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3 years ago

9

All EM waves (light waves) travel at the same speed through the vacuum of space. If the different types of waves are distinguish

ed by their frequency, what basic characteristic of the waves determines their difference

1 answer:

Rudiy273 years ago

6 0

I don’t know the answer i just need the points

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72) What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 (formula weight = 164 g/mol) in 115

Rina8888 [55]

<u>Answer:</u> The freezing point of solution is -3.34°C

<u>Explanation:</u>

Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of calcium nitrate follows:

$Ca(NO_3)_2(aq.)\rightarrow Ca^{2+}(aq.)+2NO_3^-(aq.)$

The total number of ions present in the solution are 3.

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(Ca(NO_3)_2)$ = 11.3 g

$M_{solute}$ = Molar mass of solute $(Ca(NO_3)_2)$ = 164 g/mol

$W_{solvent}$ = Mass of solvent (water) = 115 g

Putting values in above equation, we get:

$\text{Molality of }Ca(NO_3)_2=\frac{11.3\times 1000}{164\times 115}\\\\\text{Molality of }Ca(NO_3)_2=0.599m$

To calculate the depression in freezing point, we use the equation:

$\Delta T=iK_fm$

where,

i = Vant hoff factor = 3

$K_f$ = molal freezing point depression constant = 1.86°C/m.g

m = molality of solution = 0.599 m

Putting values in above equation, we get:

$\Delta T=3\times 1.86^oC/m.g\times 0.599m\\\\\Delta T=3.34^oC$

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T=\text{freezing point of water}-\text{freezing point of solution}$

$\Delta T$ = 3.34 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

$3.34^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-3.34^oC$

Hence, the freezing point of solution is -3.34°C

4 0

4 years ago

If the specific surface energy for magnesium oxide is 1.0 J/m2 and its modulus of elasticity is (225 GPa), compute the critical

alexdok [17]

Answer:

The critical stress required for the propagation of an initial crack $\sigma_{c}$ = 21.84 M pa

Explanation:

Given data

Modulus of elasticity E = 225 × $10^{9}$ $\frac{N}{m^{2} }$

Specific surface energy for magnesium oxide is $\gamma_{s}$ = 1 $\frac{J}{m^{2} }$

Crack length (a) = 0.3 mm = 0.0003 m

Critical stress is given by $\sigma_{c}^{2} }$ = $\frac{2 E \gamma}{\pi a}$ -------- (1)

⇒ 2 E $\gamma_{s}$ = 2 × 225 × $10^{9}$ × 1 = 450 × $10^{9}$

⇒ $\pi$ a = 3.14 × 0.0003 = 0.000942

⇒ Put these values in equation 1 we get

⇒ $\sigma_{c}^{2} }$ = $\frac{450 }{0.000942} 10^{9}$

⇒ $\sigma_{c}^{2} }$ = 4.77 × $10^{14}$

⇒ $\sigma_{c}$ = 2.184 × $10^{7}$ $\frac{N}{m^{2} }$

⇒ $\sigma_{c}$ = 21.84 $\frac{N}{mm^{2} }$

⇒ $\sigma_{c}$ = 21.84 M pa

This is the critical stress required for the propagation of an initial crack.

4 0

3 years ago

Hamid is going to heat sodium chloride and water in a beaker. How can he carry out his experiment as

wolverine [178]

Answer:

he can wear gogles lab coat and gloves

Explanation:

4 0

3 years ago

1.A Radio station broadcasts modern song on medium wave 350 Hz every day at ten o’clock in the morning. The velocity of radio wa

love history [14]

Answer:

$ans \: = \boxed{{4.8 \times 10}^{ - 4} Hz}$

Explanation:

$given \to \\ f_{r} = 350 \: \\ v_{r} = {3 \times 10}^{8} \\ but \to \\ v = f \gamma \to \: \gamma = \frac{v}{f} : hence \to \\ \gamma _{r} = \frac{v_{r}}{f_{r}} = \frac{3 \times 10^{8} }{350} = \boxed{857,142.85714 \: m}\\ therefore \to \\ given \to \\ f_{w} = water \: frequency = \: \boxed{ ?}\: \\ v_{w} = 14 50 \\ but \to \\ v = f \gamma \to \: \gamma = \frac{v}{f} : hence \to \\ \gamma _{w} = \frac{v_{w}}{f_{w}} = \frac{1}{100} \times \gamma _{r} = \frac{1}{100} \times 857,142.85714 \\\gamma _{w} = \boxed{8,571.4285714 \: m} : hence \to \: \\ f_{w} = \frac{v_{w}}{ \gamma _{w}} = \frac{1450}{8,571.4285714} = \boxed{0.1691666667} \\ if \: the \: number \: of \: times = \boxed{ x} \\ f_{r} (x)=f_{w} \\ (x) = \frac{f_{w}}{f_{r}} = \frac{0.1691666667}{350} = 0.0004833333 \\ hence \to \\ the \: frequency \: of \: the \: radio \: wave \: is \to \: \boxed{{4.8 \times 10}^{ - 4} }\: \\ that \: of \: the \: wave \: created \: in \: the \: water.$

♨Rage♨

8 0

3 years ago

Light bulbs are typically rated by their power dissipation when operated at a given voltage.

stira [4]

(b) At 24 W and 12 V, the bulbs can operate with the largest current of 2 A.

Answer:

Option B.

Explanation:

Power consumed in light bulbs are measured in terms of voltage and current. As it is known that power dissipated in an electric circuit is directly proportional to the square of current and also to resistance.

Power = I²R

As by Ohm's law, V = IR, then Power = VI

Here, V is the voltage and I is the current flowing through the bulb.

So here the power and voltage is given and we have to determine the largest current. So from the formula it can stated that current flowing through the bulb will be directly proportional to the power dissipated by the circuit and inversely proportional to the voltage at which the circuit is operated.

So for the given options, the current value is as follows:

(a) I = P/V = 3.2/6=0.5333 A

(b) I = P/V = 24/12 = 2 A

(c) I= P/V = 16/18 =0.9 A

(d) I = P/V = 32/24 = 1.3 A

Thus, the option B is having the maximum voltage at which the bulb can be operated. So at 24 W and 12 V, the bulbs can operate with the largest current of 2 A.

7 0

3 years ago