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3 years ago

Question 29 of 36

Physics

1 answer:

kolbaska11 [484]3 years ago

6 0

Answer:

D. Amplitude

Explanation:

The amplitude of sound waves determines how loud or soft a sound is. The loudness of a sound is a function of its amplitude.

Amplitude is defined as the maximum displacement of wave from its source.

The amplitude of a sound wave determines how loud the sound can be heard.

A wave with a large amplitude will have a large loudness. Waves of low amplitude will also show softness.

Amplitude is one of the fascinating properties of waves.

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While camping in Denali National Park in Alaska, a wise camper hangs his pack of food from a rope tied between two trees, to kee

Crank

The tension in each of the ropes is 625 N.

Draw a free body diagram for the bag of food as shown in the attached diagram. Since the bag hangs from the midpoint of the rope, the rope makes equal angles θ with the horizontal. The tensions <em>T</em> in both the ropes are also equal.

Resolve the tension T in the ropes into horizontal and vertical components T cosθ and T sinθ respectively, as shown in the figure. At equilibrium,

$2T sin\theta = 50 N$ ......(1)

Calculate the value of sinθ using the right angled triangles from the diagram.

$sin\theta =\frac{0.06 m}{1.5 m} =0.04$

Substitute the value of sinθ in equation (1) and simplify to obtain T.

$2T sin\theta = 50 N\\ T= \frac{50 N}{2 sin\theta}=\frac{50 N}{2*0.06} = 625 N$

Thus the tension in the rope is 625 N.

4 0

4 years ago

When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6

Bezzdna [24]

Answer:

Distance between the charges, r = 0.8 meters

Explanation:

Given that,

Charge 1, $q_1=+8.4\ \mu C=+8.4\times 10^{-6}\ C$

Charge 2, $q_2=+5.6\ \mu C=+5.6\times 10^{-6}\ C$

Repulsive force between charges, F = 0.66 N

Let r is the distance between charges. The formula for the electrostatic force is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$r=\sqrt{\dfrac{kq_1q_2}{F}}$

$r=\sqrt{\dfrac{9\times 10^9\times 8.4\times 10^{-6}\times 5.6\times 10^{-6}}{0.66}}$

r = 0.8009 meters

r = 0.8 meters

So, the distance between the charges i 0.8 meters. Hence, this is the required solution.

4 0

3 years ago

You need to calculate the volume of berm that has a starting cross-sectional area of 118 SF, and an ending cross-sectional area

LekaFEV [45]

6 cubic feet I’m pretty sure that’s the answer

8 0

3 years ago

Two boxes 100 kg and 200 kg are connected by a rope on a horizontal surface and are being pulled with a horizontal force of 500

crimeas [40]

Answer:

hope this helps you......

8 0

3 years ago

A gas is enclosed in a confainer fitted with a piston of cross sectional area 0.10 the pressureof the gas is maintained in 8000

Arlecino [84]

Answer:

10 Joule

Explanation:

The solution and answer are well written in the Pic above.

6 0

3 years ago