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Ilia_Sergeevich [38]
3 years ago
14

An Air Force plane lands with a velocity of 125 m/s and accelerates at a maximum rate of -6.5 m/s^2.

Physics
1 answer:
Alisiya [41]3 years ago
8 0

Answer:

a) 19.2 s

b) No

Explanation:

Given:

v₀ = 125 m/s

a = -6.5 m/s²

v = 0 m/s

a) Find: t

v = at + v₀

(0 m/s) = (-6.5 m/s²) t + (125 m/s)

t ≈ 19.2 s

b) Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (125 m/s)² + 2 (-6.5 m/s²) Δx

Δx ≈ 1200 m

An aircraft carrier that's 850 meters long won't be long enough.

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A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a
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Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]              (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°

The distance r is:

r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m

You replace the values of all parameters in the equation (1):

\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

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