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4 years ago

A wave traveling in water has a frequency of 250 Hz and a wavelength of 6.0m. What is the speed of the wave?

1 answer:

vladimir2022 [97]4 years ago

6 0

Since you are looking for the speed, you need to rearrange the formula which is f = speed / wavelength. That should give you speed = f (wavelength.) All you need to do next is to substitute the value to the following equation. speed = 250 Hz (6.0m) that should leave you with 1500 m/s which is very fast.

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3. What exerts a greater force on the table of 2 kg book lying flat or a 2 kg book on its

Darina [25.2K]

Answer:

A book on its side exerts a greater force.

Explanation:

Pressure = Force / Area

Assuming that 1kg = 10N

2kg = 20N

Area of book lying flat = 0.3m × 0.2m

= 0.6m²

Pressure of book lying flat = 20N / 0.6m²

= 30Pa (1 s.f.)

Area of book on its side = 0.2m × 0.05m

= 0.01m²

Pressure of book on its side = 20N / 0.01m²

= 2000Pa (1 s.f.)

Since 2000Pa (1 s.f.) > 30Pa (1 s.f.), a book on its side applies greater pressure than lying flat.

5 0

2 years ago

A car horn emits a frequency of 400 Hz. A car traveling at 20.0 m/s sounds the horn as it approaches a stationary pedestrian. Wh

Temka [501]

Answer:

The observed frequency by the pedestrian is 424 Hz.

Explanation:

Given;

frequency of the source, Fs = 400 Hz

speed of the car as it approaches the stationary observer, Vs = 20 m/s

Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.

The observed frequency is calculated as;

$F_s = F_o [\frac{v}{v_s + v} ] \\\\$

where;

F₀ is the observed frequency

v is the speed of sound in air = 340 m/s

$F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\$

F₀ ≅ 424 Hz.

Therefore, the observed frequency by the pedestrian is 424 Hz.

8 0

3 years ago

: The maximum theoretical efficiency of a Carnot engine operating between reservoirs at the steam point and at room temperature

Hunter-Best [27]

Answer:

The value is $\eta = 0.2145$ or 21.45%

Explanation:

From the question we are told that

The first reservoir is at steam point $T_s = 100^o C = 100 + 273 = 373 \ K$

The second reservoir is at room temperature $T_r = 20^o C = 293 \ K$

Generally the maximum theoretical efficiency of a Carnot engine is mathematically evaluated as

$\eta = 1- \frac{T_r}{T_s}$

=> $1 - \frac{ 293}{373}$

=> $\eta = 0.2145$

5 0

3 years ago

What is the volume V of a sample of 4.00 mol of copper? The atomic mass of copper (Cu) is 63.5 g/mol, and the density of copper

rusak2 [61]

Answer : The volume of a sample of 4.00 mol of copper is $28.5cm^3$

Explanation :

First we have to calculate the mass of copper.

$\text{ Mass of copper}=\text{ Moles of copper}\times \text{ Molar mass of copper}$

$\text{ Mass of copper}=(4.00moles)\times (63.5g/mole)=254g$

Now we have to calculate the volume of copper.

Formula used :

$Density=\frac{Mass}{Volume}$

Now put all the given values in this formula, we get:

$8.92\times 10^3kg/m^3=\frac{254g}{Volume}$

$Volume=\frac{254g}{8.92\times 10^3kg/m^3}=2.85\times 10^{-2}L=2.85\times 10^{-2}\times 10^3cm^3=28.5cm^3$

Conversion used :

$1kg/m^3=1g/L\\\\1L=10^3cm^3$

Therefore, the volume of a sample of 4.00 mol of copper is $28.5cm^3$

7 0

3 years ago

Read 2 more answers

It is vital to keep in mind that breaches are entirely concerned with data. No matter what physical damage a device incurs, data

Ierofanga [76]

Answer:

TRUE

Explanation:

We currently live in the digital age, where almost everything is digitized, including strategic information. Thus, each individual and especially corporations that have sensitive data must use protection mechanisms against cyber attacks. One of the measures most recommended by experts is encryption, which consists of a set of rules that aims to encode data information so that only the sender and the receiver can decipher it.

5 0

3 years ago