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Verizon [17]
3 years ago
13

ASAP! State the direction of the oscillation of a longitudinal wave?? Worth 40 P! ASAP!

Physics
1 answer:
Liula [17]3 years ago
7 0

In a longitudinal wave the particle displacement is parallel to the direction of wave propagation. ... The particles do not move down the tube with the wave; they simply oscillate back and forth about their individual equilibrium positions.Answer:

Explanation:

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Which statement is true of earths poles?
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Answer choice d is correct
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One degree Celsius indicates the same temperature change as
miss Akunina [59]
The Celsius degree is the same size as the Kelvin.
The correct choice is 'C'.
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A free positive charge released in an electric field will:
choli [55]

Answer:

accelerate in the direction in which the electric field is pointing.

Explanation:

The positive charge feels a force in the same direction as the electric field

F=Eq  

F and E are vectors, q is a scalar

(if it were a negative charge the force would be in the opposite direction)

that force will produce an acceleration in the same direction, that acceleration will cause the particle to move in the same direction, ie the direction of the electric field.

7 0
3 years ago
What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s ​2 ​​ ) towards the Eart
Wittaler [7]

Answer:

The centripetal acceleration as a multiple of g=9.8 m/s^{2} is 1.020x10^{-3}m/s^{2}

Explanation:

The centripetal acceleration is defined as:

a = \frac{v^{2}}{r}  (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

v = \frac{2 \pi r}{T}  (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude 71.9^{\circ}. Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

\cos \theta = \frac{adjacent}{hypotenuse}

\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}} (3)

Where r_{71.9^{\circ}} is the radius at the latitude of 71.9^{\circ} and r_{e} is the radius at the equator (6.37x10^{6}m).

r_{71.9^{\circ}} can be isolated from equation 3:

r_{71.9^{\circ}} = r_{e} \cos \theta  (4)

r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})

r_{71.9^{\circ}} = 1.97x10^{6} m

Then, equation 2 can be used

v = \frac{2 \pi (1.97x10^{6} m)}{24h}

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

T = 24h . \frac{3600s}{1h} ⇒ 84600s

v = \frac{2 \pi (1.97x10^{6} m)}{84600s}

v = 146.31m/s

Finally, equation 1 can be used:

a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}

a = 0.010m/s^{2}

Hence, the centripetal acceleration is 0.010m/s^{2}

To given the centripetal acceleration as a multiple of g=9.8 m/s^{2}​ it is gotten:

\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}

6 0
3 years ago
If you apply a 500N force to a car 200N car for 10 seconds but it doesn't move. How much work was done on the car?
djyliett [7]

Answer:

bbx

Explanation:

xbb

3 0
3 years ago
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