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kozerog [31]
3 years ago
9

Which statement is most likely true?

Physics
1 answer:
Bumek [7]3 years ago
3 0
Subduction occurs in both locations
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In Newton's cannonball experiment, if the velocity is equal the orbital velocity the Cannon ball
Makovka662 [10]

In Newton's cannonball experiment, if the velocity is equal to the orbital velocity then the cannonball will stay in Orbit.

Newtons cannonball experiment stated that the distance that a cannonball will travel, before being drawn into the Earth by the forces of gravity, is dependent on the initial velocity.

Therefore, if the cannonball is launched at a velocity that matches the orbital velocity, then it will not be able to be drawn in by gravity due to the Earth moving away from the cannonball at the same speed at which the cannonball itself is falling.

This means that the cannonball will continue to fall without reaching the Earth, therefore staying in orbit, much like that of the moon or planets around the sun.

To learn more visit:

brainly.com/question/22360485?referrer=searchResults

5 0
3 years ago
How is high temperature achieved by concave mirror?
tensa zangetsu [6.8K]

A concave mirror is used in the design of solar furnaces because they converge the parallel sunrays at a point. This helps to increase the temperature of the furnace.

3 0
3 years ago
Read 2 more answers
A person standing 1.20 m from a portable speaker hears its sound at an intensity of 5.50 ✕ 10−3 W/m2. HINT (a) Find the correspo
iris [78.8K]

Answer:

PART A)

L = 97.4 dB

PART B)

I = 6.11 \times 10^{-6} W/m^2

PART C)

L = 67.9 dB

Explanation:

PART A)

level of sound is given as

L = 10 Log\frac{I}{I_o}

now we have

L = 10 Log\frac{5.50\times 10^{-3}}{10^{-12}}

L = 97.4 dB

PART B)

Since source is a spherical source

so here the intensity of sound is inversely depends on the square of the distance from the source

\frac{I_2}{I_1} = \frac{r_1^2}{r_2^2}

\frac{I_2}{5.50 \times 10^-3} = \frac{1.20^2}{36^2}

I_2 = 6.11 \times 10^{-6} W/m^2

PART C)

level of sound is given as

L = 10 Log\frac{I}{I_o}

now we have

L = 10 Log\frac{6.11\times 10^{-6}}{10^{-12}}

L = 67.9 dB

3 0
3 years ago
A capacitor with capacitance (c) = 4.50 μf is connected to a 12.0 v battery. what is the magnitude of the charge on each of the
svet-max [94.6K]

<u>Answer:</u> The charge on each plates is 54\mu C .

<u>Explanation:</u>

The magnitude of charge that flows through the plates is directly dependent on the capacitance and the voltage flowing through the plates.

Mathematically,

Q=CV

where,

Q = charge flowing = ? C

C = capacitance = 4.5\mu F=4.5\times 10^{-6}F    (Conversion Factor: 1\mu F=10^{-6}F)

V = Voltage across the plates = 12 V

Putting values in above equation, we get:

Q=4.5\times 10^{-6}F\times 12V=54\times 10^{-6}C=54\mu C

Hence, the charge on each plates is 54\mu C .

8 0
4 years ago
Read 2 more answers
PLEASE HELP
Elan Coil [88]

Answer:

Explanation:

a_{x}=0 v_{xo}= v_{o}cos(delta)t a_{y}=-g v_{yo}=sinθ

X-direction                                     | Y-direction

x=x_o+v_{xo}t ⇒ x=v_{xo}cos(delta)t |

5 0
3 years ago
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