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Brrunno [24]
2 years ago
10

An Earth satellite moves in a circular orbit 680 km above Earth's surface with a period of 98.15 min. What are (a) the speed and

(b) the magnitude of the centripetal acceleration of the satellite

Physics
1 answer:
kotykmax [81]2 years ago
6 0

The solution is in the attachment

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Maru [420]
The correct answer for this problem is c
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A small cube of metal measures 19.0 mm on a side and weighs 79.6 g. What is the density of the metal in g/cm3?
Nady [450]

Answer:

density of cube =11.605 g/cm³

Explanation:

density of a substance is the mass per unit volume of that substance.

the density of a substance = \frac{mass}{volume}

volume of a cube = l³,

l = 19.0mm , lets convert mm to cm

1mm = 0.1cm, thus, 19mm =19*0.1 =1.9cm

length of cube =1.9cm

volume of cube = 1.9³

density of cube = \frac{79.6}{1.9^{3} }

density of cube =11.605 g/cm³

8 0
3 years ago
What is the difference between a physical change and a chemical change
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A physical change in something doesn't change what the it is. For example, if you break glass, it will still be glass. In a chemical change where there is a chemical reaction, a new thing is formed and energy is either given off or absorbed. For example, when you burn a log. The carbon in the log is reacting to the oxygen to create ashe and smoke
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3 years ago
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What causes earthquake s to occur?​
Sliva [168]

Answer:

tilte in the earth

Explanation:

6 0
2 years ago
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A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she
lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

f_{(1  \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}}  }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}  = 1.3225 \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz  = 740.6 \  Hz

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, L_{new} = 2/3·L

The value of the fundamental frequency will then be given as follows;

f_{(1  \, new)} =  \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \times \dfrac{2 \times L}{3} }  =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz =  840 \ Hz

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0
3 years ago
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