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sleet_krkn [62]

2 years ago

13

The weight of an object on the surface of mass is 1850 Newton and its mass is 500kg.Find the acceleration due to gravity on mars

1 answer:

Fudgin [204]2 years ago

3 0

F = m.a

a = F/m

a = 1850/500

a = 3.7 m/s²

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If you adjusted the frequency to be lower than the resonant frequency, would the voltage have been leading ahead of or lagging b

tatyana61 [14]

Answer:

current will lead the voltage

Explanation:

From the bellow figure we can see the plot between the frequency and impedance

From the figure it is clear that when the frequency is less than the resonance frequency then the impedance decreases as the frequency increases it is possible only when impedance is capacitive in nature.

As the impedance is capacitive in nature so current will lead than the voltage

4 0

4 years ago

Problem 6.056 Air enters a compressor operating at steady state at 15 lbf/in.2, 80°F and exits at 275°F. Stray heat transfer and

vova2212 [387]

To solve this process it is necessary to consider the concepts related to the relations between pressure and temperature in an adiabatic process.

By definition the relationship between pressure and temperature is given by

$(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}$

Here

P = Pressure

T = Temperature

$\gamma =$ The ratio of specific heats. For air normally is 1.4.

Our values are given as,

$P_1 = 15lb/in^2\\T_1= 80\°F = 299.817K\\T_2 =400\°F = 408.15K$

Therefore replacing we have,

$(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}$

$(\frac{P_2}{15})=(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}$

Solving for $P_2,$

$P_2 = 15*(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}$

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5 0

4 years ago

You lift a 10-kg box to a height of 1m. How much work do you do on the box when you lift it from the ground? (g= 9.81 m/s2)

Andreyy89

Answer:

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Explanation:

Solution,

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⇒Distance(d)=1m

Now,

Work done=F×d

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