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3 years ago

9

A rocket is launched straight up from the earth's surface at a speed of 1.60×104 m/s . what is its speed when it is very far awa

y from the earth?

1 answer:

AlladinOne [14]3 years ago

7 0

16,000 m/s
Since it’s speed, and the distance is unknown. Gravity isn’t applying a noticeable force too on the rocket, as if it were, then the rocket would be accelerating negatively.

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What is the wavelength of visible light emitted at 4.2 x 10^8 Answer in units of M

Ber [7]

Answer:

0.714 m

Explanation:

From the question,

v = λf......................... Equation 1

Where v = velocity of visible light, f = frequency of visible light, λ = wave length of visible light.

make λ the subject of the equation

λ = v/f.................. Equation 2

Given: f = 4.2×10⁸ Hz

Note: Light is an electromagnetic wave, and all electromagnetic waves travels with the same speed (3×10⁸ m/s)

Constant: v = 3×10⁸ m/s

Substitute these values into equation 2

λ = 3×10⁸/4.2×10⁸

λ = 0.714 m

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A wire carries a current of 4.1 A. How many electrons per second are passing any cross sectional area of the wire? Enter your an

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To solve this problem it is necessary to apply the concepts related to Current and Load.

The current in terms of the charge of an electron can be expressed as

$i = \frac{q}{t}$

Where,

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At the same time the Charge is the amount of electrons multiplied by the amount of these, that is

q = ne

Replacing in the first equation we have to

$i = \frac{q}{t}$

$i = \frac{ne}{t}$

Clearing n,

$n = \frac{it}{e}$

Here the time is one second then

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2 years ago

A ball is thrown directly downward with an initial speed of 7.70 m/s, from a height of 30.2 m. After what time interval does it

KatRina [158]

Answer:

$t = 1.82$

Explanation:

Given

$u = 7.70m/s$ -- initial velocity

$s = 30.2m$ --- height

Required

Determine the time to hit the ground

This will be solved using the following motion equation.

$s = ut + \frac{1}{2}gt^2$

Where

$g = 9,8m/s^2$

So, we have:

$30.2 = 7.70t + \frac{1}{2} * 9.8 * t^2$

$30.2 = 7.70t + 4.9 * t^2$

Subtract 30.2 from both sides

$30.2 -30.2 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9t^2 - 30.2$

$7.70t + 4.9t^2 - 30.2 = 0$

$4.9t^2 + 7.70t - 30.2 = 0$

Solve using quadratic formula:

$t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

Where

$a = 4.9;\ b = 7.70;\ c = -30.2$

$t = \frac{-7.70\±\sqrt{7.70^2 - 4*4.9*-30.2}}{2*4.9}$

$t = \frac{-7.70\±\sqrt{651.21}}{9.8}$

$t = \frac{-7.70\±25.52}{9.8}$

Split the expression

$t = \frac{-7.70+25.52}{9.8}$ or $t = \frac{-7.70-25.52}{9.8}$

$t = \frac{17.82}{9.8}$ or $t = -\frac{33.22}{9.8}$

Time can't be negative; So, we have:

$t = \frac{17.82}{9.8}$

$t = 1.82$

Hence, the time to hit the ground is 1.82 seconds

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