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gizmo_the_mogwai [7]
3 years ago
14

The velocity of sound on a particular day outside is 331 meters/second. What is the frequency of a tone if it has a wavelength o

f 0.6 meters?
5.52 × 102 hertz

2.56 × 102 hertz

459 × 102 hertz

675 × 102 hertz

990 × 102 hertz
Physics
1 answer:
Leya [2.2K]3 years ago
4 0
Answer:
frequency = 5.52 * 10² Hz

Explanation:
the equation that relates velocity, frequency and wavelength is:
velocity = frequency * wavelength

We are given that:
velocity = 331 m/sec
wavelength = 0.6 m

Substitute with the givens in the equation to get the frequency as follows:
velocity = frequency * wavelength
331 = frequency * 0.6
frequency = 331 / 0.6
frequency = 5.52 * 10² Hz

Hope this helps :)
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5 0
3 years ago
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(I) In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second pro
Kipish [7]

Answer:

The second projectile was 1.41 times faster than the first.

Explanation:

In the ballistic pendulum experiment, the speed (v) of the projectile is given by:  

v = \frac{m + M}{m} \cdot \sqrt{2gh}

<em>where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum.   </em>

To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:    

\frac{v_{2}}{v_{1}} = \frac{\frac{m_{2} + M}{m_{2}} \cdot \sqrt{2gh_{2}}}{\frac{m_{1} + M}{m_{1}} \cdot \sqrt{2gh_{1}}}           (1)

<em>where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively.  </em>

Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):

\frac{v_{2}}{v_{1}} = \frac{\sqrt{h_{2}}}{\sqrt{h_{1}}}  

\frac{v_{2}}{v_{1}} = \frac{\sqrt{5.2 cm}}{\sqrt{2.6 cm}}

\frac{v_{2}}{v_{1}} = 1.41  

Therefore, the second projectile was 1.41 times faster than the first.  

I hope it helps you!

8 0
3 years ago
The voltage across the terminals of a generator is 5.7 v when it supplies a current of 0.3 A. It becomes 5.1 V when I=0.9A. Find
snow_tiger [21]

Answer:

  • The emf of the generator is 6V
  • The internal resistance of the generator is 1 Ω

Explanation:

Given;

terminal voltage, V = 5.7 V, when the current, I = 0.3 A

terminal voltage, V = 5.1 V, when the current, I = 0.9 A

The emf of the generator is calculated as;

E = V + Ir

where;

E is the emf of the generator

r is the internal resistance

First case:

E = 5.7   + 0.3r -------- (1)

Second case:

E = 5.1 + 0.9r -------- (2)

Since the emf E, is constant in both equations, we will have the following;

5.1 + 0.9r = 5.7   + 0.3r  

collect similar terms together;

0.9r - 0.3r = 5.7 - 5.1

0.6r = 0.6

r = 0.6/0.6

r = 1 Ω

Now, determine the emf of the generator;

E = V + Ir

E = 5.1 + 0.9x1

E = 5.1 + 0.9

E = 6 V

6 0
2 years ago
3. Identify three elements and three compounds. How are they<br> similar? How are they different?
Basile [38]

Answer:  A compound is a substance formed when two or more elements are chemically joined. Water, salt, and sugar are examples of compounds. When the elements are joined, the atoms lose their individual properties and have different properties from the elements they are composed of.

6 0
3 years ago
A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi
Tatiana [17]

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L)  ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L

Finally, using the Parallel Axis Theorem, we calculate I_B:

I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4}  ML^{2} =\frac{1}{3} ML^{2}

5 0
3 years ago
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