How can resistance be useful and where can they be useful?
1 answer:
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Answer:
Explanation:
First of all we shall calculate electric field near charge 2Q .
electric field due to charge Q = K x Q / (5 x 10⁻² )²
E₁ = KQ / 25 x 10⁻⁴ = KQ x 10⁴ / 25 . It is acting along positive x axis
E₁ = KQ x 10⁴ i / 25
Similarly electric field due to charge 3Q near 2Q
= 3KQ x 10⁴ i / 25 . It is acting along y-axis
E₂ = 3KQ x 10⁴ j / 25
Similarly electric field due to charge 4Q near 2Q
= 4KQ x 10⁴ j / (25 x 2 )
= 2 KQ x 10⁴ / 25 . It is acting acting along north east direction
unit vector in north east direction = ( i + j )/ √2
So E₃ can be represented by
E₃ = 2 KQ x 10⁴ ( i + j ) / 25 x √2
Total field = KQ x 10⁴ i / 25 + 3KQ x 10⁴ j / 25 + 2 KQ x 10⁴ ( i + j ) / ( 25 x √2 )
= KQ x 10⁴ [ i + 3 j + √2 i + √2 j ) / 25
= 400 KQ ( 2.414 i + 4.414 j ) N / C
Force on 2Q = Field x charge = 400 KQ ( 2.414 i + 4.414 j ) x 2Q N
= 800 KQ² ( 2.414 i + 4.414 j ) N
= 800 x 9 x 10⁹ x ( 2.5 x 10⁻⁶ )² x 2.414 x ( i + 2 j ) N
= 108.63 ( i + 2 j ) N .
Magnitude of this force
= 108.63 x √5
= 243 N approx .
Answer:
7g/cm³
Explanation:
The formula for density is . Let's apply the formula to the question:
= 7
g/cm³
Answer:
D) The worker will accelerate at 2.17 m/s² and the box will accelerate at 1.08 m/s² , but in opposite directions.
Explanation:
Newton's third law
Newton's third law or principle of action and reaction states that when two interaction bodies appear equal forces and opposite directions. in each of them.
F₁₂= -F₂₁
F₁₂: Force of the box on the worker
F₂₁: Force of the worker on the box
Newton's second law
∑F = m*a
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Formula to calculate the mass (m)
m = W/g
Where:
W : Weight (N)
g : acceleration due to gravity (m/s²)
Data
W₁ =1.8 kN : box weight
W₂ = 0.900 kN : worker weight
F₂₁ = 0.200 kN
F₁₂ = - 0.200 kN
g = 9.8 m/s²
Newton's second law for the box
∑F = m*a
F₂₁ = m₁*a₁ m₁=W₁/g
0.2 kN = (1.8kN)/(9.8 m/s² ) *a₁
a₁= 1.08 m/s² : acceleration of the box
Newton's second law for the worker
∑F = m*a
F₁₂ = m₂*a₂ , m₂=W₂/g
- 0.2 kN =( (0.9 kN) /(9.8 m/s² ) )*a₂
a₂= -2.17 m/s² : acceleration of the worker