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A circuit diagram with a power source labeled 12 V is connected to 3 resistors in series. The resistors are labeled 1.0 Ohms, 2.

nevsk [136]

Answer:

there it is fella u were right with ur answer

7 0

3 years ago

[1] The assembly starts from rest and reaches an angular speed of 150 rev/min under the action of a 20-N force T applied to the

ExtremeBDS [4]

Answer:

t = 5.89 s

Explanation:

To calculate the time, we need the radius of the pulley and the radius of the sphere which was not given in the question.

Let us assume that the radius of the pulley ( $r_p$ ) = 0.4 m

Let the radius of the sphere (r) = 0.5 m

w = angular speed = 150 rev/min = (150 × 2π / 60) rad/s = 15.708 rad/s

Tension (T) = 20 N

mass (m) = 3 kg each

$\int\limits^0_t {Tr_p} \, dt=H_2-H_1\\( Tr_p)t=4rm(rw)\\( Tr_p)t=4r^2mw$

$t = \frac{4r^2mw}{Tr_P}$

Substituting values:

$t = \frac{4r^2mw}{Tr_P}= \frac{4*(0.5)^2*3*15.708}{20*0.4}=5.89s$

7 0

3 years ago

Read 2 more answers

What is the amount of thermal energy needed to make 5 kg of ice at - 10 °C to

agasfer [191]

Answer:

The amount of thermal energy needed is 15167500 joules.

Explanation:

By First Law of Thermodynamics, we see that amount of thermal energy ( $Q$ ), in joules, is equal to the change in internal energy. From statement we understand that change in internal energy consisting in two latent components ( $U_{l,ice}$ , $U_{l,steam}$ ), in joules, and two sensible component ( $U_{s,w}$ ), in joules, that is:

$Q = U_{l,ice} + U_{s, w} + U_{s,ice} + U_{l,steam}$ (1)

By definitions of Sensible and Latent Heat, we expanded the formula:

$Q = m\cdot (h_{f,w}+h_{v,w}+c_{ice}\cdot \Delta T_{ice}+c_{w}\cdot \Delta T_{w})$ (2)

Where:

$m$ - Mass, in kilograms.

$h_{f,w}$ - Latent heat of fussion of water, in joules per kilogram.

$h_{v,w}$ - Latent heat of vaporization of water, in joules per kilogram.

$c_{ice}$ - Specific heat of ice, in joules per kilogram per degree Celsius.

$c_{w}$ - Specific heat of water, in joules per kilogram per degree Celsius.

$\Delta T_{ice}$ - Change in temperature of ice, measured in degrees Celsius.

$\Delta T_{w}$ - Change in temperature of water, measured in degrees Celsius.

If we know that $m = 5\,kg$ , $h_{f,w} = 3.34\times 10^{5}\,\frac{J}{kg}$ , $h_{v,w} = 2.26\times 10^{6}\,\frac{J}{kg}$ , $c_{ice} = 2.090\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}$ , $c_{w} = 4.186\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}$ , $\Delta T_{ice} = 10\,^{\circ}C$ and $\Delta T_{w} = 100\,^{\circ}C$ , then the amount of thermal energy is: