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weqwewe [10]
3 years ago
10

2. How many grams of water can be heated from 20.0°C to 75°C using 12500.0 Joules?

Physics
1 answer:
bogdanovich [222]3 years ago
3 0

Answer:

Grams of water can be heated from 20.0°C to 75°C using  12500.0 Joules = 54.3 g

Explanation:

Heat required to increase temperature

                 H = mcΔT

m = mass of material

C = specific heat of material

ΔT = Change in temperature.

Here we need to find  how many grams of water can be heated from 20.0°C to 75°C using  12500.0 Joules

That is

            H = 12500 J

            Specific heat of water, C = 4186 J/kg°C

            ΔT = 75 - 20 = 55

Substituting

            12500 = m x 4186 x 55

              m = 0.0543 kg

              m = 54.3 g

Grams of water can be heated from 20.0°C to 75°C using  12500.0 Joules = 54.3 g

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Answer:

a) the oscillation of this field is in phase, when the magnetic field goes in the negative direction of y, the elective field goes in the positive direction of the z axis

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Explanation:

a) the polarization the determined wave oscillates the electric field, which is the z axis

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the oscillation of this field is in phase, when the magnetic field goes in the negative direction of y, the elective field goes in the positive direction of the z axis

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2 years ago
How does the function of the parts of the system contribute to its function as a whole
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2 years ago
Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a –80-nC charge at x = –4 m on
leva [86]

Answer:48 V

Explanation:

Given

Three charged particle with charge

q_1=50\ nC at y=6\ m

q_2=-80\ nC at x=-4\ m

q_3=70\ nC at y=-6\ m

Electric Potential is given by

V=\frac{kQ}{r}

Distance of q_1 from x=8\ m

d_1=\sqrt{6^2+8^2}

d_1=\sqrt{36+64}

d_1=10\ m

similarly d_2=8-(-4)

d_2=12\ m

d_3=\sqrt{(-6)^2+8^2}

d_3=\sqrt{36+64}

d_3=10\ m

Potential at x=8\ m is

V_{net}=\frac{kq_1}{d_1}+\frac{kq_2}{d_2}+\frac{kq_3}{d_3}

V_{net}=k[\frac{q_1}{d_1}+\frac{q_2}{d_2}+\frac{q_3}{d_3}]

V_{net}=9\times 10^9[\frac{50}{10}-\frac{80}{12}+\frac{70}{10}]\times 10^{-9}

V_{net}=9\times 5.33

V_{net}=47.97\approx 48\ V

5 0
3 years ago
Brainliest!
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Answer:

C) 40,000 Joules

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3 years ago
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