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3 years ago

How is taring accomplished?

Physics

1 answer:

attashe74 [19]3 years ago

8 0

Answer:

by taring a balance the process of weighing by difference is done automatically. When a balance is tared with an object, on the balance pan, the weight of the object will be automatically subtracted from reading until the balance is re-tared or zeroed

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How can accuracy be limited?

lozanna [386]

Answer:

accuracy refers to the deviation of a measurement from a standard or true value of the quantity being measured

5 0

3 years ago

2. What biotic and abiotic factors might influence the wolf and moose population numbers? List 1 biotic factors 1 abiotic factor

Ulleksa [173]

Answer:

Abiotic - sun Biotic- Plants

Water supply, climate, shape of the land, vegetation, soils and availability of natural resources.

Explanation:

Abiotic means non living so the sun is non living. The sun gives us warmth and the ability to survive

Biotic means alive or living- plants are living and give off oxygen and take in carbon to help us live umm ok

6 0

3 years ago

Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

Ulleksa [173]

A) $v=\sqrt{\frac{2qV}{m}}$

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

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

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

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

$K=\frac{1}{2}mv^2$

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0

3 years ago

Identifying What are five things

g100num [7]

Answer:

1. Naturally occuring

2. Solid

3. Inorganic

4. Crystalline

5. Specific Chemical Compostion

Explanation:

Minerals are inorganic, crystalline solids that occur during biogeochemical processes in nature like in cooled lava or evaporated sea water. Minerals are not rocks, but are actually the components that make up rocks. Though they vary in color and shape, each mineral has a distinct chemical composition.

1. Minerals are formed by natural geological processes. Most minerals form from molten lava, sea evaporation or hot liquids in caves or cracks. Laboratory-generated minerals like synthetic gems made for commercial purposes are not considered actual minerals.

2. Though minerals vary in shape, color, luster (the way a mineral reflects light) and hardness, all minerals are a solid at a given temperature. If a substance is not in its solid state, it is not currently a mineral. For example, ice is a mineral, but liquid water is not. The Mohr scale, rates a minerals hardness from one to 10, 10 being the hardest. Diamond is the hardest mineral. Talc is a very soft mineral with a Mohr rating of one.

3. Minerals are wholly inanimate, inorganic compounds. But there are exceptions to this qualifier. There are rare organic substances with definitive chemical compositions that are labeled as “organic minerals." The most famous of this oxymoronic exception is whewellite. Whewellite is a component of kidney stones and coal deposits.

4. Most minerals will grow into a crystal shape, space permitting. Mineral deposits are often small because there is usually a variety of minerals in the same vicinity competing for the same room to grow. A mineral’s crystalline structure determines its hardness, cleavage (how it breaks) and color. There are six different crystal shapes: cubic, tetragonal, orthohombic, hexagonal, monoclinic and triclinic.

5. A mineral is defined by its chemical composition. A rock, on the other hand, does not have a specific chemical composition because it is a composite of a variety of minerals. Minerals are classified based on their anionic group. The major mineral groups are native elements, sulfides, sulfosalts, oxides and hydroxides, halides, carbonates, nitrates, borates, sulfates, phosphates and silicates. Silica is abundant in the Earth’s crust, so silicates are the most common group of mineral.

4 0

2 years ago

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

5 0

3 years ago

Read 2 more answers