Answer:
6.78 X 10³ N/C
Explanation:
Electric field near a charged infinite plate
= surface charge density / 2ε₀
Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.
Field due to charge density of +95.0 nC/m2
E₁ = 95 x 10⁻⁹ / 2 ε₀
Field due to charge density of -25.0 nC/m2
E₂ = 25 x 10⁻⁹ / 2ε₀
Total field
E = E₁ + E₂
= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ / 2ε₀
= 6.78 X 10³ N/C
The answer would be C. It will decrease with descent. Hope this helps!
L = 1.00 m, the original length
A = 0.5 mm² = 0.5 x 10⁻⁶ m², the cross sectional area
E = 2.0 x 10¹¹ n/m², Young's modulus
P = 1500 N, the applied tension
Calculate the stress.
σ = P/A = (1500 N)/(0.5 x 10⁻⁶ m²) = 3 x 10⁹ N/m²
Let δ = the stretch of the string.
Then the strain is
ε = δ/L
By definition, the strain is
ε = σ/E = (3 x 10⁹ N/m²)/(2 x 10¹¹ N/m²) = 0.015
Therefore
δ/(1 m) = 0.015
δ = 0.015 m = 15 mm
Answer: 15 mm
The two systems that work together to deliver oxygen are D, respiratory and cardiovascular
Answer: Change in ball's momentum is 1.5 kg-m/s.
Explanation: It is given that,
Mass of the ball, m = 0.15 kg
Speed before the impact, u = 6.5 m/s
Speed after the impact, v = -3.5 m/s (as it will rebound)
We need to find the change in the magnitude of the ball's momentum. It is given by :
So, the change in the ball's momentum is 1.5 kg-m/s. Hence, this is the required solution.
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