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Andreas93 [3]
3 years ago
12

In a Joule experiment, a mass of 6.51 kg falls through a height of 66.8 m and rotates a paddle wheel that stirs 0.68 kg of water

. The water is initially at 15 degrees C. By how much does its temperature rise (in degrees C)
Physics
1 answer:
MatroZZZ [7]3 years ago
5 0

Explanation:

As the given data is as follows.

         m = 6.51 kg,   g = 9.8 m/s^{2},     h = 66.8 m

mass of water (M) = 0.68 kg,    Specific heat of water = 4200 J/kg^{o}C

    T_{1} = 15^{o}C

According to the given situation, the decrease in potential energy of mass will be equal to heat energy gained by water.

Therefore,

                mgh = m \times C \times (T_{f} - T_{i})

    6.51 kg \times 9.8 \times 66.8 m = 0.68 kg \times 4200 J/kg^{o}C \times (T_{f} - 15)^{o}C

        4261.7064 = 2856 (T_{f} - 15)^{o}C

              1.492 = T_{f} - 15

              T_{f} = 16.492^{o}C

So,    \Delta T = (16.492 - 15)^{o}C

                       = 1.492^{o}C

Therefore, we can conclude that rise in temperature will be 1.492^{o}C.

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A rigid container holds 0.30g of hydrogen gas.
Strike441 [17]

Answer:

Part A:    \mathbf{Q =94 \ J} to two significant figures

Part B:    \mathbf{Q =160  \ J} to two significant figures

Part C:    \mathbf{Q =220  \ J} to two significant figures

Explanation:

Given that :

mass of the hydrogen = 0.30 g

the molar mass of hydrogen gas molecule = 2 g/mol

we all know that:

number of moles = mass/molar mass

number of moles = 0.30 g /2 g/mol

number of moles = 0.15 mol

For low temperature between the range of 50 K to 100 K, the specific heat at constant volume for a diatomic gas molecule = C_v=\dfrac{3}{2}R

For Part A:

Q = mC_v\Delta T

Q= 0.15 \ mol (\dfrac{3}{2})(8.314 \ J/mole.K )(100-50)K

Q= 0.15 \times (\dfrac{3}{2}) \times (8.314 \ J )\times (50)

Q=93.5325 \ J

\mathbf{Q =94 \ J} to two significant figures

Part B. For hot temperature, C_v=\dfrac{5}{2}R

Q = mC_v\Delta T

Q= 0.15 \ mol (\dfrac{5}{2})(8.314 \ J/mole.K )(300-250)K

Q= 0.15 \times (\dfrac{5}{2}) \times (8.314 \ J )\times (50)

Q=155.8875 \ J

\mathbf{Q =160  \ J} to two significant figures

Part C. For an extremely hot temperature, C_v=\dfrac{7}{2}R

Q = mC_v\Delta T

Q= 0.15 \ mol (\dfrac{7}{2})(8.314 \ J/mole.K )(2300-2250)K

Q= 0.15 \times (\dfrac{7}{2}) \times (8.314 \ J )\times (50)

Q=218.2425 \ J

\mathbf{Q =220  \ J} to two significant figures

6 0
3 years ago
A 1 kg flashlight is dropped from rest and falls to the floor without air resistance . At the point during its fall, when it is
alexgriva [62]

Answer:

The speed of the flashlight at that point is 3.7 m/s

Explanation:

When an object of mass M is at a height H above the ground, the potential energy of the object is:

U = M*H*g

Where g is the gravitational acceleration, g = 9.8 m/s^2

And for an object with velocity v, the kinetic energy is:

K = (M/2)*v^2

We know that when the flashlight of mass  1kg is 0.7 meters above the ground, the potential energy is equal to the kinetic energy, then:

M = 1kg

H = 0.7m

g = 9.8 m/s^2

Replacing these in the equations, we get:

U = K

(1kg)*(0.7m)*(9.8 m/s^2) = ((1kg)/2)*v^2

As the mass factor appears in both sides, we can remove it:

(0.7 m)*(9.8 m/s^2) = (v^2)/2

Now we can multiply both sides by 2:

2*(0.7 m)*(9.8 m/s^2) = v^2

Now let's apply the square root to both sides:

√(2*(0.7 m)*(9.8 m/s^2)) = v = 3.7 m/s

8 0
2 years ago
A daring stunt woman sitting on a tree limb wishes to drop vertically onto a horse gallop ing under the tree. The constant speed
Shkiper50 [21]

Answer:

0.67 seconds

8.576 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow 2.23=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2.23\times 2}{9.8}}\\\Rightarrow t=0.67\ s

Time taken by the stunt woman to drop to the saddle is 0.67 seconds which is the time she will stay in the air.

Speed of the horse = 12.8 m/s

Distance = Speed × Time

⇒Distance = 12.8×0.67

⇒Distance = 8.576 m

Hence, the distance between the horse and stunt woman should be 8.576 m when she jumps.

3 0
3 years ago
The change in pitch of a train's horn as it passes while you are standing still can be explained by:________
Mazyrski [523]
D.the Doppler effect
3 0
3 years ago
How high would a projectile go if it was launched from ground level with an initial speed of 26 m/s at an angle of 30 degrees ab
tigry1 [53]

Answer:

Vy = 26 m/s sin 30 = 13 m/s      vertical speed

t = Vy / a = 13 m/s / 9.80 m/s^2 = 1.33 sec     time to reach Vy = 0

H = Vy t + 1/2 g t^2

H = 13 m/s * 1.33 sec - 1.33^2 * 9.8 / 2 m = 8.62 m

4 0
10 months ago
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