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erica [24]
3 years ago
14

When you strike two tuning forks simultaneously, you hear three beats per second. The frequency of the first tuning fork is 440

Hz. What is the frequency of the second tuning fork? 1. 446 Hz 2. 437 Hz 3. 443 ㎐ 4. not enough information is given to decide
Physics
2 answers:
tekilochka [14]3 years ago
5 0

Answer:4

Explanation:

Given

Frequency of beats is 3 beats per second

Frequency of first tuning fork is f_1=440\ Hz

Beat frequency is difference in the frequency of tuning forks i.e.

either f_1-f_2 or f_2-f_1 =3

so f_2=3+440=443\ Hz

or

f_2=440-3=437\ Hz

so there is not enough information given to decide.

AlladinOne [14]3 years ago
3 0

Answer:

The frequency of the second tuning fork is 437 Hz.

(2) is correct option.

Explanation:

Given that,

Frequency of first tuning fork = 440 Hz

Frequency of beat = 3 Hz

We need to calculate the frequency of the second tuning fork

Using formula of beat frequency

beat\ frequency=f_{1}-f_{2}

Put the value in to the formula

3=440-f_{2}

f_{2}=440-3

f_{2}=437\ Hz

Hence, The frequency of the second tuning fork is 437 Hz.

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kykrilka [37]

The horse's position on the ground at time <em>t</em> is

<em>x</em> = (20 m/s) <em>t</em>

The baboon's height from the ground at time <em>t</em> is

<em>y</em> = 3 m - 1/2 <em>g</em> <em>t</em>²

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The baboon falls and lands on the horse, so that the two animals meet when the baboon's height is 2 m from the ground, which happens after

2 m = 3 m - 1/2 <em>g</em> <em>t</em>²

1/2 <em>g</em> <em>t</em>² = 1 m

<em>t</em>² = (2 m) / (9.80 m/s²)

<em>t</em> ≈ 0.452 s

In this time, the horse reaches the tree, so its distance from it is

(20 m/s) * (0.452 s) ≈ 9.04 m

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4 years ago
In a parachute, falling steadily at 7.5m/s, how long does it take to reach the ground from 3,000 m high?
photoshop1234 [79]

Answer:

25 minutes

Explanation:

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3 years ago
Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor
fenix001 [56]

Answer:

a) F_{r}= -583.72MN i + 183.47MN j + 6.05GN k

b) E=3.04 \frac{GN}{C}

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}

which yields:

r_{AP}^{2}= 43 m^{2}

(I will assume the positions are in meters)

Next, we can make use of the force formula:

F=k_{e}\frac{q_{1}q_{2}}{r^{2}}

so we substitute the values:

F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}

which yields:

F_{AP}=418.14 MN

Now we can find its components:

F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i

F_{APx}=191.30 MNi

F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j

F_{APy}=191.30MN j

F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k

F_{APz}=318.83 MN k

And we can now write them together for the first force, so we get:

F_{AP}=(191.30i+191.30j+318.83k)MN

We continue with the next force. The procedure is the same so we get:

r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}

which yields:

r_{BP}^{2}= 33 m^{2}

Next, we can make use of the force formula:

F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}

which yields:

F_{BP}=2.18 GN

Now we can find its components:

F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i

F_{BPx}=758.98 MNi

F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j

F_{BPy}=758.98MN j

F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k

F_{BPz}=1.897 GN k

And we can now write them together for the second, so we get:

F_{BP}=(758.98i + 758.98j + 1897k)MN

We continue with the next force. The procedure is the same so we get:

r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}

which yields:

r_{CP}^{2}= 30 m^{2}

Next, we can make use of the force formula:

F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}

which yields:

F_{CP}=4.20 GN

Now we can find its components:

F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i

F_{CPx}=-1.534 GNi

F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j

F_{CPy}=-766.81 MN j

F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k

F_{CPz}=3.83 GN k

And we can now write them together for the third force, so we get:

F_{CP}=(-1.534i - 0.76681j +3.83k)GN

So in order to find the resultant force, we need to add the forces together:

F_{r}=F_{AP}+F_{BP}+F_{CP}

so we get:

F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN

So when adding the problem together we get that:

F_{r}=(-0.583.72i + 0.18347j +6.05k)GN

which is the answer to part a), now let's take a look at part b).

b)

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}

which yields:

F_{r}=6.08 GN

and now we take the formula for the electric field which is:

E=\frac{F_{r}}{q}

so we go ahead and substitute:

E=\frac{6.08GN}{2C}

E=3.04\frac{GN}{C}

7 0
4 years ago
A light beam is traveling through an unknown substance. When it exits that substance and enters into air, the angle of reflectio
bogdanovich [222]

Answer:

0.79

Explanation:

Using Snell's law, we have that:

n(1) * sin θ1 = n(2) * sinθ2

Where n(1) = refractive index of air = 1.0003

θ1 = angle of incidence

n(2) = refractive index of second substance

θ2 = angle of refraction

The angle of reflection through the unknown substance is the same as the angle of incidence of air. This means that θ1 = 32°

=> 1.0003 * sin32 = n(2) * sin42

n(2) = (1.0003 * sin32) / sin42

n(2) = 0.79

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3 years ago
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klasskru [66]

Answer:

There is net loss of gravitational energy .

Explanation:

When Xanaxa is on the ground , her potential energy is assumed to be zero . When she leaps to a height of 153 m , she gains gravitational energy . When she dives and reaches the surface , she loses potential energy and on reaching the ground her potential energy becomes zero . When she further goes down inside ground to a depth of 17.5 m , she loses potential energy further . Her potential energy becomes less than zero or negative .

Ultimately her potential energy changes from zero to negative in the whole process . So there is net loss of potential energy .

8 0
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