Consider the two-body system at the right. A 22.7-N block is placed upon an inclined plane which is inclined at a 17.2 degree an
gle. The block is attached by a string to a 34.5-N block which is suspended over a pulley. The coefficient of friction is 0.219. Determine the acceleration of the block and the tension in the stringAnswers:
1 answer:
Answer:
Given that
For A weight Wt= 22.7 N
m₁ = 22.7/10 = 2.27 kg
Force alone inclined plane
Wt₁ = m₁ g sin θ
Wt₁ =22.7 sin 17.2°
Wt₁ = 6.7 N
For B weight Wt₂= 34.5-N
m₂ = 3.45 kg
coefficient of friction ,μ= 0.219
θ = 17.2 degree
The friction force on the block A
Fr= μ m₁ g cos θ
Fr= 0.216 x 22.7 x cos 17.2°
Fr= 4.68 N
Lets take acceleration of system is a m/s²
Tension = T
From Newtons law
Wt₂ - Wt₁ - Fr = (m₁ +m₂) a
34.5 - 6.7 - 4.68 = (2.27 + 3.45 ) a
a= 4.05 m/s²
Block B
Wt₂ - T = m₂ a
T = 34.5 - 4.05 x 3.45
T= 20.52 N
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Answer:
8451.62109367671 Newtons
Explanation:
It is known that
So,
The force in Newtons is 8451.6210936765 Newtons
Answer:
The right solution is "165.8 nm".
Explanation:
Given:
Index of refraction,
n = 1.81
Wavelength,
λ = 600 nm
We know that,
⇒
By putting the values, we get
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:
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