Red, yellow, and blue are

Electrical systems are governed by Ohm’s law, which states that V = IR, where V is the voltage, I is the current, and R is the r

ella [17]

Answer:

$\frac{dR(t)}{dt}=0.06\Omega$

Explanation:

Since $R(t)=\frac{V}{I(t)}$ , we calculate the resistance rate by deriving this formula with respect to time:

$\frac{dR(t)}{dt}=\frac{d}{dt}(\frac{V}{I(t)})=V\frac{d}{dt}(\frac{1}{I(t)})$

Deriving what is left (remember that $(\frac{1}{f(x)})'=-\frac{1}{f(x)^2}f'(x)$ ):

$\frac{d}{dt}(\frac{1}{I(t)})=-\frac{1}{I(t)^2}\frac{dI(t)}{dt}$

So we have:

$\frac{dR(t)}{dt}=-\frac{V}{I(t)^2}\frac{dI(t)}{dt}$

Which for our values is (the rate of <em>I(t)</em> is decreasing so we put a negative sign):

$\frac{dR(t)}{dt}=-\frac{24V}{(56A)^2}(-8A/s)=0.06\Omega$

8 0

3 years ago

Suppose snow is falling in Utah. What type of air mass is responsible for this weather?

Viefleur [7K]

Maritime polar air mass comes from the northern part of the pasific ocean and that air mass goes to Utah.

5 0

3 years ago

In a hydraulic lift, if the pressure exerted on the liquid by one piston is increased by 100 N/m2 , how much additional weight c

shepuryov [24]

Answer:

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

Explanation:

By means of the Pascal's Principle, the hydraulic lift can be modelled by the following two equations:

Hydraulic Lift - Before change

$P = \frac{F}{A}$

Hydraulic Lift - After change

$P + \Delta P = \frac{F + \Delta F}{A}$

Where:

$P$ - Hydrostatic pressure, measured in pascals.

$\Delta P$ - Change in hydrostatic pressure, measured in pascals.

$A$ - Cross sectional area of the hydraulic lift, measured in square meters.

$F$ - Hydrostatic force, measured in newtons.

$\Delta F$ - Change in hydrostatic force, measured in newtons.

The additional weight is obtained after some algebraic handling and the replacing of all inputs:

$\frac{F}{A} + \Delta P = \frac{F}{A} + \frac{\Delta F}{A}$

$\Delta P = \frac{\Delta F}{A}$

$\Delta F = A\cdot \Delta P$

Given that $\Delta P = 100\,Pa$ and $A = 25\,m^{2}$ , the additional weight is:

$\Delta F = (25\,m^{2})\cdot (100\,Pa)$

$\Delta F = 2500\,N$

The additional mass needed for the additional weight is:

$\Delta m = \frac{\Delta F}{g}$

Where:

$\Delta F$ - Additional weight, measured in newtons.

$\Delta m$ - Additional mass, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

If $\Delta F = 2500\,N$ and $g = 9.807\,\frac{m}{s^{2}}$ , then:

$\Delta m = \frac{2500\,N}{9.807\,\frac{m}{s^{2}} }$

$\Delta m = 254.92\,kg$

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

3 0

3 years ago

A 4.50-kg wheel that is 34.5 cm in diametet rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 1

lisabon 2012 [21]

Answer:

$\alpha =10.93radian/sec^2$

Explanation:

We have given given the final angular velocity $\omega _{final}=13.5rad/sec$

And $\omega _{initial}=22rad/sec$

Displacement $\Theta =13.8radian$

We have to find the angular acceleration $\alpha$

According to law of motion $\omega _{final}^2=\omega _{initial}^2+2\alpha \Theta$

So $13.5^2=22^2+2\times \alpha \times 13.8$

$\alpha =-10.93radian/sec^2$

In question we have tell about magnitude only so $\alpha =10.93radian/sec^2$

4 0

3 years ago

Why do we use scatter plots for most data in Physical Science? Question 1 options: because the teacher said to use scatter plots

SVEN [57.7K]

Answer:

scatter plots show the relationship between the independent and dependent variables

Explanation:

A scatter plot is a graph which shows two variables plotted along two axes (usually the x and y axes). Scatter plots are useful in establishing any form of correlation between the dependent and independent variables in any study.

Correlation simply means the degree of relationship between variables, that is, how much does one variable affect the other? When scatter plots are almost a straight line graph, there is a high correlation between the variables. When the points in a scatter plot are isolated, there is little (sometimes zero) correlation between the variables.

8 0

3 years ago