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3 years ago

1.A wave has a period of 20 seconds. What is the frequency?

1 answer:

6 0

1). Frequency = (1/period) = 1/(20sec) = (1/20) per sec .

2). Frequency= (3 waves) / (4 sec) = (3/4) (per sec)
Period = (1/freq) = 1/(3/4 per sec) = (4/3) sec

3). Speed= (freq) x (wavelength) (2 per sec) x (5 m) = (2 x 5) ( m per sec)
Speed doesn't depend on amplitude.

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A 15g football was kicked to an acceleration of 55m/s2. What was the force applied to it? 40N 3.7N 825N

Pepsi [2]

Answer:

f =ma = 0.015 * 55 = 0.825 N

so yeah that's ur ans

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3 years ago

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The momentum of an object is proportional to its weight and speed.<br> a. true<br> b. false

Crank

It would be A.. True

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4 years ago

We know that the law of conservation of energy states that energy can not be created or destroyed. It only changes form. Conside

Pavel [41]

The conservation of energy always holds true even when not clearly observable in machines that are less than 100% efficient. More often than not a machine will suffer energy losses (e.g. consider for a cooling fan: friction between the rotating blades, drag resistance in the air the fan is pushing around, resistance in the wire, and heat radiating/conducting away from the circuitry).

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4 years ago

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For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate

faust18 [17]

Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue= $h_1$ =0.200 m

Metal volume flow rate,Q=0.03 $m^3/min$

Q= $\frac{0.03}{60}=5\times 10^{-4}m^3/s$ because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

$d_=0.030 m$

$h_2=0$

$A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}$

$A_1=7.065\times 10^{-4} m^2$

$v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}$

$v_1=0.708 m/s$

Pressure at the top and bottom of the sprue is atmospheric

$h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}$

Substitute the values

$0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}$

$v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264$

$v_2=\sqrt{4.421264}=2.1 m/s$

$Q=A_2v_2$

$5\times 10^{-4}=A_2\times 2.1$

$A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2$

Reynolds number= $\frac{v_2D\rho}{\eta}$

$\eta=0.004 N.s/m^2$

$\rho=2700 kg/m^3$

Substitute the values then we get

Reynolds number= $\frac{2.1\times 0.03\times 2700}{0.004}$

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

8 0

3 years ago

A communications channel has a bandwidth of 4,000 hz and a signal-to-noise ratio (snr of 30 db. what is the maximum possible dat

levacccp [35]

Through Shannon's Theorem, we can calculate the capacity of the communications channel using the value of its bandwidth and signal-to-noise ratio. The capacity, C, can be expressed as

C = B × log₂(1 + S/N)

where B is the bandwidth of the channel and S/N is its signal-to-noise ratio.

Since the given SN ratio is in decibels, we must first express it as a ratio with no units as

SN (in decibels) = 10 × log (S/N)
30 = 10log(S/N)
log(S/N) = 3
S/N = 10³ = 1000

Now that we have S/N, we can solve for its capacity (in bits per second) as

C = 4000 × log₂(1 + 1000)
C = 39868.91 bps

Thus, the maximum capacity of the channel is 39868 bps or 40 kbps.

Answer: 40 kbps

4 0

3 years ago