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zavuch27 [327]
2 years ago
15

When a 440-Hz tuning fork and a piano key are struck together, five beats are heard. If the pitch of the note on the piano is lo

wer than the tuning fork, what is the frequency of the note?
Physics
2 answers:
vovangra [49]2 years ago
7 0
The frequency would also be lower
guapka [62]2 years ago
6 0
440Hz < n2
n2 - 440 = 5
n2 = 445hz
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A ball of mass 0.4 kg is initially at rest on the ground. It is kicked and leaves the kicker's foot with a speed of 5.0 m/s in a
yawa3891 [41]

Answer:

the answer the correct  is 3

Explanation:

Let's use the relationship between momentum and momentum

         I = Δp

         I = m v_{f} - m v₀

     

Let's calculate

         I = 0.4 5.0 - 0

         I = 2.0 N s

By Newton's law of action and reaction the force on the ball is equal to the force that the ball exerts on the foot, therefore the impulse on the foot of equal magnitude, but in the opposite direction

        I = 2.0 Ns with 60°

When reviewing the answer the correct  is 3

4 0
2 years ago
89. An electron is moving in a straight line with a velocity of 4.0×105 m/s. It enters a region 5.0 cm long where it undergoes a
ddd [48]

Explanation:

Given that,

Initial speed of the electron, u=4\times 10^5\ m/s

Distance, s = 5 cm = 0.05 cm

Acceleration of the electron, a=6\times 10^{12}\ m/s^2  

(a) Let v is the electron's velocity when it emerges from this region. It can be calculated as :

v^2=u^2+2as

v^2=(4\times 10^5)^2+2\times 6\times 10^{12}\times 0.05

v = 871779.788 m/s

or

v=8.71\times 10^5\ m/s

(b) Let t is the time for which the electron take to cross the region. It can be calculated as:

t=\dfrac{v-u}{a}

t=\dfrac{8.71\times 10^5-4\times 10^5}{6\times 10^{12}}

t=7.85\times 10^{-8}\ s

Hence, this is the required solution.

4 0
2 years ago
1. Two identical bowling balls of mass M and radius R roll side by side at speed v0 along a flat surface. Ball 1 encounters a ra
UNO [17]

Answer:

1/2

Explanation:

We need to make a couple of considerations but basically the problem is solved through the conservation of energy.  

I attached a diagram for the two surfaces and begin to make the necessary considerations.

Rough Surface,

We know that force is equal to,

F_r = mgsin\theta

F_r = \mu N

F_r = \mu mg cos\theta

Matching the two equation we have,

\mu N = \mu mg cos\theta

\mu = tan\theta

Applying energy conservation,

\frac{1}{2}mv^2_0+\frac{1}{2}I_w^2 = F_r*d+mgh_1

\frac{1}{2}mv^2_0+\frac{2}{5}mR^2\frac{V_0^2}{R^2} = F_r*d+mgh_1

\frac{1}{2}mv^2_0+\frac{mv_0^2}{5} = mgsin\theta \frac{h_1sin\theta}+mgh_1

\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_1+gh_1

h_1 = \frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})

Frictionless surface

\frac{1}{2}mv_0^2+\frac{1}{2}I\omega^2 = mgh_2

\frac{1}{2}m_v^2+\frac{1}{2}\frac{2}{5}mR^2\frac{v_0^2}{R^2} =mgh_2

\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_2

h_2 = \frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})

Given the description we apply energy conservation taking into account the inertia of a sphere. Then the relation between h_1 and h_2 is given by

\frac{h_1}{h_2} = \frac{\frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})}{\frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})}

\frac{h_1}{h_2} = \frac{1}{2}

8 0
2 years ago
The average mass of an automobile in the United States is about 1.440x10^6 g express this mass in kilograms
-BARSIC- [3]
From the problem statement, this is a conversion problem. We are asked to convert from units of grams to units of kilograms. To do this, we need a conversion factor which would relate the different units involved. We either multiply or divide this certain value to the original measurement depending on what is asked. From literature, we will find that 1000 grams is equal to 1 kilogram. We use this as follows:

<span> 1.440x10^6 g ( 1 kg / 1000 g ) = 1440 kg</span><span>
</span>
8 0
2 years ago
Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to owat a constant rate of 6L/min.
lbvjy [14]

Answer:

T = 693.147 minutes

Explanation:

The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.

Therefore, the total salt in the tank at time t = 1000x kg

Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L

Hence, the amount of salt that is added to the tank per minute = (6\times0.1)kg/min=0.6kg/min

Also, there is a continuous outflow from the tank at a rate of 6 L/min.

Hence, amount of salt subtracted from the tank per minute = 6x kg/min

Now, the rate of change of salt concentration in the tank = \frac{dx}{dt}

So, the rate of change of salt in the tank can be given by the following equation,

1000\frac{dx}{dt} =0.6-6x

or, \int\limits^{0.05}_0 {\frac{1000}{0.6-6x} } \, dx =\int\limits^T_0 {} \, dt

or, T = 693.147 min      (time taken for the tank to reach a salt concentration

of 0.05 kg/L)

3 0
2 years ago
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