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Scorpion4ik [409]

2 years ago

12

Which amplitude of the following longitudinal waves has the greatest energy?

2 answers:

Rashid [163]2 years ago

8 0

Which amplitude of the following longitudinal waves has the greatest energy?

amplitude = 10 cm; wavelength = 6 cm; period = 4 seconds

sukhopar [10]2 years ago

8 0

Answer: A longitudinal wave has the greatest energy for amplitude 10 cm.

Explanation:

We know that,

The energy is directly proportional to the square of the amplitude.

$E \propto a^2$

(I). Amplitude = 10

then, the energy will be

$E = 100$

(II). Amplitude = 6

then, the energy will be

$E = 36$

(II). Amplitude = 4

then, the energy will be

$E = 16$

Hence, A longitudinal wave has the greatest energy for amplitude 10 cm.

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1 year ago

If two waves with identical crests and troughs meet, what is happening? The wave is reflecting. Constructive interference is occ

Kazeer [188]

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1 year ago

Read 2 more answers

A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.

Pie

#1

Volume of lead = 100 cm^3

density of lead = 11.34 g/cm^3

mass of the lead piece = density * volume

$m = 100 * 11.34 = 1134 g$

$m = 1.134 kg$

so its weight in air will be given as

$W = mg = 1.134* 9.8 = 11.11 N$

now the buoyant force on the lead is given by

$F_B = W - F_{net}$

$F_B = 11.11 - 11 = 0.11 N$

now as we know that

$F_B = \rho V g$

$0.11 = 1000* V * 9.8$

so by solving it we got

V = 11.22 cm^3

(ii) this volume of water will weigh same as the buoyant force so it is 0.11 N

(iii) Buoyant force = 0.11 N

(iv)since the density of lead block is more than density of water so it will sink inside the water

#2

buoyant force on the lead block is balancing the weight of it

$F_B = W$

$\rho V g = W$

$13* 10^3 * V * 9.8 = 11.11$

$V = 87.2 cm^3$

(ii) So this volume of mercury will weigh same as buoyant force and since block is floating here inside mercury so it is same as its weight = 11.11 N

(iii) Buoyant force = 11.11 N

(iv) since the density of lead is less than the density of mercury so it will float inside mercury

#3

Yes, if object density is less than the density of liquid then it will float otherwise it will sink inside the liquid

3 0

1 year ago

(a) when rebuilding her car's engine, a physics major must exert 300 n of force to insert a dry steel piston into a steel cylind

Vilka [71]

There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03

Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
$F_f = \mu N$
where $\mu$ is the coefficient of friction, while N is the normal force. So we have:
$F=\mu N$
since we know that F=300 N and $\mu=0.3$ , we can find N, the magnitude of the normal force:
$N= \frac{F}{\mu}= \frac{300 N}{0.3}=1000 N$

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is $\mu=0.03$ due to the presence of the oil. Therefore, we have:
$N= \frac{F}{\mu}= \frac{300 N}{0.03}=10000 N$

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1 year ago

Suppose that a rectangular toroid has 2,000 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i

andrey2020 [161]

Answer:

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Explanation:

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$i=\sqrt{\frac{2U}{L}}$

$i=\sqrt{\frac{2\times 4\times 10^{-6}}{0.06}}=0.01154\ A$

So the current flowing through rectangular toroid will be 0.01154 A

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1 year ago