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11 months ago

solve the following system of equations using substitution x+y+z=7y=32x+y-z=5enter your answer in the form of (x,y,z)

Physics

1 answer:

murzikaleks [220]11 months ago

5 0

Given equations:

$x+y+z=7\ldots(1)$ $y=3\ldots(2)$ $2x+y-z=5\ldots(3)$

Substitute 3 for y in equation (1);

$\begin{gathered} x+3+z=7 \\ x+z=7-3 \\ x+z=4\ldots(4) \end{gathered}$

Substitute 3 for y in equation (3);

$\begin{gathered} 2x+3-z=5 \\ 2x-z=5-3 \\ 2x-z=2\ldots(5) \end{gathered}$

Adding equation (4) and (5);

$\begin{gathered} (x+z)+(2x-z)=4+2 \\ x+z+2x-z=6 \\ 3x=6 \\ x=\frac{6}{3} \\ x=2 \end{gathered}$

Substitute 2 for x in equation (4);

$\begin{gathered} 2+z=4 \\ z=4-2 \\ z=2 \end{gathered}$

Therefore, the values of (x,y,z) is (2,3,2).

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Suppose you have two identical sheets of paper and you crumple one of the sheets of paper into a ball. If you drop the crumpled

sammy [17]

Answer:

Because it can easily resist air resistance.

Explanation:

Since air resistance is not negligible, the crumpled paper will reach the ground first because it can easily resist air resistance surrounding it compare to the un-crumpled one that will be influenced by the air thereby causing the un-crumpled paper to spend more time in the air

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2 years ago

Which form of energy is equal to the sum of an object’s kinetic and potential energy?

Natali5045456 [20]

Answer:

Mechanical Energy

Explanation:

The sum of kinetic energy and potential energy of an object is its total mechanical energy.

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3 years ago

Think about a typical school day. In the space provided below, describe how each of the different forms of energy we have learne

Nezavi [6.7K]

Answer:

During a typical school day all forms of eneergy is being utilised and also transfer of energy takes place from one form to another.

Explanation:

Chemical energy- A bunsen burner burning a beaker filled with water.

Heat energy- The water in the beaker absorbing the heat from the burner.

Electrical energy- Running Fans and lights in a classroom by switches.

Solar energy- Solar energy harnessed by solar panels to run the fans and lights by converting it into electrical energy.

Potential energy- A ball being held by a student at a certain height possesses energy due to gravity.

Kinetic energy- The same ball being left by the boy from a certain height produces kinetic energy

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2 years ago

Read 2 more answers

A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha

Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3 s

the formula for the time period of the pendulum is given by

$T = 2\pi \sqrt{\frac{I}{mgd}}$ .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

$I = \frac{mL^{2}}{12}+ md^{2}$

Substituting the values in equation (1)

$3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}$

$9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )$

12d² -26.84 d + 2.25 = 0

$d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}$

$d=\frac{26.84\pm 24.75}{24}$

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

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2 years ago

Physics B 2020 Unit 3 Test

weqwewe [10]

Answer:

When a charge is in motion in a magnetic field, the charge experiences a force of magnitude

$F=qvB sin \theta$

where here:

For the proton in this problem:

$q=1.602\cdot 10^{-19}C$ is the charge of the proton

v = 300 m/s is the speed of the proton

B = 19 T is the magnetic field

$\theta=65^{\circ}$ is the angle between the directions of v and B

So the force is

$F=(1.602\cdot 10^{-19})(300)(19)(sin 65^{\circ})=8.28\cdot 10^{-16} N$

The magnetic field produced by a bar magnet has field lines going from the North pole towards the South Pole.

The density of the field lines at any point tells how strong is the magnetic field at that point.

If we observe the field lines around a magnet, we observe that:

- The density of field lines is higher near the Poles

- The density of field lines is lower far from the Poles

Therefore, this means that the magnetic field of a magnet is stronger near the North and South Pole.

The right hand rule gives the direction of the force experienced by a charged particle moving in a magnetic field.

It can be applied as follows:

- Direction of index finger = direction of motion of the charge

- Direction of middle finger = direction of magnetic field

- Direction of thumb = direction of the force (for a negative charge, the direction must be reversed)

In this problem:

- Direction of motion = to the right (index finger)

- Direction of field = downward (middle finger)

- Direction of force = into the screen (thumb)

The radius of a particle moving in a magnetic field is given by:

$r=\frac{mv}{qB}$

where here we have:

$m=6.64\cdot 10^{-22} kg$ is the mass of the alpha particle

$v=2155 m/s$ is the speed of the alpha particle

$q=2\cdot 1.602\cdot 10^{-19}=3.204\cdot 10^{-19}C$ is the charge of the alpha particle

B = 12.2 T is the strength of the magnetic field

Substituting, we find:

$r=\frac{(6.64\cdot 10^{-22})(2155)}{(3.204\cdot 10^{-19})(12.2)}=0.366 m$

The cyclotron frequency of a charged particle in circular motion in a magnetic field is:

$f=\frac{qB}{2\pi m}$

where here:

$q=1.602\cdot 10^{-19}C$ is the charge of the electron

B = 0.0045 T is the strength of the magnetic field

$m=9.31\cdot 10^{-31} kg$ is the mass of the electron

Substituting, we find:

$f=\frac{(1.602\cdot 10^{-19})(0.0045)}{2\pi (9.31\cdot 10^{-31})}=1.23\cdot 10^8 Hz$

When a charged particle moves in a magnetic field, its path has a helical shape, because it is the composition of two motions:

1- A uniform motion in a certain direction

2- A circular motion in the direction perpendicular to the magnetic field

The second motion is due to the presence of the magnetic force. However, we know that the direction of the magnetic force depends on the sign of the charge: when the sign of the charge is changed, the direction of the force is reversed.

Therefore in this case, when the particle gains the opposite charge, the circular motion 2) changes sign, so the path will remains helical, but it reverses direction.

The electromotive force induced in a conducting loop due to electromagnetic induction is given by Faraday-Newmann-Lenz:

$\epsilon=-\frac{N\Delta \Phi}{\Delta t}$

where

N is the number of turns in the loop

$\Delta \Phi$ is the change in magnetic flux through the loop

$\Delta t$ is the time elapsed

From the formula, we see that the emf is induced in the loop (and so, a current is also induced) only if $\Delta \Phi \neq 0$ , which means only if there is a change in magnetic flux through the loop: this occurs if the magnetic field is changing, or if the area of the loop is changing, or if the angle between the loop and the field is changing.