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Whitepunk [10]
3 years ago
11

A plane is heading south with a velocity of 150 kilometers/hour. It experiences a tallwind with a velocity of 20 kilometers/hour

. Find the resultant
velocity of the plane.

A.120 kilometers/hour to the south

B.148 kilometers/hour to the north

C.170 kilometers/hour to the south

D.130 kilometers/hour to the north
E.150 kilometers/hour to the south
Physics
1 answer:
pogonyaev3 years ago
6 0

Answer: 170km/hr to the South

Explanation:

The resultant velocity of the plane ilwill be gotten by adding the value of the velocities of the plane with the tallwind. It should be noted that the direction of the wind is.takwn into consideration.

Therefore, the resultant velocity of the plane will be:

= 20km/hr + 150km/hr

= 170km/hr to the South

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the less motion the more potential energy there is

6 0
3 years ago
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Physics // how i solve?
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V=IR

180mA=V/R

I=2V/3R

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7 0
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A 0.600 m long pendulum is used to determine the acceleration due to gravity on a distant plane. If 20 oscillations are complete
katrin2010 [14]

Answer:

7.50 m/s^2

Explanation:

The period of a pendulum is given by:

T=2\pi \sqrt{\frac{L}{g}} (1)

where

L = 0.600 m is the length of the pendulum

g = ? is the acceleration due to gravity


In this problem, we can find the period T. In fact, the frequency is equal to the number of oscillations per second, so:

f=\frac{N}{t}=\frac{20}{35.5 s}=0.563 Hz

And the period is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{0.563 Hz}=1.776 s

And by using this into eq.(1), we can find the value of g:

g=\frac{4 \pi^2 L}{T^2}=\frac{4 \pi^2 (0.600 m)}{(1.776 s)^2}=7.50 m/s^2

6 0
3 years ago
Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius r = 5.92×10−11 m around a stationary p
DaniilM [7]

Answer:

2.068 x 10^6 m / s

Explanation:

radius, r = 5.92 x 10^-11 m

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charge of electron, q = 1.6 x 10^-19 C

As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.

centripetal force = \frac{mv^{2}}{r}

Electrostatic force = \frac{kq^{2}}{r^{2}}

where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2

So, balancing both the forces we get

\frac{kq^{2}}{r^{2}}=\frac{mv^{2}}{r}

v=\sqrt{\frac{kq^{2}}{mr}}

v=\sqrt{\frac{9\times 10^{9}\times1.6\times 10^{-19}\times 1.6\times 10^{-19}}{9.1\times 10^{-31}\times 5.92\times10^{-11}}}

v = 2.068 x 10^6 m / s

Thus, the speed of the electron is give by  2.068 x 10^6 m / s.

6 0
3 years ago
How long does it take for 4 coulombs of charge to pass through a cross
Galina-37 [17]

Answer: 2 seconds

Explanation:

Given that,

Time (T) = ?

Charge (Q) = 4 coulombs

current (I) = 2 Amps

Since charge depends on the amount of current flowing through the wire in a given time, hence

Charge = Current x Time

Q = IT

4 coulombs = 2 Amps x Time

Time = 4 coulombs / 2 Amps

Time = 2 seconds

Thus, it takes 2 seconds for the current to flow through the wire

4 0
3 years ago
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