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3 years ago

Why is the term minority group no longer effective

1 answer:

8 0

The term minority group is no longer effective because these groups now make up significant percentages of the total population

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A mass of 80 g of KNO3 is dissolved in 100 g of water at 50 ºC. The solution is heated to 70ºC. How many more grams of potassium

slava [35]

Hi um i think its At 30 ° C your welcome

7 0

3 years ago

If the output current for the transformer is 1.8 A, find the input current.

Iteru [2.4K]

The answer is 3.16 B

3 0

3 years ago

Read 2 more answers

If the velocity of blood flow in the aorta is normally about 0.32 m/s, what beat frequency would you expect if 4.40-MHz ultrasou

dusya [7]

Answer:

The beat frequency is 0.0019 MHz.

Explanation:

Given that,

Velocity = 0.32 m/s

Frequency = 4.40 MHz

Speed of wave = 1540 m/s

We need to calculate the frequency

Case (I),

Observer is moving away from the source

Using Doppler's effect

$f'=\dfrac{v-v'}{v}f$

Where, v' = speed of observer

Put the value into the formula

$f'=\dfrac{1540-0.32}{1540}\times4.40$

$f'=4.399\ MHz$

Case (II),

Cell is as the source of sound of frequency f' and it moving away from the observer.

Using formula of frequency

$f''=\dfrac{v-v_{s}}{v+v_{s}}\times f$

$f''=\dfrac{1540-0.32}{1540+0.32}\times4.399$

$f''=4.3971\ MHz$

We need to calculate the beat frequency

$\Delta f= f'-f''$

$\Delta f=4.399-4.3971=0.0019\ MHz$

Hence, The beat frequency is 0.0019 MHz.

4 0

3 years ago

What properties of Pluto might make scientists think that it is a Kuiper Belt Object?

Nataliya [291]

It is the smallest planet with a bluish color that can probably make it look like a comet

6 0

3 years ago

Suppose an oven's radiation wavelength is 0.125 m. a container with 350.00 g of water was placed in the oven, and the temperatur

ivann1987 [24]

The heat (energy) needed to raise the temperature of the water is given by
$Q=m C_S (T_f - T_i)=(350.0 g)(4.18 J/gC)(80C-20C)=87780 J$

The wavelength of the radiation of the oven is $\lambda=0.125 m$ , so the energy of a single photon of this radiation is
$E=h \frac{c}{\lambda}=(6.6 \cdot 10^{-34}J) \frac{3\cdot 10^8 m/s}{0.125 m}=1.6 \cdot 10^{-24} J$

So, the number of photons required to heat the water is the total energy absorbed by the water divided by the energy of a single photon:
$N= \frac{Q}{E}= \frac{87780 J}{1.6\cdot 10^{-24}J}= 5.5 \cdot 10^{28}$ photons

8 0

3 years ago