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Velocity which stone gains when falling from height of 80 m is approximately equal to *

jekas [21]

Answer:

40

Explanation:

3 0

2 years ago

Short wavelengths, from high-pitched sounds, cause displacement of the basilar membrane near the oval window. true false

lakkis [162]

Answer:

True

Explanation:

Short wavelengths, from high-pitched sounds, cause displacement of the basilar membrane near the oval window, is a True statement.

Stapes bones transmit movement to the oval windows. Stapes cause the round window membrane to move out. This turn allows the movement of fluid within the cochlea. This leads to motion of inner cochlear inner hair and thus hearing is possible.

8 0

2 years ago

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During a fireworks display, a shell is shot into the air at an angle of theta degrees above the horizontal. The fuse is timed to

Klio2033 [76]

Answer:

θ = 83.6º

Explanation:

Once the shell is in the air, the only influence acting on it (assuming that the resistance of the air is negligible) is gravity.
Gravity acts on the shell accelerating it in the vertical direction with a=g=-9.8 m/s2 (taking the upward direction as the positive one).
Since movements in perpendicular directions are independent each other, we conclude that in the horizontal direction, speed must be constant.
So, we can find the speed in the horizontal direction (the projection of the velocity vector along the horizontal axis) applying simply the definition of average velocity, as follows:

$v_{avg} =\frac{\Delta x}{\Delta t} (1)$

Taking the launch site as the origin, and making the horizontal direction coincident with the x-axis (positive to the right), we know that the horizontal displacement is 81.7 m, and the time at the explosion, 8.6 s, so we can write the horizontal component of the velocity vector (its projection along the x-axis) as follows:

$v_{x} = \frac{81.7m}{8.6s} = v_{o} * cos \theta = 9.5 m/s (2)$

In the vertical direction, since we know that the shell will explode when it reaches to the maximum height, at this point just before exploding, the vertical component of the velocity is just zero, even though it continues being accelerated downward.
Applying the definition of acceleration, replacing a by the value of g, we can write the following expression:

$v_{fy} = v_{oy} - g*t = v_{o} * sin \theta -g*t = 0 (3)$

⇒ v₀* sin θ = g*t = 9.8m/s2* 8.6s = 84.3 m/s (4)

Dividing (4) by (2), we can find tg θ (the angle of the initial velocity with the x-axis), as follows:

$\frac{v_{o}* sin \theta}{v_{o}* cos \theta} = tg \theta= \frac{84.3m/s}{9.5m/s} = 8.87 (5)$

⇒ θ = tg⁻¹ (8.87) = 83.6º

7 0

2 years ago

The speed is represented by

nordsb [41]

Answer:

Instantaneous velocity is equal to speed of the object at that particular instant.

Explanation:

Instantaneous velocity is the velocity of the object at that particular instant. It is also equal to speed of the object at that instant. It can be calculated by drawing a tangent to the position-time graph at that point and finding the tangent’s slope.

The first option ‘The ratio of change in position to the time interval during that change’ gives the average velocity of an object and not speed. Similarly the second option ‘the absolute value of the slope of position time graph’ gives the average speed.

3 0

2 years ago

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Oceanographers use submerged sonar systems, towed by a cable from a ship, to map the ocean floor. In addition to their downward

KATRIN_1 [288]

Answer:

Tension in the cable is T = 16653.32 N

Explanation:

Give data:

Cross section Area A = 1.3 m^2

Drag coefficient CD = 1.2

Velocity V = 4.3 m/s