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inna [77]
3 years ago
6

Muốn đun sôi 200g nước từ 30 độ cần cung cấp nhiệt lượng bao nhiêu :

Physics
1 answer:
Aleks [24]3 years ago
7 0

Answer:

cần cung cấp 70 độ vì nước sôi ở 100°C

Explanation:

You might be interested in
The smallest unit of charge is − 1.6 × 10 − 19 C, which is the charge in coulombs of a single electron. Robert Millikan was able
vovangra [49]

Answer:

-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C

Explanation:

<u>Charge of an Electron</u>

Since Robert Millikan determined the charge of a single electron is

q_e=-1.6\cdot 10^{-19}\ C

Every possible charged particle must have a charge that is an exact multiple of that elemental charge. For example, if a particle has 5 electrons in excess, thus its charge is 5\times -1.6\cdot 10^{-19}\ C=-8 \cdot 10^{-19}\ C

Let's test the possible charges listed in the question:

-8.0 \times 10 ^{-19 }. We have just found it's a possible charge of a particle

-3.2 \times 10 ^{-19 }. Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets

-1.2 \times 10 ^{-19 } this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge

-5.6 \times 10 ^{-19 },\ -9.4 \times 10 ^{-19 } cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6

Finally, the charge -4.8 \times 10 ^{-19 }\ C is four times the charge of the electron, so it is a possible value for the charge of an oil droplet

Summarizing, the following are the possible values for the charge of an oil droplet:

-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C

5 0
3 years ago
Suppose a constant net force of 345 N is applied to slide a heavy stationary couch across the
ira [324]

Answer:

517.5Ns

Explanation:

F=(MV - MU)/t

where MV - MU is the change in momentum,

therefore, MV - MU = Ft

= 345 X 1.

= 517.5Ns

4 0
3 years ago
A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region
zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with  due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

K_{1} = K_{2} + W_{f}

Where:

K_{1}, K_{2} are the initial and final translational kinetic energies of the tobbogan, measured in joules.

W_{f} - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})

Where:

f - Friction force, measured in newtons.

\Delta s - Distance travelled by the toboggan in the rough region, measured in meters.

m - Mass of the toboggan, measured in kilograms.

v_{1}, v_{2} - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}

If m = 375\,kg, v_{1} = 4.50\,\frac{m}{s}, v_{2} = 1.20\,\frac{m}{s} and \Delta s = 5.40 \,m, then:

f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}

f = 653.125\,N

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%

\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%

\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%

\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%

\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%

If v_{1} = 4.50\,\frac{m}{s} and v_{2} = 1.20\,\frac{m}{s}, then:

\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%

\%K_{loss} = 92.889\,\%

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%

\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%

If v_{1} = 4.50\,\frac{m}{s} and v_{2} = 1.20\,\frac{m}{s}, then:

\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%

\%v_{loss} = 73.333\,\%

The speed of the toboggan is reduced in 73.333 %.

5 0
3 years ago
A train travels 120 km In 2 hours and 30 minutes what’s its average speed
dangina [55]
Formula for distance is d=st
so for speed is s=d/t 
 48 km per hour
8 0
3 years ago
Read 2 more answers
At 9:00 on Saturday morning, two bicyclists heading in opposite directions pass each other on a bicycle path.The bicyclist headi
notka56 [123]
Between 9:00 am and 10:45 am, there have been 1 hour and 45 minutes or 1.75 hours have passed. Let x be the speed of the slower cyclist and x+ 5 be the rate of the second cyclist. The given situation is best represented through the equation below,
                                 x(1.75) + (x + 5)(1.75) = 47.25 km
The value of x from the equation is 11. Thus, the two bicyclists' rates are 11 km/h and 16 km/h. 
3 0
3 years ago
Read 2 more answers
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