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Vaselesa [24]
3 years ago
6

Help pleaseeeeeeeeeeeeee

Physics
1 answer:
77julia77 [94]3 years ago
8 0

Answer:

hello the answer is 47m/s

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A satellite is to be launched into an orbit of radius,r Show that v = 2gr, where V is the
Snezhnost [94]

Explanation:

We start by using the conservation law of energy:

\Delta{K} + \Delta{U} = 0

or

\dfrac{1}{2}mv^2 - G\dfrac{mM}{r} = 0

Simplifying the above equation, we get

v^2 = 2G\dfrac{M}{r}

We can rewrite this as

v^2 = 2\left(G\dfrac{M}{r^2}\right)r

Note that the expression inside the parenthesis is simply the acceleration due to gravity g so we can write

v^2 = 2gr

where v is the launch velocity.

6 0
3 years ago
A heated piece of metal cools according to the function c(x) = (.5)^(x _ 11), where x is measured in hours. A device is added th
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You just need to replace x with 5 in each function

.5^5 - 11
-5-3

.5 ^-6
-8


64 - 8 = 56 A Celcius

Hope this helps
3 0
3 years ago
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Eddie and Val observed the picture of an athlete running in a race.
dedylja [7]

Answer:

Your answer would be C <u><em>Hope this helps</em></u>

8 0
3 years ago
Can an object be moving downward and have an upward acceleration
dusya [7]

Answer: no

Explanation:

because the object is moving downwards so it will be  called decceleration . which is the opposite of acceleration .

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3 years ago
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Consider the standing wave pattern below created by a string fixed at both ends. The string is under a tension of 0.98 N and has
Shalnov [3]

Answer:

11.07Hz

Explanation:

Check the attachment for diagram of the standing wave in question.

Formula for calculating the fundamental frequency Fo in strings  is V/2L where;

V is the velocity of the wave in string

L is the length of the string which is expressed as a function of its wavelength.

The wavelength of the string given is 1.5λ(one loop is equivalent to 0.5 wavelength)

Therefore L = 1.5λ

If L = 3.0m

1.5λ = 3.0m

λ = 3/1.5

λ = 2m

Also;

V = √T/m where;

T is the tension = 0.98N

m is the mass per unit length = 2.0g = 0.002kg

V = √0.98/0.002

V = √490

V = 22.14m/s

Fo = V/2L (for string)

Fo = 22.14/2(3)

Fo = 22.14/6

Fo = 3.69Hz

Harmonics are multiple integrals of the fundamental frequency. The string in question resonates in 2nd harmonics F2 = 3Fo

Frequency of the wave = 3×3.69

Frequency of the wave = 11.07Hz

3 0
3 years ago
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