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Virty [35]
3 years ago
5

If the mass of a 1.8 g paperclip was able to be completely converted to energy, how much energy would you obtain?

Physics
1 answer:
Anton [14]3 years ago
7 0

Answer:

E=1.62\times 10^{14}\ J

Explanation:

Given that,

The mass of the paperclip, m = 1.8 g = 0.0018 kg

We need to find the energy obtained. The relation between mass and energy is given by :

E=mc^2

Where

c is the speed of light

So,

E=0.0018\times (3\times 10^8)^2\\\\E=1.62\times 10^{14}\ J

So, the energy obtained is 1.62\times 10^{14}\ J.

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The heating element in a kettle behaves like a resistor. A particular kettle needs to operate at 230 V, with a power of 1500 W.
dedylja [7]

Answer:

R = 35.27 Ohms

Explanation:

Given the following data;

Voltage = 230V

Power = 1500W

To find the resistance, R;

Power = V²/R

Where:

V is the voltage measured in volts.

R is the resistance measured in ohms.

Substituting into the equation, we have;

1500 = 230²/R

Cross-multiplying, we have;

1500R = 52900

R = 52900/1500

R = 35.27 Ohms.

Therefore, the resistance which the heating element needs to have​ is 35.27 Ohms.

4 0
3 years ago
Glycerin (C3H8O3) is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 154 g of glycerin to 316 mL o
garri49 [273]

Answer:

P_{sol}=50.4\ mm.Hg

Explanation:

According to given:

  • molecular mass of glycerin, M_g=3\times 12+8+3\times 16=92\ g.mol^{-1}
  • molecular mass of water, M_w=2+16=18\ g.mol^{-1}
  • ∵Density of water is 0.992\ g.cm^{-3}= 0.992\ g.mL^{-1}
  • ∴mass of water in 316 mL, m_w=316\times 0.992=313.5 g
  • mass of glycerin, m_g=154\ g
  • pressure of mixture, P_x=55.32\ torr= 55.32\ mm.Hg
  • temperature of mixture, T_x=40^{\circ}C

<em>Upon the formation of solution the vapour pressure will be reduced since we have one component of solution as non-volatile.</em>

<u>moles of water in the given quantity:</u>

n_w=\frac{m_w}{M_w}

n_w=\frac{313.5}{18}

n_w=17.42 moles

<u>moles of glycerin in the given quantity:</u>

n_g=\frac{m_g}{M_g}

n_g=\frac{154}{92}

n_g=1.674 moles

<u>Now the mole fraction of water:</u>

X_w=\frac{n_w}{n_w+n_g}

X_w=\frac{17.42}{17.42+1.674}

X_w=0.912

<em>Since glycerin is non-volatile in nature so the vapor pressure of the resulting solution will be due to water only.</em>

\therefore P_{sol}=X_w\times P_x

\therefore P_{sol}=0.912\times 55.32

P_{sol}=50.4\ mm.Hg

7 0
3 years ago
Which of the following shows the conservation of mass during cellular respiration?
otez555 [7]
C, hope this helps!!!!!!!                                                                                  

7 0
4 years ago
Read 2 more answers
Which items do NOT describe possible solutions to air pollution?
bonufazy [111]

Answer:

2, 6

Explanation:

2 because if you cut down more trees you will have less items to help collect co2

6 because if you have more manufracturing more gasses will be release and moe carbon dioxcide in the air so it will slowly kill the ozone layer.

6 0
3 years ago
A spherical capacitor contains a charge of 3.40 nC when connected to a potential difference of 240.0 V. Its plates are separated
I am Lyosha [343]

Answer:

A) 1.4167 × 10^(-11) F

B) r_a = 0.031 m

C) E = 3.181 × 10⁴ N/C

Explanation:

We are given;

Charge;Q = 3.40 nC = 3.4 × 10^(-9) C

Potential difference;V = 240 V

Inner radius of outer sphere;r_b = 4.1 cm = 0.041 m

A) The formula for capacitance is given by;

C = Q/V

C = (3.4 × 10^(-9))/240

C = 1.4167 × 10^(-11) F

B) To find the radius of the inner sphere,we will make use of the formula for capacitance of spherical coordinates.

C = (4πε_o)/(1/r_a - 1/r_b)

Rearranging, we have;

(1/r_a - 1/r_b) = (4πε_o)/C

ε_o is a constant with a value of 8.85 × 10^(−12) C²/N.m

Plugging in the relevant values, we have;

(1/r_a - 1/0.041) = (4π × 8.85 × 10^(−12) )/(1.4167 × 10^(-11))

(1/r_a) - 24.3902 = 7.8501

1/r_a = 7.8501 + 24.3902

1/r_a = 32.2403

r_a = 1/32.2403

r_a = 0.031 m

C) Formula for Electric field just outside the surface of the inner sphere is given by;

E = kQ/r_a²

Where k is a constant value of 8.99 × 10^(9) Nm²/C²

Thus;

E = (8.99 × 10^(9) × 3.4 × 10^(-9))/0.031²

E = 3.181 × 10⁴ N/C

3 0
3 years ago
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