HELP PLSSSSSSSSSjjdndnsnsj

Answer:

its temperature stays constant

Explanation:

Answer:

ω = 15275.25 rad/s

Explanation:

Given that,

Radial acceleration of an ultracentrifuge is, [tex]a=5\times 10^5g[/tex]

Distance from the axis, r = 2.1 cm = 0.021 m

g is the free-fall acceleration such that g = 9.8 m/s²

We need to find the angular speed of an ultracentrifuge. The formula that is used to find the angular speed is given by formula as follows :

[tex]a=r\omega^2[/tex]

Putting all the values,

[tex]\omega=\sqrt{\dfrac{a}{r}} \\\\\omega=\sqrt{\dfrac{5\times 10^5\times 9.8}{0.021}} \\\\\omega=15275.25\ rad/s[/tex]

So, the required angular speed, ω, of an ultracentrifuge is 15275.25 rad/s.

Answer:

Rotation frequency is 0.377 hertz.

Explanation:

After a careful reading of statement, we need to apply the concept of radial acceleration due to uniform circular motion, whose formula is:

[tex]a_{r} = \omega^{2}\cdot L[/tex] (Eq. 1)

Where:

[tex]a_{r}[/tex] - Radial acceleration, measured in meters per square second.

[tex]\omega[/tex] - Angular velocity, measured in radians per second.

[tex]L[/tex] - Length of the beam, measured in meters.

Now we clear the angular velocity within:

[tex]\omega = \sqrt{\frac{a_{r}}{L} }[/tex]

If [tex]a_{r} = 29.9\,\frac{m}{s^{2}}[/tex] and [tex]L = 5.33\,m[/tex], the angular velocity is:

[tex]\omega = \sqrt{\frac{29.9\,\frac{m}{s^{2}} }{5.33\,m} }[/tex]

[tex]\omega \approx 2.368\,\frac{rad}{s}[/tex]

The frequency is the number of revolutions done by device per second and can be found by using this expression:

[tex]f = \frac{\omega}{2\pi}[/tex] (Eq. 2)

Where [tex]f[/tex] is the frequency, measured in hertz.

If we know that [tex]\omega \approx 2.368\,\frac{rad}{s}[/tex], then rotation frequency is:

[tex]f = \frac{2.368\,\frac{rad}{s} }{2\pi}[/tex]

[tex]f = 0.377\,hz[/tex]

Rotation frequency is 0.377 hertz.

Answer:

43

Explanation:

you divide 2600 by 60 and your answer is 43.3333333

I just took the quiz and got it right!

Answer:

neutron beta particle and proton (last option in the list)

Explanation:

The neutron beta particle and proton inside a neutron is a clear example of a negative particle (beta particle) and a positive particle (proton) experiencing electromagnetic force (attraction between positive and negative charges) at a very short distance.

Answer:

I'm pretty sure it's the flashlight because electromagnetic force produces electricity.

Answer:   Mechanical energy is the energy that is possessed by an object due to its motion or due to its position.

(The energy acquired by the objects upon which work is done)

Answer:

Kenetic Energy?

Explanation:

I need more context of what you are studying here...

Answer:

The heat will travel through the wire to your hand.

Explanation:

Thats the one that makes the most sense.

Complete Question

A parallel plate capacitor contains a positively charged plate on the left, and a negatively charged plate on the right. An electron in between the plates is moving to the right. Which statement is true?

Group of answer choices

a The potential energy of the electron is decreasing and it is moving to a region having a higher potential

b The potential energy of the electron is decreasing and it is moving to a region having a lower potential.

c The potential energy of the electron is increasing and it is moving to a region having a lower potential

d  The potential energy of the electron is increasing and it is moving to a region having a higher potential.  

Answer:

The correct option is  C

Explanation:

From the question we are told that

  An electron in between the plates is moving to the right

Generally the potential energy of the electron is mathematically represented as  

          [tex]PE  =  e *  V[/tex]

Here  e is the charge on the electron and  V  is the electric potential of the electron

Generally the left side with the positive charge has a higher potential than the right side with the negative  charge

Now when the potential energy of the electron increases it will move toward the plate with the lower potential  which is the left plate

Answer:

53.85m

Explanation:

If we are assuming that d1 is one side of a triangle, and d2 is the other, and we are looking for the magnitude, which is essentially the hypotenuse, we use Pythagorean Theorem. a^2+b^2=c^2. 20^2 + 50^2 = 2900.

The square root of 2900 is 53.85164. Therefore, that is your hypotenuse.

Answer:

Not gonna lie. I actually had to think about this a lot XD

Running in the rain can cause you to slip and get hurt. Which I think isn't what you would want.

Walking in the rain (although it is sad), is actually a better idea than running.

Hope this helps!

Answer:

Walk

Explanation:

Running might stop you from being outside for as long, but running on wet ground will most likely result in a not-so-graceful faceplant.

Answer:

Explanation:

given:

distance = 270 km

speed = 85 km/h

find:

how long does it take to get into his audition in hours?

solution:

velocity = distance / time

85 km/h =  270 km  

                      t

85 (t) = 270

t = 270 / 85

t = 3.176 hours

Answer:

Explanation:

Ideal gas equation- The volume (V) occupied by the n moles of any gas has pressure(P) and temperature (T) Kelvin,

the relationship for these variables PV=nRT where R gas constant is called the ideal gas law

Derivation of the Ideal Gas Equation

Let us consider the pressure exerted by the gas to be ‘p,’

The volume of the gas be – ‘v’  

Temperature be – T  

n – be the number of moles of gas

Universal gas constant – R

According to Boyle’s Law,

it constant n & T, the volume bears an inverse relation with the pressure exerted by a gas.

i.e. v∝1p ………………………………(i)

According to Charles’ Law,

When p & n are constant, the volume of a gas bears a direct relation with the Temperature.

i.e. v∝T ………………………………(ii)

According to Avogadro’s Law,

When p & T are constant, then the volume of a gas bears a direct relation with the number of moles of gas.

i.e. v∝n ………………………………(iii)

Combining all the three equations, we have-

v∝nTp

or pv=nRT

where R is the Universal gas constant, which has a value of 8.314 J/mol-K

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

[tex]\Sigma F = -f = m\cdot a[/tex] (Eq. 1)

Where:

[tex]f[/tex] - Kinetic friction force, measured in newtons.

[tex]m[/tex] - Mass of the box, measured in kilograms.

[tex]a[/tex] - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ([tex]W = m\cdot g[/tex]) and uniform accelerated motion ([tex]v = v_{o}+a\cdot t[/tex]), we expand the previous expression:

[tex]-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)[/tex]

And the magnitude of the friction force exerted on the box is calculated by this formula:

[tex]f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)[/tex] (Eq. 1b)

Where:

[tex]W[/tex] - Weight, measured in newtons.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]v_{o}[/tex] - Initial speed, measured in meters per second.

[tex]v[/tex] - Final speed, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

If we know that [tex]W = 52.4\,N[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{o} = 1.37\,\frac{m}{s}[/tex], [tex]v = 0\,\frac{m}{s}[/tex] and [tex]t = 2.8\,s[/tex], the magnitud of the kinetic friction force exerted on the box is:

[tex]f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s} }{2.8\,s} \right)[/tex]

[tex]f = 2.614\,N[/tex]

The magnitude of the friction force exerted on the box is 2.614 newtons.

Answer:

[tex]B=1.61\times 10^{-4}\ T[/tex]

Explanation:

The attached figure shows the path followed by an electron in the semicircular path from A to B.

Velocity of the electron is, [tex]v=1.42\times 10^6\ m/s[/tex]

It can be seen from the figure that the radius of thenpath, r = 5 cm or 0.05 m

The magnetic force acting on the electron is balanced by the centripetal force acting on it. It means,

[tex]Bqv=\dfrac{mv^2}{r}[/tex]

B is the magnitude of the magnetic field

[tex]B=\dfrac{mv}{rq}\\\\\text{Putting all the values}\\\\B=\dfrac{9.1\times 10^{-31}\times 1.42\times 10^6}{0.05\times 1.6\times 10^{-19}}\\\\B=1.61\times 10^{-4}\ T[/tex]

So, the magnitude of the magnetic field is [tex]1.61\times 10^{-4}\ T[/tex].

The magnitude of the magnetic field that will cause the electron to follow the semicircular path from A to B will be  [tex]B=1.61\times 10^-{4]\ T[/tex]

The magnetic field is defined as when the current passes through the wire, then the magnetic field is generated around the wire in a circular pattern.

The attached figure shows the path followed by an electron in the semicircular path from A to B.

The velocity of the electron is, [tex]v=1.42\times 10^6\ \frac{m}{s}[/tex]

It can be seen from the figure that the radius of then path, r = 5 cm or 0.05 m

The magnetic force acting on the electron is balanced by the centripetal force acting on it. It means,

[tex]Bqv=\dfrac{mv^2}{r}[/tex]

B is the magnitude of the magnetic field

[tex]B=\dfrac{mv}{rq}[/tex]

[tex]B=\dfrac{9.1\times 10^{-31}\times1.42\times 10^6}{0.05\times 1.6\times 10^{-19}}[/tex]

[tex]B=1.61\times 10^{-4}\ T[/tex]

So, the magnitude of the magnetic field is  [tex]B=1.61\times 10^-{4]\ T[/tex]

To know more about the magnetic field, follow

https://brainly.com/question/26257705

Answer: 143 N

Explanation:

Answer:

The reason behind the given instance is summarized below.

Explanation:

Answer:

Explanation:

10.00 5

Answer:

the reason why is that if it is hit with the same force the moon has a lot less gravity than the earth dose so the golf ball will hit the ground on earth first before the golf ball on moon dose

Answer:

10. B

13. It's actually 30 seconds but 20 comes closest so B

14. B, any type of stretching is good before you perform any type of physical activity because it loosens up the muscles

15. D, all of the above

Explanation:

I have done many sports in my life and am still very active, I grew to know most of these things through the years.

Complete question:

A scallop forces open its shell with an elastic material called abductin, whose Young's modulus is about 2.0×10⁶ N/m2 .

If this piece of abductin is 3.1 mm thick and has a cross-sectional area of 0.49 cm2 , how much potential energy does it store when compressed 1.5 mm ?

Answer:

The elastic potential energy of the material is 0.036 J

Explanation:

Given;

Young's modulus, E = 2.0×10⁶ N/m²

Thickness of the abductin, l = 3.1 mm = 0.0031 m

compression of the abductin, x = 1.5 mm = 0.0015 m

area, A = 0.49 cm² = 0.49 x 10⁻⁴ m²

Young's modulus for elastic material is given by;

[tex]E = \frac{stress}{strain} = \frac{Fl}{Ax} \\\\ E = \frac{F}{x}*\frac{l}{A}\\\\ E = k*\frac{l}{A}\\\\k = \frac{AE}{l}\\\\k = \frac{(0.49 x10^{-4})(2*10^6)}{0.0031}\\\\ k = 31,612.9 \ N/m[/tex]

The elastic potential energy of the material is given by;

U = ¹/₂kx²

U = ¹/₂(31,612.9)(0.0015)²

U = 0.036 J

Therefore, the elastic potential energy of the material is 0.036 J

Answer:

(a) the speed of the electron is 1.237 x 10⁷ m/s

(b) the radius of its path in the magnetic field is 4.21 x 10⁻⁴ m

Explanation:

Given;

potential difference, V = 435 V

magnetic field, B = 167 mT = 0.167 T

(a) the speed of the electron

eV = ¹/₂mv²

[tex]v = \sqrt{\frac{2eV}{m} }\\\\ v = \sqrt{\frac{2(1.6*10^{-19})(435)}{(9.1*10^{-31})}}\\\\v = 1.237*10^7 \ m/s[/tex]

(b) the radius of its path in the magnetic field

[tex]r = \frac{mv}{eB}\\\\r = \frac{(9.1*10^{-31})(1.237*10^7)}{(1.6*10^{-19})(0.167)}\\\\r = 4.21 *10^{-4} \ m[/tex]

Answer:

The viscosity is  [tex]\eta  = 0.76243 \ kg/ m \cdot s [/tex]  

Explanation:

From the question we are told that

  The  density is  [tex]\rho  =  7.80 *10^{3} \  kg/m^3[/tex]

  The diameter is  [tex]d =  3.0 \  mm =0.003 \ m[/tex]

    The  time taken is  [tex]t  =  12 \ s[/tex]

   The  distance covered is  [tex]d =  0.60 \ m[/tex]

Generally the velocity of the ball is  

      [tex]v  =  \frac{d}{t}[/tex]

=>    [tex]v  =  \frac{0.60}{12}[/tex]

=>    [tex]v  =  0.05 \ m/s [/tex]

Generally the mass of the steel ball is  

    [tex]m  =  \rho  *  V[/tex]

Here  V  is the volume and this is mathematically represented as

     [tex]V  =  \frac{4}{3} *  \pi  * [\frac{d}{2}  ]^3[/tex]

=>    [tex]V  =  \frac{4}{3} *  3.142  * [\frac{0.003}{2}  ]^3[/tex]

=>      [tex]V  = 1.414 *10^{-8} \  m^3[/tex]

So

     [tex]m  = 7.80 *10^{3}  *   1.414 *10^{-8}[/tex]

     [tex]m  = 0.00011 \  kg [/tex]

Generally the viscosity is mathematically represented as  

     [tex]\eta  =  \frac{m  *  g}{6\pi  *  r  *  v }[/tex]

Here  r is the radius represented as

      [tex]r =  \frac{d}{2}[/tex]

=>   [tex]r =  \frac{0.003}{2}[/tex]

[tex]r =  0.0015  \  m [/tex]

So

  [tex]\eta  =  \frac{0.00011  *  9.8}{6 *  3.142   * 0.0015  * 0.05 }[/tex]    

=> [tex]\eta  = 0.76243 \ kg/ m \cdot s [/tex]