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Andrews [41]
4 years ago
12

Falculate the speed of a wave with frequency of 10KHz and wave length of 2m​

Physics
1 answer:
aleksklad [387]4 years ago
8 0

Answer:20000m/s

Explanation:

Frequency=10khz

Frequency=10x1000

Frequency=10000hz

Wavelength=2m

Speed= frequency x wavelength

Speed=10000 x 2

Speed=20000

speed=20000m/s

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A small block of mass M = 0.10 kg is released from rest at point 1 at a height H = 1.8 m above the bottom of a track, as shown i
Ede4ka [16]

Answer:

D

Explanation:

A) is not correct, because the gravitation potential energy will depend on the height the block is located at. It will be calculated with the formula:

U=mgh.

If we take the ground as a zero height reference, then on point 2 the potential energy will be:

U_{2} = 0.10kg(9.81 m/s^{2})(0.6m)

U_{2}=0.59 J

While on point 3, the potential energy will be greater.

U_{3}=0.10kg(9.81 m/s^{2})(1.2m)

U_{3}=1.18 J

B) is not the right answer because the kinetic energy will vary with the height the block is located at in the fact that the energy is conserved (this is if we don't take friction into account or air resistance) so in this case:

U_{2}+K_{2}= U_{3}+K_{3}

We already know the potential energy at point 2. We can calculate the kinetic energy at point 3 like this:

K_{3} =\frac{1}{2}mv_{3}^{2}

K_{3} =\frac{1}{2}(0.10kg)(2.5 m/s)^{2}

K_{3} =0.31 J

So the kinetic energy at point 2 is given by the equation:

K_{2}  =U_{3}-U_{2}+K_{3}

so:

K_{2} = (1.18J)-(0.59J)+0.31J

K_{2} =0.9J

As you may see the kinetic energy at point 2 is greater than the kinetic energy at point 3.

C) Is not correct because according to the first law of thermodinamics, energy is not lost, only transformed. So, since we are not taking into account friction or any other kind of loss, then we can say that the amount of mechanical energy at point 1 is exactly the same as the mechanical energy at point 3.

D) Because of what we talked about on part C, this will be the true situation, because the mechanical energy of the block will be the same no matter theh point you measure it at.

7 0
3 years ago
Using carson's rule, what is the required bw to transmit an fm signal if the maximum deviation is 60 khz and intelligence freque
VashaNatasha [74]

Carson's Rule says:

FM occupied bandwidth =

(2) · (Peak deviation + Highest modulating frequency)

so

FM bw = (2) · (60 kHz + 15 kHz)

FM bw = (2) · (75 kHz)

FM bw = 150 kHz

(I used to eat this stuff for lunch, but it's been almost 40 years.  Thanks for taking me back.  Those were the good old days.)

5 0
3 years ago
Consider an insulated tank with a volume V = 2 L is separated into two equal-volume parts by a thin wall. On the left is an idea
steposvetlana [31]

Answer

given,

V = 2 L

the left is an ideal gas at  P = 100 k Pa and T = 500 K

mass is constant

 m_1 = m_2

\dfrac{P_1V_1}{RT_1} = \dfrac{P_2V_2}{RT_2}

Pressure is same because it's not changing due to process

\dfrac{V}{500} = \dfrac{2 V}{T_2}

T_2 = 1000\ K

\Delta S_{univ} = \Delta S_{sys} + (\Delta S)_{surr}

\Delta S_{univ} =m(C_v ln (\dfrac{T_2}{T_1}))+ R ln (\dfrac{V_2}{V_1})

m = \dfrac{P_1V_1}{RT_1}

m = \dfrac{100 \times 10^3 \times 2 \times 10^{-3}}{287\times 500}

m = 1.39 x 10⁻³ Kg

\Delta S_{univ} =1.39\times 10^{-3}(0.718 ln\ 2+ 0.287 ln (2)

\Delta S_{univ} =0.968\times 10^{-3}\ kJ/K

5 0
3 years ago
Which items or activities should a pregnant woman avoid to maintain her health and the health of her fetus? Check all that apply
Fynjy0 [20]

Answer:

Alcohol and drugs are a no no. you should put the answers in so people can answer correctly.

6 0
3 years ago
A 2000 kg car moving at 100 km/h crosses the top of a hill with a radius of curvature of 100 m. What is the normal force exerted
Tpy6a [65]

Answer:

The normal force the seat exerted on the driver is 125 N.

Explanation:

Given;

mass of the car, m = 2000 kg

speed of the car, u = 100 km/h = 27.78 m/s

radius of curvature of the hill, r = 100 m

mass of the driver, = 60 kg

The centripetal force of the driver at top of the hill is given as;

F_c = F_g - F_N

where;

Fc is the centripetal force

F_g is downward force due to weight of the driver

F_N is upward or normal force on the drive

F_N = F_g-F_c\\\\F_N = mg - \frac{mv^2}{r} \\\\F_N = (60 \times 9.8) -\frac{60 \ \times \ 27.78^2 \ }{100} \\\\F_N = 588 \ N - 463 \ N\\\\F_N = 125 \ N

Therefore, the normal force the seat exerted on the driver is 125 N.

6 0
3 years ago
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