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Alinara [238K]
2 years ago
13

Convert 700kg into grams​

Physics
2 answers:
Hitman42 [59]2 years ago
8 0

Answer:

700 kilograms is equal to 700,000 grams

Explanation:

One kilogram has 1000 grams, or if you want the other way around, one gram is 1/1000th of one kilogram. If you follow this logic, we want to use the 'one kilogram has 1000 grams' "rule", so we would do some math here.

700 kg x 1000 g = ______ g

700*1000 = 700,000

700 kg = 700,000 g

Please let me know if I misunderstood and I hope this helped!

Verdich [7]2 years ago
5 0

Answer:

700,000g

Explanation:

1kg = 1000 Grams (or any kg x 1000)

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At the moment a hot cake is put in a cooler, the difference between the cakes and the cooler's
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Answer: D(t)= 50(4/5)^t

Explanation: If 1/5 of the temperature difference is lost each minute, that means 4/5 of the difference remains each minute. So each minute, the temperature difference is multiplied by a factor of 4/5 (or 0.8).

If we start with the initial temperature difference, 50° Celsius, and keep multiplying by 4/5, this function gives us the temperature difference t minutes after the cake was put in the cooler.

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Which simple machine is a doorknob?
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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 %.

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yaroslaw [1]

The AREA of the shaded region is the moving object's displacement.

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TRANSFORMA: 765 mm Hg a atm
-Dominant- [34]

Presión

765

=

ATMÓSFERA

1,00658

Answer:

Explanation:

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