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kaheart [24]
3 years ago
7

A flute plays a note with a frequency of 266 Hz. What is the speed of sound if the wavelength is 1.3 m?

Physics
2 answers:
hjlf3 years ago
6 0

Wave speed  =  (frequency)  x  (wavelength)

                       =  (266 /sec)  x  (1.3 meters)

                       =     345.8 meters/sec
ivolga24 [154]3 years ago
3 0

345.8 meters/sec  is the correct answer


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When the Apollo 11 lunar module Eagle lands on the moon it comes to a stop 10m above the surface of the moon. The last 10m it fr
Scrat [10]

Answer:

i) 3.514 s, ii) 5.692 m/s

Explanation:

i) We can use Newton's second law of motion to find out how long does it take for the Eagle to touch down.

as the equation says for free-falling

h = ut +0.5gt^2

Here, h = 10 m, g = acceleration due to gravity = 1.62 m/s^2( on moon surface)

initial velocity u = 0

10 = 0.5×1.62t^2

t = 3.514 seconds

Therefore, it takes t = 3.514 seconds for the Eagle to touch down.

ii) use Newton's 1st equation of motion to calculate the velocity of the lunar module when it hits the surface of the moon

v = u + gt

v = 0+ 1.62×3.514

v= 5.692 m/s

7 0
3 years ago
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attashe74 [19]
We can’t see the attachment :(
8 0
3 years ago
Read 2 more answers
1. Describe your egg drop apparatus and at least three (3) design features you implemented in order to try to help your egg surv
Olegator [25]

Answer:

Explanation:

There are three basic ways to increase the likelihood of safely dropping an egg:

Slow down the descent speed.

Parachutes are an obvious method for slowing the decent speed, as long as the design includes a way to keep the parachute open.

Cushion the egg so that something other than the egg itself absorbs the impact of landing.

The largest end of the egg has an area of air trapped between the egg's two membranes. This air space forms when the contents of the egg cool and contract after the egg is laid. It accounts for the crater you often see at the end of a hard-cooked egg. Upon impact the heavier spherical yolk continues moving towards the ground. The compression of the airspace acts like an air bag for the eggs' valuable contents. Building an artificial cushioning device will also help absorb the impact of landing.

The largest end of the egg has an area of air trapped between the egg's two membranes. This air space forms when the contents of the egg cool and contract after the egg is laid. It accounts for the crater you often see at the end of a hard-cooked egg. Upon impact the heavier spherical yolk continues moving towards the ground. The compression of the airspace acts like an air bag for the eggs' valuable contents. Building an artificial cushioning device will also help absorb the impact of landing.

Orient the egg so that it lands on the strongest part of the shell.

The arch structure at either end of the egg is stronger than its sides. Pressure is distributed down (or up) the arches so that less pressure acts on any one point. Orienting the arch downwards will increase the egg's survival.

Hope this helps you

5 0
2 years ago
What are two examples of observations you might make, and two examples of inferences you might make, when looking at logs burnin
Sedbober [7]

Answer:

One observation would be the change in odor when observing the logs burning in a campfire and because of that a inference could be because a change in odor is occurring, a chemical change is happening to the logs (combustion).

A second observation would be that the wood being burned gave off smoke at first but then stopped and because of that a inference would be that the compound that were being burned creating the smoke was all evaporated from the wood.

5 0
3 years ago
A fluid, with a density of rho = 1165 kg/m3, flows in a horizontal pipe. In one segment of the pipe the flow speed is v1 = 4.53
Elza [17]

Answer:

The pressure difference between two pipe is 1.01 \times 10^{4} Pa

Explanation:

Density \rho = 1165 \frac{kg}{m^{3} }

Speed in one pipe v_{1} = 4.53 \frac{m}{s}

Speed in second pipe v_{2} = 1.77 \frac{m}{s}

According to the bernoulli equation,

The pressure difference is given by,

     P = \frac{1}{2} \rho v^{2}

P_{2} - P_{1} = \frac{1}{2}  \rho (v_{1}^{2}  - v_{2}^{2}  )

P_{2} - P_{1} = \frac{1}{2} \times 1165 \times[ (4.53)^{2}- (1.77)^{2}]

P_{2} - P_{1} = 10128.51

P_{2} - P_{1} = 1.01 \times 10^{4} Pa

Therefore, the pressure difference between two pipe is 1.01 \times 10^{4} Pa

6 0
3 years ago
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