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4 years ago

Is F=mxa expressed in equation in newtons second law of motion?

1 answer:

6 0

Yes. That equation is exactly how we write Newton's second law nowadays. Note that the 'x' just means 'times'. It's not another variable. The equation could be written as simply. F = m a .

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In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their component

JulijaS [17]

Answer:

The calculated vectors are:

$\vec{A}-\vec{B}=(3,-5,-4)$

$\vec{B}-\vec{C}=(-5,4,0)$

$-\vec{A}+\vec{B}-\vec{C}=(-6,4,3)$

$3\vec{A}-2\vec{C}=(-3,-2,-11)$

Explanation:

To operate with vectors, you sum or rest component to component. To multiply scalars with vectors, you distribute the scalar with each component of the vector. These are the following rules you must apply in these cases:

$\vec{V}+\vec{W}=(V_1,V_2,V_3)+(W_1,W_2,W_3)=(V_1+W_1,V_2+W_2,V_3+W_3)$ (1)

$\vec{V}-\vec{W}=(V_1,V_2,V_3)-(W_1,W_2,W_3)=(V_1-W_1,V_2-W_2,V_3-W_3)$ (2)

$\alpha\cdot\vec{V}=\alpha\cdot(V_1,V_2,V_3)=(\alpha\cdot V_1,\alpha\cdot V_2,\alpha\cdot V_3)$ (3)

The operations in these cases are:

$\vec{A}-\vec{B}=(1,0, -3)-(-2,5,1)=(3,-5,-4)$

$\vec{B}-\vec{C}=(-2,5,1)-(3,1,1)=(-5,4,0)$

$-\vec{A}+\vec{B}-\vec{C}=-(1,0, -3)+(-2,5,1)-(3,1,1)=(-6,4,3)$

$3\vec{A}-2\vec{C}=3(1,0, -3)-2(3,1,1)=(3,0, -9)-(6,2,2)=(-3,-2,-11)$

3 0

3 years ago

Read 2 more answers

What is the energy dissipated as a function of time in a circular loop of N turns of wire having a radius of r0 and a resistance

Snezhnost [94]

Explanation:

Let N be the number of turns in a circular loop having a radius of r₀ and a resistance R. The magnetic field is given by :

$B(t)=B_oe^{-\dfrac{t}{\tau}}$

The induced emf in the circular coil is given by :

$\epsilon=\dfrac{-d\phi}{dt}$

Where

$\phi=BA$

$\epsilon=N\dfrac{-d(BA)}{dt}$

$\epsilon=NAB_o\dfrac{-d(e^{-\dfrac{t}{\tau}})}{dt}$

$\epsilon=\dfrac{NAB_o}{\tau}e^{-t/\tau}$

Power dissipated is given by :

$P=\dfrac{\epsilon^2}{R}$

$P=\dfrac{(\dfrac{NAB_o}{\tau}e^{-t/\tau})^2}{R}$

$P=(\dfrac{NAB_o}{\tau})^2\dfrac{e^{-2t/\tau}}{R}$

Energy dissipated in the circuit is given by :

$E=\int\limits^T_0 {P.dt}$

$E=\int\limits^T_0 {(\dfrac{NAB_o}{\tau})^2\dfrac{e^{-2t/\tau}}{R}.dt}$

$E=\dfrac{1}{R}(\dfrac{NAB_o}{\tau})^2\int\limits^T_0 {e^{-2t/\tau}.dt}$

$E=\dfrac{(N\pi r_o^2B_o^2)^2}{2R}{\tau(e^{2t/\tau}-1)}{}$

Hence, this is the required solution.

7 0

3 years ago

The continuity equation shows that the ratio of fluid velocities within different openings is inversely proportional to the cros

Stolb23 [73]

Answer:

A typical example of where the continuity equation can be applied, is in finding the cross sectional area of a laminar falling stream of water from a pipe at an elevation, at the point where it hits the ground.

The speed of the water through the discharge opening of the pipe can be measured with a flow-meter, and the cross sectional opening of the pipe is known.

As the water leaves the pipe, and falls towards the ground, the water stream is accelerated under gravity, and the speed of the water stream increase. The increase in the speed of the water stream causes the cross sectional area to decrease in obedience of the continuity law and equation. If the speed of can be measured, then the area of the water stream just before the water touches the ground can be calculated from the continuity equation.

If the speed of the water stream can not be measured, then it can be calculated using Newton' equation of motion

$v^{2} = u^{2} + 2gz$

$v = \sqrt{u^{2} + 2gz}$

where

v = the velocity of the water stream at the point where it hits the ground

u = the velocity of the water at the pipe discharge (measured with a flow meter in the pipe)

g = acceleration due to gravity

z = the elevation of the pipe above the ground.

After calculating the speed of the water stream at the point where it hits the ground, the cross sectional area of the the stream can be calculated from the continuity equation.

$uA = va$