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zepelin [54]
3 years ago
12

Umar has two copper pans, each containing 500cm3 of water. Pan A has a mass of 750g and pan B has a mass of 1.5kg. Which pan wil

l require the least energy to heat the water to 100°C?
Physics
1 answer:
Olin [163]3 years ago
3 0

Answer:

heat required in pan B is more than pan A

Explanation:

Heat required to raise the temperature of the substance is given by the formula

Q = ms\Delta T

now we know that both pan contains same volume of water while the mass of pan is different

So here heat required to raise the temperature of water in Pan A is given as

Q_1 = (m_w s_w + m_ps_p)\delta T

Q_1 = (0.5(4186) + 0.750(s))\Delta T

Now similarly for other pan we have

Q_2 = (m_w s_w + m_ps_p)\delta T

Q_2 = (0.5(4186) + 1.50(s))\Delta T

So here by comparing the two equations we can say that heat required in pan B is more than pan A

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Please someone help me !!
sergij07 [2.7K]

Answer:

D

Explanation:

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2 years ago
At what height h above the ground does the projectile have a speed of 0.5v?
dedylja [7]

Answer:height above ground at which projectile have velocity

0.5v is (0.0375v^2)

Explanation:

Using Vf = Vi - gt

Where Vf is final velocity

Vi is initial velocity

g is the acceleration due to gravity

t is the time taken

So, 0.5v = v - gt

t = 0.05v

Therefore height h = vt - 0.5gt^2

Subtitute t

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3 0
2 years ago
QUESTION 1
lisabon 2012 [21]

Answer:

2 * 10^5 pa

Explanation:

Pressure = Force / Area

Each thigh bone has a cross sectional area of 10cm²

Both thigh bones :

2 * 10cm² = 20cm²

To m² : 20 * (0.01)²

20 * 0.0001 m² = 0.002 m²

Force = mass * acceleration due to gravity(g)

g = 10m/s² ;

Force = 40 * 10 = 400N

Pressure = 400 N / 0.002 m²

Pressure = 200,000 N/m² = 2 * 10^5 pascal

8 0
2 years ago
g calculate the effectiveness radiation dosage in sieverts for a 79 kg person who is exposed 6.8x10^9
Elza [17]

Answer:

The answer is "\bold{dosage = 0.031 rem}"

Explanation:

please find the complete question in the attached file.

Given value:

m = 79\  kg  \\\\n = 3.4 \times  10^9 \\\\E = 5.5  \times  10^{-13} \\\\ RBE = 15

\to E = n E\\

        = 3.4  \times  10^9  \times  5.5  \times  10^{-13} \\\\      = 1.87 \times  10^{-3}

\to E(absorbed) = 1.87  \times 10^{-3}  \times  0.87 = 1.63  \times  10^{-3}

calculating the radiation absorbed per kg:

= \frac{1.63  \times  10^{-3}}{79}  \\\\ = 2.06  \times  10^{-5} \\\\ = 0.00206 \  rad

\to Dosage = 0.00206  \times  15 \\

                 = 0.031 \ \ rem

4 0
2 years ago
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