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Paladinen [302]
2 years ago
6

Which term below describes a measurement of how hard an object pushes against a surface?

Physics
2 answers:
Scilla [17]2 years ago
8 0

I hope the answer is D. Pressure

expeople1 [14]2 years ago
3 0

Answer:

like the other guy, the answer is pressure

Explanation:

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The width of a particular microwave oven is exactly right to support a standing-wave mode. Measurements of the temperature acros
Serjik [45]

Answer:

Explanation:

In standing wave pattern we find region of nodes where vibration is minimum or cold spots . The distance between any two consecutive node is half the wave length . There are 5 cold spot or node in between which is equal to 4 half wave length .

width of oven = 4 x half wave length

= 4 x (12 / 2 )

= 24 cm

3 0
3 years ago
You stand at the top of a deep well. To determine the depth, D, of the well you drop a rock from the top of the well and listen
Paladinen [302]

Answer:

(A)

\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s  \right )D+t_t^2v_s^2=0

<em>(B)  D=54.71 m</em>

Explanation:

<u>Free Fall</u>

When a particle is dropped in free air, it starts falling to the ground with an acceleration equal to the gravity. If one wanted to know the height of launching, it can indirectly be measured by the time it takes to reach the ground by the formula

\displaystyle D=\frac{gt^2}{2}

Solving for t

\displaystyle t=\sqrt{\frac{2D}{g}}

If we are taking into consideration the time we can hear the sound it makes when hitting the ground (or water in this case), we must also consider the speed of the sound for the time it takes to reach back our ears. That time can be computed from the basic equation for the speed

\displaystyle t=\frac{D}{v_s}

(A)

The total measured time is the sum of both times and it's given as t_t=3.5\ seconds

\displaystyle t_t=\sqrt{\frac{2D}{g}}+\frac{D}{v_s}

From this equation we'll manage to compute D

First, we isolate the square root

\displaystyle \sqrt{\frac{2D}{g}}=t_t-\frac{D}{v_s}

Let's square both sides

\displaystyle \frac{2D}{g}=t_t^2-2t_t\frac{D}{v_s}+\frac{D^2}{v_s^2}

Multiplying by v_s^2

\displaystyle \frac{2Dv_s^2}{g}=t_t^2v_s^2-2t_tDv_s+D^2

Rearranging and factoring

\boxed{\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s\right )D+t_t^2v_s^2=0}

Now, let's put in numbers:

g=9.8\ m/s^2,\ v_s=345\ m/s,t_t=3.5\ sec

\displaystyle D^2-\left (\frac{2(345)^2}{9.8}+2(3.5)(345)\right )D+(12.25)345^2=0

Computing all the coefficients:

\displaystyle D^2-26,705.82D+1,458,056.25=0

Solving for D, we have two possible solutions:

D=54.71,\ D=26,651.11

The second solution is called "extraneous", since it comes from squaring an equation, which can introduce non-valid (or external) solutions. It's impossible, given the conditions of the problem, that the well could be 26.5 km deep. So we'll keep the only solution as.

<em>D=54.71 m</em>

Let's prove our calculations by computing both times:

\displaystyle t_1=\sqrt{\frac{2(54.71)}{9.8}}=3.34\ sec

\displaystyle t_2=\frac{54.71}{345}=0.16\ sec

We can see their sum is 3.5 seconds, 3.34 of which were taken to reach the bottom of the well, and 0.16 sec took the sound to reach the top.

3 0
3 years ago
Fill in the blanks to complete the sentence.
mina [271]

The answer is:

Fill in the blanks to complete the sentence.

Light acts like a  PARTICAL  when it bounces off surfaces,

and acts like a  WAVE  when it bends around objects.

I hope this helps you :>

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Which excerpt are you talking about?
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A boy is pedaling his bicycle at a velocity of 0.20km/ minute . How far will he travel in 2.5 hours
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30km. 24 the first two hours and 6 the half hour
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