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3 years ago

HELPPPPP MEEEEE PLEASE I NEED TO SUBMIT IN LESS THAN 10 MINSS

Physics

1 answer:

DerKrebs [107]3 years ago

6 0

Answer:

X = 2146.05 m

Explanation:

We need to understand first what is the value we need to calculate here. In this case, we want to know how far from the starting point the package should be released. This is the distance.

We also know that the plane is flying a certain height with an specific speed. And the distance we need to calculate is the distance in X with the following expression:

X = Vt (1)

However we do not know the time that this distance is covered. This time can be determined because we know the height of the plain. This time is referred to the time of flight. And the time of flight can be calculated with the following expression:

t = √2h/g (2)

Where g is gravity acceleration which is 9.8 m/s². Replacing the data into the expression we have:

t = √(2*2500)/9.8

t = 22.59 s

Now replacing into (1) we have:

X = 95 * 22.59

This is the distance where the package should be released.

Hope this helps

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The acceleration due to gravity on the Moon's surface is

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Answer:

50 lb

Explanation:

Given,

The weight of astronaut's life support backpack on Earth (w) = 300 lb

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A wire of length 6cm makes an angle of 20° with a 3 mT

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Answer:

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Explanation:

Let $\theta$ denote the angle between the wire and the magnetic field.

Let $B$ denote the magnitude of the magnetic field.

Let $l$ denote the length of the wire.

Let $I$ denote the current in this wire.

The magnetic force on the wire would be:

$F = B \cdot l \cdot I \cdot \sin(\theta)$ .

Because of the $\sin(\theta)$ term, the magnetic force on the wire is maximized when the wire is perpendicular to the magnetic field (such that the angle between them is $90^\circ$ .)

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$\theta = 20^\circ$ (or, equivalently, $(\pi / 9)$ radians, if the calculator is in radian mode.)
$B = 3\; \rm mT = 3 \times 10^{-3}\; \rm T$ .
$l = 6\; \rm cm = 6 \times 10^{-2}\;\rm m$ .
$F = 1.5\times 10^{-4}\; \rm N$ .

Rearrange the equation $F = l \cdot I \cdot \sin(\theta)$ to find an expression for $I$ , the current in this wire.

$\begin{aligned} I &= \frac{F}{l \cdot \sin(\theta)} \\ &= \frac{3\times 10^{-3}\; \rm T}{6 \times 10^{-2}\; \rm m \times \sin \left(20^{\circ}\right)} \\ &\approx 7.3 \times 10^{-3}\; \rm A = 7.3 \; \rm mA\end{aligned}$ .

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