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atroni [7]
3 years ago
5

A plucked guitar string produces a sound wave with a wavelength of 0.15 m and a velocity of 10.5 m/s, what is the frequency of t

he wave?
Physics
1 answer:
Leno4ka [110]3 years ago
8 0
We know,
V= f× wavelength
10.5= f×0.15
f=10.5/0.15
f= 70 Hz
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A jet transport has a weight of 1.87 x 10⁶ N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel
qwelly [4]

Answer:

a)  R(fw)  = 46.75*10⁶ (N)

b)  R(rwi) = 70.125*10⁶ [N]

Explanation: See Attached file (the rectangle stands for a jet)

The diagram shows forces acting on the jet

Let R(fw)    Reaction of front wheels and

R(rw)   Reaction of rear wheels

Now we apply the Stevin relation, for R(fw)  and a jet weight as follows

R(fw)/ 4   =  187*10⁶ / 16

Then :

R(fw)  = ( 1/4) *187*10⁶           ⇒   R(fw)  = 46.75*10⁶ (N)

And e do the same for the reaction on rear wheels

R(rw) / 12  = 187*10⁶ /16   ⇒     R(rw) =(3/4)*187*10⁶

R(rw) = 140,25*10⁶ [N]

The last expression is for the whole reaction, and must be devide by 2

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5 0
3 years ago
A rocket, initially at rest, is fired vertically with an upward acceleration of 10 m/s2. At an altitude of 0.50 km, the engine o
noname [10]

Answer:

The maximum altitude reached with respect to the ground = 0.5 + 0.510 = 1.01 km

Explanation:

Using the equations of motion,

When the rocket is fired from the ground,

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The engine cuts off at y = 0.5 km = 500 m

The velocity at that point = v

v² = u² + 2ay

v² = 0² + 2(10)(500) = 10000

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The velocity at this point is the initial velocity for the next phase of the motion

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v = final velocity = 0 m/s (at maximum height, velocity = 0)

y = vertical distance travelled after the engine shuts off beyond 0.5 km = ?

g = acceleration due to gravity = - 9.8 m/s²

v² = u² + 2gy

0 = 100² + 2(-9.8)(y)

- 19.6 y = - 10000

y = 510.2 m = 0.510 km

So, the maximum altitude reached with respect to the ground = 0.5 + 0.510 = 1.01 km

Hope this helps!!!

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3 years ago
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