Answer:
Force of friction, f = 751.97 N
Explanation:
it is given that,
Mass of the car, m = 1100 kg
It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.
From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.
f = 751.97 N
So, the force of friction on the car is 751.97 N. Hence, this is the required solution.
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The magnitude of the net force that the rods exert after an electron is released at point P is 2.885 × 10⁻¹⁵ N.
Given values:
Length of non-conducting rod, l = 1.20 m
Charge on positive rod, +Q = +2.50 μC = +2.50 × 10⁻⁶ C
Charge on negative rod, -Q = -2.50 μC = -2.50 × 10⁻⁶ C
Distance from point P of each rod, x = 60 cm = 0.60 m
Calculation of Net electric force exerted on point P:
Consider an electron released at point P, then the net electric force exerted will be given as:
F = e. E_net - ( 1 )
Step 1:
The net electric field value is given as:
E_net = E₁ cos Φ + E₂ cos Φ
= 2E₁ cos Φ -( 2 )
where, E₁ & E₂ are electric fields due to positive and negative rod
respectively.
Φ is phase angle
Step 2:
The electric field due to positive rod is given as:
E₁ = k (λ/r) - ( 3 )
where, k is Coulomb's force constant
λ is linear charge density
r is distance between point P and half of the rod.
Now, the linear charge density is given as:
λ = Charge/length = Q/x
The value of r is given as:
r = √x²-a²
where, x is length of rod
a is half length of rod
Applying values in above equation, we get:
r = √x²-(x/2)²
r = √(1.20 m)²-(1.20/2)²
= √1.08
= 1.04 m
Substituting all the determined values in equation 3 we get:
E₁ = k (λ/r)
= k [(Q/x)/r]
= k [ Q/xr ]
= (9×10⁹ Nm²/C²) [ |+2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]
= 1.803×10⁴ N/C
Step 3:
Similarly, the electric field due to negative rod is given as:
E₂ = k [ Q/xr ]
= (9×10⁹ Nm²/C²) [ |-2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]
= 1.803×10⁴ N/C
Step 4:
Consider equation 2:
E_net = 2E₁ cos Φ
From the figure we get the phase angle as:
Φ = tan⁻¹ (0.60 m/0.60 m)
= tan⁻¹ ( 1 )
= π/4
Now, the electric field produced due to each rod is equal and mutually perpendicular. Thus, the net electric field after applying values can be calculated as:
E_net = 2(1.803×10⁴ N/C) cos π/4
= 2(1.803×10⁴ N/C) (0.5)
= 18030 N/C
Step 5:
Consider equation 1 :
F = e. E_net
where, e is charge on an electron
Applying values in above equation we get:
F = (1.6 × 10⁻¹⁹ C)(18030 N/C)
= 2.885 × 10⁻¹⁵ N
Therefore, the magnitude of the net force that the rods exert after an electron is released at point P is 2.885 × 10⁻¹⁵ N.
Learn more about electric force here:
<u>brainly.com/question/1634182</u>
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