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3 years ago

A particle with mass 1.8110^-3kg and a charge of 1.2210^-8C has, at a given instant, a velocity v= (3.0*10^4 m/s)j.

Physics

1 answer:

charle [14.2K]3 years ago

7 0

Answer:

a = -0.33 m/s² k^

Direction: negative

Explanation:

From Newton's law of motion, we know that;

F = ma

Now, from magnetic fields, we know that;. F = qVB

Thus;

ma = qVB

Where;

m is mass

a is acceleration

q is charge

V is velocity

B is magnetic field

We are given;

m = 1.81 × 10^(−3) kg

q = 1.22 × 10 ^(−8) C

V = (3.00 × 10⁴ m/s) ȷ^.

B = (1.63T) ı^ + (0.980T) ȷ^

Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;

a = qVB/m

a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))

From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^

Thus;

a = -0.33 m/s² k^

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A rescue pilot drops a survival kit while her plane is flying at an ultitude of 2500m with a forward velocity of 95m. If the air

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Answer:

Approximately $2.1\; \rm km$ , assuming that $g = -9.8\; \rm m \cdot s^{-2}$ .

Explanation:

Let $t$ denote the time required for the package to reach the ground. Let $h(\text{initial})$ and $h(\text{final})$ denote the initial and final height of this package.

$\displaystyle h(\text{final}) = \frac{1}{2}\, g\, t^2 + h(\text{initial})$ .

For this package:

Initial height: $h(\text{initial}) = 2500\; \rm m$ .
Final height: $h(\text{final}) = 0\; \rm m$ (the package would be on the ground.)

Solve for $t$ , the time required for the package to reach the ground after being released.

$\displaystyle t^{2} = \frac{2\, (h(\text{final}) - h(\text{initial}))}{g}$ .

$\begin{aligned} t &= \sqrt{\frac{2\, (h(\text{final}) - h(\text{initial}))}{g} \\ &\approx \sqrt{\frac{2\times (0\; \rm m - 2500\; \rm m)}{(-9.8\; \rm m \cdot s^{-2})}} \approx 22.588\; \rm s\end{aligned}$ .

Assume that the air resistance on this package is negligible. The horizontal ("forward") velocity of this package would be constant (supposedly at $95\; \rm m \cdot s^{-1}$ .) From calculations above, the package would travel forward at that speed for about $22.588\; \rm s$ . That corresponds to approximately: $95\; \rm m \cdot s^{-1} \times 22.588\; \rm s \approx 2.1 \times 10^{3}\; \rm m = 2.1\; \rm km$ .