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alexira [117]
3 years ago
12

Suppose an object is moving 2.1 m/s north on a river, but the river is flowing to the east at a velocity of 1.2 m/s. What is the

magnitude of the resultant velocity?
Physics
1 answer:
rewona [7]3 years ago
8 0

Answer:

2.4 m/s

Explanation:

Given:

Velocity of the object moving north = 2.1 m/s

Velocity of the river moving eastward = 1.2 m/s

The resultant velocity is the vector sum of the velocities of object and river.

Since the directions of velocity of object and river are perpendicular to each other, the magnitude of the resultant velocity is obtained using Pythagoras Theorem.

The velocities are the legs of the right angled triangle and the resultant velocity is the hypotenuse.

The magnitude of the resultant velocity (R) is given as:

R^2=2.1^2+1.2^2\\\\R^2=4.41+1.44\\\\R=\sqrt{5.85}\\\\R=2.4\ m/s

Therefore, the resultant velocity has a magnitude of 2.4 m/s.

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The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This
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To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

g = \frac{GM}{(d-R_{CM})^2}

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

R_{CM} = Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

g = \frac{GM}{(d-R_{CM})^2}

g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}

g = 3.4*10^{-5}m/s^2

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

\omega = \frac{2\pi}{T}

At the same time we have that centripetal acceleration is given as

a_c = \omega^2 r

Replacing

a_c = (\frac{2\pi}{T})^2 r

a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)

a_c = 3.34*10^{-5}m/s^2

3 0
3 years ago
Pls help! True or False?
jasenka [17]

Answer:

True

Explanation:

i hope it helps

please follow me

7 0
3 years ago
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4.5 billion km, the average separation between the sun and Neptune (report answer in hours). How long does it take light to trav
Liula [17]

Answer:

t = 4.17 hours

Explanation:

given,

The distance between Sun and Neptune, d = 4.5 billion Km

                                                                         = 4.5 x 10⁹ Km

                                                                          = 4.5 x 10¹¹ m

The velocity of light, c = 3 x 10⁸ m/s

The velocity is always equal to displacement by the time.

                                           <em>V = d / t    m/s</em>

∴                                           t = d / V

                                               = 4.5 x 10¹¹ m / 3 x 10⁸ m/s

                                               = 15,000 s

                                               = 4.17 h

Hence, the time taken by the light rays to reach the Neptune is, t = 4.17 h

4 0
3 years ago
Starting circuit One battery. 2 light bulbs in parallel; switch What is the voltage across the battery? What is the voltage acro
balu736 [363]

Answer:

The voltage across light bulb 1 and light bulb 2 is the the same i.e V

Explanation:

In a parallel circuit, the Voltage is same across all the components of the circuit and the current flowing through each component is added to get the total current across the circuit.

Let us say, the voltage across the circuit is V. The voltage across light bulb 1 and light bulb 2 is the the same i.e V

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3 years ago
es la atmósfera que se percibe en un cuento de acuerdo a la manera de pensar y sentir de los personajes no se ve sólo se siente​
masya89 [10]

Answer:

can you please ask in english

8 0
3 years ago
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