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3 years ago

An engine develops 500 Nm of torque at 3900 rev/min. The power developed by the engine is ...

Physics

1 answer:

Oduvanchick [21]3 years ago

3 0

Answer:

204kW

Explanation:

P = t x ω = 500 x (2πx3900/60) = 204203W = 204kW

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You have a 3.00-liter container filled with N₂ at 25°C and 4.45 atm pressure connected to a 2.00-liter container filled with Ar

LuckyWell [14K]

Answer : The final pressure in the two containers is, 2.62 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

Thus, the expression for final pressure in the two containers will be:

$PV=P_1V_1+P_2V_2$

$P=\frac{P_1V_1+P_2V_2}{V}$

where,

$P_1$ = pressure of N₂ gas = 4.45 atm

$P_2$ = pressure of Ar gas = 2.75 atm

$V_1$ = volume of N₂ gas = 3.00 L

$V_2$ = volume of Ar gas = 2.00 L

P = final pressure of gas = ?

V = final volume of gas = (4.45 + 2.75) L = 7.2 L

Now put all the given values in the above equation, we get:

$P=\frac{(4.45atm)\times (3.00L)+(2.75atm)\times (2.00L)}{7.2L}$

$P=2.62atm$

Thus, the final pressure in the two containers is, 2.62 atm

8 0

3 years ago

HELP PLEASE...

skad [1K]

Ya because they’re are both 50 liters

7 0

3 years ago

F = 2.0*10^20 N,

natka813 [3]

$F=\dfrac{Gm_1m_2}{r^2}$

With the given values of $F,G,m_1,r$ , we have

$2.0\times10^{20}\,\mathrm N=\dfrac{\left(6.67\times10^{-11}\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98\times10^{24}\,\mathrm{kg}\right)m_2}{\left(3.8\times10^8\,\mathrm m\right)^2}$

Try dealing with the powers of 10 first: On the right, we have

$\dfrac{10^{-11}\times10^{24}}{(10^8)^2}=\dfrac{10^{24-11}}{10^{16}}=10^{-3}$

Meanwhile, the other values on the right reduce to

$\dfrac{6.67\times5.98}{3.8^2}\approx2.76$

Then taking units into account, we end up with the equation

$2.0\times10^{20}\,\mathrm N=\left(2.76\times10^{-3}\,\dfrac{\mathrm N}{\mathrm{kg}}\right)m_2$

Now we solve for $m_2$ :

$m_2=\dfrac{2.0\times10^{20}\,\mathrm N}{2.76\times10^{-3}\,\frac{\mathrm N}{\mathrm{kg}}}\approx0.725\times10^{20-(-3)}\,\mathrm{kg}$

$m_2=7.25\times10^{22}\,\mathrm{kg}$

or, if taking significant digits into account,

$m_2=7.3\times10^{22}\,\mathrm{kg}$

5 0

3 years ago

Jeff’s father is installing a do-it-yourself security system at his house. He needs to get a device from his workshop that conve

Lilit [14]

Answer: buzzer.

The working principle of a buzzer is the conversion of electrical energy to sound energy.

The switch just cuts or permits the flow of current, the motor convertes electrical or other kind of energy into mechanical energy, a bulb converts electrical energy into light and a battery converts chemical energy into electrical energy.

4 0

3 years ago

Read 2 more answers

A railroad freight car of mass 3.34 × 104 kg collides with a stationary caboose car. They couple together, and 19.0% of the init

Tresset [83]

Answer:

4.123 * 10∧4 kg

Explanation:

mass of car M' = 3.34 * 10∧4 kg

energy loss E = 19/100 K

from law of conservation of momentum ,

M' V' = ( M' + M'' ) v

V = M' V' / ( M' + M'' )

Initial kinetic energy is K' = (M' V')² / 2

final kinetic energy K" = 81 K' /100

= (M' V')² 81 / (200) = ( M' + M'' ) v²/ 2

therefore , M' / ( M' + M" ) = 0.81

mass of caboose is , M" = 1.234 M' - M'

M" = .234 M'

= 0,234 ( 3.34 * 10∧4 kg)

= 4.123 * 10∧4 kg

8 0

3 years ago