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3 years ago

To change from one SI prefix to another, we use?

Physics

2 answers:

Kruka [31]3 years ago

5 0

Explanation :

To change from one SI prefix to another, we use conversion factors. It is used to convert the unit of a quantity without changing its value.

For example,

To convert 1 inch into cm, firstly we have to know the relation between inch and cm.

$1\ inch=2.54\ cm$

Hence, using conversion factor inch is converted into cm.

MrMuchimi3 years ago

4 0

conversion factors is the answer!

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One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large

Eva8 [605]

Answer:

6666.67 Newtons

Explanation:

The formula F=ma (force is equal to mass multiplied by acceleration) can be used to calculate the answer to this question.

In this case:

mass= 0.1mg= 1*10^-7 kg
velocity= 4.00*10^3 m/s
time= 6.00*10^-8 s

Using velocity and time, acceleration can be calculated as:

a= 6.667*10^10 m/s²

Substituting these values into the formula F=ma, the answer is:

F= (1*10^-7)kg * (6.667*10^10) m/s²
F= 6666.67 Newtons of force

5 0

4 years ago

help please what is the heat required in kilocalories to convert 2 kg of ice at 0°c completely into steam at 100°c? a. 80 calori

ivann1987 [24]

<span>Q = mL
</span>
We need to know the latent heat of fusion, L, for 0 degrees.

L = 336 kJ/kg, m = 2kg

where L is the latent heat for vaporization

Q = 2 * 336 = 672 kJ

For conversion between 0 and 100 degree Celsius

Q = mcθ, specific heat capacity = 4.2 kJ/kgK

Q = 2*4.2* (100 - 0) = 840 kJ

For conversion to steam at 100 degrees Celsius

<span>Q = mL , L = latent heat of vaporization = 2256 kj/kg
</span>
Q = 2 * 2256 = 4512 kJ

Total heat = 672 + 840 + 4512 = 6024 kJ

= 6 024 000 J

But 1 calorie = 4.2 J

Therefore 6 024 000 J will be: 6 024 000/4.2 ≈ 1 434 286 Calories

≈ 1 434 kCalories

4 0

3 years ago

Which type of energy increases when an object’s atoms move faster? A.nuclear B.mechanical C.chemical D.thermal

erica [24]

When an object's atoms move faster, its thermal energy increases and the object becomes warmer.

4 0

4 years ago

Read 2 more answers

Air is compressed from 14.7 psia and 60°F to a pressure of 150 psia while being cooled at a rate of 10 Btu/lbm by circulating wa

iren2701 [21]

Answer:

A. 6.36 lbm/s

b. $T_2=341\textdegree F$

Explanation:

a. Given the following information;

#Compressor inlet:

Air pressure, $P_1=14.7psia,T_1=60\textdegree F$

#Compressor outlet:

$Air \ Pressure, P_2=150psia\\\\Air \ volume \ flow \ rate, \dot V_1=5000ft^3/min$

#Cooling rate,

$q_{out}=10Btu/lbm, \dot W=700hp$

# From table A-1E

Gas constant of air $R=0.3704\ psia.ft^3/lbm.R$

Specific enthalpy at $P_1=520R-h_1=124.27Btu/lbm$

Using the mass balance:

$\dot m_{in}-\dot m_{out}=\bigtriangleup \dot m_{sys}=0.0\\\\\dot m_{in}=\dot m_{out}\\\\v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{0.3704\times520}{14.7}\\\\\\v_1=13.1026ft^3/lbm\\\\\dot m=\frac{5000/60}{13.1025}\\\\\dot m=6.36\ lbm/s$

Hence, the mass flow rate of the air is 6.36lbm/s

b The specific enthalpy at the exit is defined as the energy balance on the system:

$\dot m_{in}-\dot m_{out}=\bigtriangleup \dot m_{sys}=0.0\\\\\dot m_{in}=\dot m_{out}\\\\\dot W_{in}+ \dot mh_1=\dot Q_{out}+\dot m_2\\\\h_2=\frac{\dot W_{in}-\dot Q_{out}}{\dot m}+h_1\\\\h_2=\frac{700\times 0.7068-10\times 6.36}{6.36}+124.27\\\\h_2=192.06\ Btu/lbm\\\\#at \ h_2=182.06Btu/lbm, T_2=801R=341\textdegree F$

Hence, the temperature at the compressor exit $T_2=341\textdegree F$

5 0

4 years ago

g Let the orbital radius of a planet be R and let the orbital period of the planet be T. What quantity is constant for all plane

Wittaler [7]

Explanation:

Kepler's third law gives the relationship between the orbital radius and the orbital period of the planet. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

Here,

G is gravitational constant

M is mass of sun

It means that the mass of Sun is constant for all planets orbiting the sun, assuming circular orbits.

7 0

3 years ago