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2 years ago

ANSWER ASAP!!!If a wave machine produces 10 waves in 2 seconds, what is the frequency of the machine in Hertz?

Physics

1 answer:

Rufina [12.5K]2 years ago

4 0

Answer:

5Hz

Explanation:

Frequency = No. of waves per second

10 waves are produced in 2 seconds

in one second 10/2 waves are produced

(ie) 5 waves are produced in 1 second

Therefore the frequency of wave is 5 Hz

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A scientist directs monochromatic light toward a single slit in an opaque barrier. The light has a wavelength of 580 nm and the

vredina [299]

Answer:

a) 9.72 mm

b) 4.86 mm

Explanation:

wave length of light λ is 580 nm = 580 \times 10⁻⁹ m

Width of slit d = 0.215\times 10⁻³ m

Distance of screen D = 1.8 m.

Width of one fringe = $\frac{\lambda\times D}{d}$

Putting the values we get fringe width

= $\frac{580\times10^{-9}\times1.8}{.000315}$

=4.86 mm.

a) Width of central maxima = 2 times width of one fringe

= 2 times 4.86

=9.72 mm

b) width of each fringe except central fringe is same , no matter what the order is.Only brightness changes .

So width of either of the two first order bright fringe will be same and it will be

= 4.86 mm.

3 0

2 years ago

A transformer has a secondary voltage of 140 volts and a secondary current of 3.5 amps. if the primary current is 10 amps, what

Lynna [10]

For an ideal transformer power loss is assumed to be zero

i.e. the power in primary coil due to input voltage must be equal to power in secondary coil due to output voltage

this can be written in form of equation

$V_1 i_1 = V_2 i_2$

here we know that

$V_2 = 140 volts$

$i_2 = 3.5 A$

$i_1 = 10 A{/tex]now we will use above equation[tex]140*3.5 = 10 * V_1$

$V_1 = 49 volts$

So primary coil voltage is 49 Volts

7 0

2 years ago

An electric furnace is to melt 40 kg of aluminium/hour. The initial temperature of aluminium is 32°C. Given that aluminium has s

gizmo_the_mogwai [7]

Answer:

Part a)

P = 13.93 kW

Part b)

R = 8357.6 Cents

Explanation:

Part A)

heat required to melt the aluminium is given by

$Q = ms\Delta T + mL$

here we have

$Q = 40(950)(680 - 32) + 40(450 \times 10^3)$

$Q = 24624 kJ + 18000 kJ$

$Q = 42624 kJ$

Since this is the amount of aluminium per hour

so power required to melt is given by

$P = \frac{Q}{t}$

$P = \frac{42624}{3600} kW$

$P = 11.84 kW$

Since the efficiency is 85% so actual power required will be

$P = \frac{11.84}{0.85} = 13.93 kW$

Part B)

Total energy consumed by the furnace for 30 hours

$Energy = power \times time$

$Energy = 13.93 kW\times 30 h$

$Energy = 417.9 kWh$

now the total cost of energy consumption is given as

$R = P \times 20 \frac{Cents}{kWh}$

$R = 417.9 kWh\times 20 \frac{cents}{kWh}$

$R = 8357.6 Cents$

3 0

3 years ago

Fill in the blank: An object is most likely to sink in water if _________________.

drek231 [11]

High density
random words to fill up 20 character minimum for answering questions :P

4 0

3 years ago

Read 2 more answers

……………………………………………………………..?

kotegsom [21]

Answer:

Explanation:

First find the electrical wattage

W = I^2 * R

R = 12 ohms

I = 2 amps

Wattage = 2^2 * 12

Wattage = 4* 12

Wattage = 48 watts.

Now you need to use the power formula

Work = Power * Time

Work = ?

Power = 48 watts

Time = 3 minutes = 3 * 60 = 180 seconds.

Work = 48 * 180

Work = 8640 J

That's C

3 0

2 years ago