A liquid at rest in a fixed container exerts a force perpendicular to the wall of the container. Two students make claims about the microscopic cause of this force. Student A says that the force exerted by individual molecules as they bounce off the wall is always perpendicular to the wall. Student B says that the molecules may strike the wall at angles that are not perpendicular. Which student is correct and why?A. Student A . If the molecules hit at nonperpendicular angles, there would be a component of force parallel to the wall exerted on them. The net result would be macroscopic motion of the fluid.B. Student A . The molecules in a liquid are in random motion at the microscopic scale, so they move in every direction. Near the wall the molecules collide with each other in such a way that the components of their motion not parallel to the wall cancel.C. Student B . The molecules in a liquid are in random motion at the microscopic scale. For every atom that hits at an angle to one side of perpendicular, there is likely to be another atom hitting at the same speed at the same angle on the other side of perpendicular. On average the nonperpendicular components cancel.D. Student B . The walls of the container cannot move, so near the walls the parallel component of the motion of the molecules is cancelled by the net force the wall exerts on the liquid.

Answer:

A) 4.035 × 10^(9) J

B) 9.29 × 10^(9) J

Explanation:

We are given;

Capacitance of the original capacitor; C = 1.33 F

Potential difference given to the original capacitor; V = 77.9 kV = 77.9 × 10³ V

A) The formula for Potential energy (U) for the original capacitor is given as:

U = ½CV²

Plugging in the relevant values, we have;

U = ½ × 1.33 × (77.9 × 10³)²

U = 4.035 × 10^(9) J

B) We are told that the capacitor with dielectric constant of 427, was replaced with one possessing a dielectric constant of 983.

Thus;

U = ½ × 1.33 × (983/427) × (77.9 × 10³)²

U = 9.29 × 10^(9) J

Answer:

Lower

Lower

gsintheta (gsinθ)

Explanation:

The sum of forces resolved parallel to the inclined plane is given by;

F - mgsinθ = 0

ma - mgsinθ = 0

ma = mgsinθ

a = gsinθ

Acceleration is proportional to angle of inclination, thus the lower the angle of the slope, lower the acceleration along the ramp.

therefore, the speed at the bottom of a slope will be lower, (velocity is directly proportional to acceleration) and, consequently, the control will be better.

The acceleration along the ramp, is gsintheta (gsinθ)

Answer:

0.65 m

Explanation:

See attachment for calculation

Answer:

C

Explanation:

Measuring your weight with a scale that measures to the nearest

tenth of a kg but adds 2 kg to each measurement

The initial angular acceleration is 5.78 rad/s^2.

Since

A gymnast on the uneven parallel bars is at rest, tipped at a 45∘ angle from the vertical. The distance from her hands to her feet is 1.8 m.

So,

[tex]Torque \tau = r \times F\\\\ \tau = l\alpha\\\\so, r \times F = l\alpha\\\\mg\ cos\ 45 \times r = (1/3)mR^2 \times \alpha\\\\9.8\times cos\ 45 \times 0.9 = (1/3)\times (1.8)^2 \times \alpha\\\\\alpha = 5.78\ rad/s^2[/tex]

hence, The initial angular acceleration is 5.78 rad/s^2.

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Answer:

no

Explanation:

ok

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5  x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

[tex]E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}[/tex]

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

[tex]r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \ \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm[/tex]

The magnitude of the electric field is now given as;

[tex]E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m[/tex]

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Answer:

A that the answer I think I'm not tryna do u bad thi

Explanation:

but try it

Answer:

13.72 m/s

Explanation:

Here, we are to determine the final velocity of the coconut. Applying first equation of free fall;

v = u + gt

where: v is the final velocity of the object, u is the initial velocity, g is the force of gravity and t is the time.

Since the coconut falls in the direction of gravity, then g = 9.8 m/[tex]s^{2}[/tex].

t = 1.4 s, u = 0 m/s

So that;

v = 0 + 9.8(1.4)

  = 13.72

The coconut hits Chief Boolie's toe with a velocity of 13.72 m/s

Answer:

The answer is D

Explanation:

Because it is big and can fit the truck.

Answer:

It is gaining potential energy

Answer: it could also be gravitational potential energy

This is because as it rises in the air it will gain more energy to fall back down

Answer:

C requires the most work and A requires the least work

Answer:

Which scenario requires the most work?

✔ C

Which scenario requires the least work?

✔ A

Answer:

a)     Eₓ = 0  ,   b)    c)    Ex_total = 3.96 10⁻⁶ N

Explanation:

For this problem we will look for the expression for the electric field of a ring of charge, at a point on its central axis.

From the symmetry of the ring, the resulting field is in the direction of the axis, we suppose that it corresponds to the x and y axis and the perpendicular direction but since the two sides of the annulus are offset.

       dE = k dq / r²

The component of this field in the direction to x is

       d Eₓ = dE cos θ

the distance of ring to the point

        r² = x² + a²

where a is the radius of each ring

the cosine is the adjacent leg between the hypotenuse

       Cos θ = x /r

we substitute

   d Ex = k / (x² + a²)  dq      x /√ (x² + a²)

we integrate

      Ex = ∫  k / (x² + a²)^{3/2}  x dq

      Ex = k x/ (x² + a²)^{1.5}   Q

the total field is the sum of the fields created by each ring, since they are on the same line (x-axis), you can perform an algebraic sum

     Eₓ_total = Eₓ₁ + Eₓ₂

a) at the origin x = 0

           in field is zero, since the numerator e makes zero

              Eₓ = 0

b) for point x = 40.0cm = 0.400 m

we substitute the values ​​in our equation

       Ex_total = k x [Q (x² + 0.1²)^{1.5} + 2Q / (x² + 0.2²2) ^3/2]

        Ex_total = k Q x [1 / (x² + 0.01)^1.5 + 2 / (x² + 0.04)^1.52]

let's calculate

       Ex_total = 9 10⁹ 30 10⁻⁹ 0.40 [1 / (0.4 2 + 0.01) 3/2 + 2 / (0.4 2 + 0.04) 3/2] 10⁻⁹

       Ex_total = 108 [14.2668 + 22.36] 10⁻⁹

       Ex_total = 3.96 10⁻⁶ N

Answer:

a) F= 0,19  [N]   according to problem statement

b) F = 0,19*10⁹ [N]  using the right value of K

Explanation:

The force between two electric charges is according to Coulomb´s law is:

F = K * q₁*q₂ / d²    where  q₁  and q₂ are the charges on body one and body 2 respectively, d is the distance between the two bodies and K is a constant  K = 8,988100*10⁹ N.m²/C². The problem establishes to use        K = 8,988100 N.m²/C².

NOTE: To value of is :  K = 8,988100*10⁹ N.m²/C². I am going to solve the problem using K = 8,988100 N.m²/C² if that information was an error, all we need to get the right answer is multiply the result by 10⁹

Then:

F = 8,988100 * 1,2* 0,36 / (4,5)²     [ N*m²/C² ] * [ C*C*/m²]

F = 3,882859/ 20,25  [N]

F= 0,19  [N]

The force is of repulsion since the two charges are positive and in the direction of the straight line which passes through the centers of the bodies

Answer:

2

Explanation:

Answer:

The answer is B

Explanation:

Answer:

The ratio of the magnitude of the electric force on the bee to the magnitude of the bee's weight is 2.35 x 10⁻⁶

Explanation:

Given;

mass of the honeybee, m = 0.1 g

charge acquired by the honeybee, Q = 23pC = 23 x 10⁻¹² C

the electric field near the earth's surface, E = 100 N/C

The magnitude of the electric force on the bee is given by;

F = QE

F = (23 x 10⁻¹²)(100) = 23 x 10⁻¹⁰ N

The weight of the bee is given by;

W = mg

W = 0.1 x 10⁻³ x 9.8

W = 9.8 x 10⁻⁴ N

The the ratio of the magnitude of the electric force on the bee to the magnitude of the bee's weight is given by;

[tex]\frac{F}{W} = \frac{23*10^{-10}}{9.8*10^{-4}} = 2.35 *10^{-6}[/tex]

Therefore, the ratio of the magnitude of the electric force on the bee to the magnitude of the bee's weight is 2.35 x 10⁻⁶

Answer:

Same

Explanation:

Mass is the quantity of matter in a certain object.

WHEREVER you take a 2kg object, the mass will remain 2kg. All that changes is the Weight ..Weight the force which the centre of a Planet uses to pull everything towards itself.

On earth, it is 9.81 whereas on the Moon it is 1.6

Here, we are required to determine the mass of a 2kg object on the Earth and on the moon.

The mass of an object is simply defined as the quantity of matter present in the object and this quantity remains constant.

The mass of an object is simply defined as the quantity of matter present in the object and this quantity remains constant.In essence, if the mass of an object is 2kg, this mass remains constant on planet earth and the moon.

PS: The weight of an object which is a measure of the acceleration due to gravity which differs with location and position is the constant which may be different about the object on the earth and the moon respectively.

The acceleration due to gravity on earth and the moon is 9.8m/s² and 1.6m/s² respectively.

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Answer:

27%

Explanation:

15.999 divided by 58.32 = .27433128

Move the decimal place over 2 places.

27%

Make use of percentages. Given that the molecular mass of magnesium hydroxide (MOH.) is 58.32 amu and the atomic mass of an oxygen atom is 15.999 amu, oxygen accounts for 27% of this compound.

Chemical substances known as molecular compounds assume the shape of distinct molecules. Examples include common compounds like carbon dioxide (CO2) and water (H2O) (Figure 3.1.1 ).

These substances differ significantly from ionic substances like sodium chloride (NaCl). When metal atoms give up one or more of their electrons to non-metal atoms, ionic compounds are created. The cations and anions that result

Therefore,

58.32 / 15.999 (27.3%) =.27433128

Add two places to the decimal place.

27%

The molar mass, which is stated in g/mol, is defined as the mass of a specific material divided by the amount of a substance. In light of this, the molar mass is average. Amu, short for atomic mass units, is the unit used to quantify molecular mass.

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Answer:

more/ faster

Explanation:

I got it right on my test

Particles in a warmer block will have more kinetic energy and move faster than the particles in a colder block option (D) more/ faster is correct.

The energy which an item has as a function of motion is determined by the initial energy. It is described as the effort required to propel a material body from rest to the given velocity.

As we know,

As the temperature rises, the particles' kinetic energy rises and they begin to move more quickly. Particles have higher kinetic energy in gases than in solids.

Thus, particles in a warmer block will have more kinetic energy and move faster than the particles in a colder block option (D) more/ faster is correct.

The question is incomplete.

The complete question is:

Particles in a warmer block will have ________ kinetic energy and move ______than the particles in the colder block.

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Answer:

F

Explanation: Its common sence

Answer:

It’s false

Explanation:

2020 (WORST YEAR EVER) on Edge ;-)

Answer:

How big your foot is. (Big Brain)

Explanation:

Answer:

θ = 720º

Explanation:

Let's use the rotational kinematics relations for this exercise, remember that all angles must be given in radians.

Let's reduce the magnitudes to the SI system

         θ = 2 rev (2π rad / 1 rev) = 4π rad

we assume that the pulley has a radius r

linear and angular coordinates are related

         y = θ r

let's calculate

         y = 4π r

to convert the distance in angles to radians

        θ = y / r

        θ = 4π

let's change this angle to degrees

        π rad = 180º

         θ = 4π rad (180º /π rad)

         θ = 720º

Answer:

a

[tex]P =0.1094 \ m[/tex] past the goal post

Yes the Flyers won the match

Explanation:

Generally the distance covered by the pluck during the last 1.25 seconds is mathematically represented as

     [tex]D = ut + \frac{1}{2} * a * t^2[/tex]

=>  [tex]D =30  * 1.25 + \frac{1}{2} * (-0.5) * (1.25)^2[/tex]

=>   [tex]D =37.1094 \ m  [/tex]

Generally the position of the puck when the game clock reaches zero is mathematically represented as

     [tex]P = 37.1094 -37[/tex]

       [tex]P =0.1094 \ m[/tex] past the goal post

Given that the Bruins where one point down and Flyers scored another goal  it means that the flyer are now two point up hence they won the match

Answer:

The acceleration of the car is -4.166 m/s²

The time between applying the brakes and hitting the obstacle is 6.4 s

Explanation:

Given;

initial velocity of the car when the brake was applied, u = 31.3 m/s

final velocity of the car after hitting an obstacle, v = 4.65 m/s

distance traveled before stopping, s = 115 m

The acceleration of the car is given by;

v² = u² + 2as

2as = v² - u²

a = (v² - u²) / 2s

a = (4.65² - 31.3²) / (2 x 115)

a = -4.166 m/s²

The time between applying the brakes and hitting the obstacle;

v = u + at

at = v - u

t = (v - u) / a

t = (4.65 - 31.3) / -4.166

t = 6.4 s

Answer:

If Earth were to stop spinning on its axis, gradually the oceans would migrate towards the poles from the equator.  The atmosphere would still be in motion with the Earth's original 1100 mile per hour rotation speed at the equator. And this means, rocks, topsoil, trees, buildings,  and so on, would be swept away into the atmosphere.

Hope that helped <3

Answer:

Yellow

Explanation:

It is for sure not red

Blackbody radiation graph for Star A, which has a surface temperature of 5,700 degrees Celsius. At the point where the curve peaks, the wavelength is approximately 550 nanometers, in the middle of the visible light spectrum. The color will Star A be Yellow.

A waveform signal that is carried in space or down a wire has a wavelength, which is the separation between two identical places (adjacent crests) in the consecutive cycles. This length is typically defined in wireless systems in meters (m), centimeters (cm), or millimeters (mm).

A blackbody's spectrum is continuous (it emits some light at all wavelengths), and it peaks at a particular wavelength. For hotter objects, the peak of the blackbody curve shifts to shorter wavelengths in a spectrum. If you conceive of a blackbody in terms of visible light, the hotter it is, the bluer its peak emission wavelength will be. For instance, the temperature of the sun is about 5800 Kelvin. With this temperature, a blackbody's peak is roughly 500 nanometers away, or the wavelength of yellow.

The color will Star A be Yellow.

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Answer:

a) 3.66 s

b) 124.4 m

c) 3.12s

Explanation:

Given that

Speed of the Red Car, v₁ = 34 m/s

Speed of the Blue Car, v₂ = 28 m/s

Distance between the two cars, d = 22 m

The difference between the speed of the cars is: 34 - 28 = 6 m/s

From this, we can deduce that the red car will be catching up to the blue car at a speed of 6 m/s.

1)

If we divide the distance by the difference in speed. This becomes

d/v = 22/6 = 3.66 s. Which means, It takes 3.66 seconds for the red car to meet up with the blue car.

2

From the previous part, we were able to confirm that it took 3.66 seconds for the red car to meet up the blue car. Also, the speed with which it were traveling was travelling at, was constant, so we only need to multiply it by the time from (1) to get the distance.

v = d * t = 34 * 3.66 = 124.4

Therefore the red car travels at 124.4 m before catching up to the blue car.

3

If the red car starts to accelerate the moment we see it, the time it will take, to get to the blue car will be less than what we had gotten. We can find this using one of the equations of motion.

S = ut + ½gt², where

S = 22

u = 6

t = ?

g = 2/3

22 = 6t + 1/3t²

By using the quadratic formula, we find out the two answers listed below

t1 = 3.12 s

t2 = - 21.12 s

We all know that negative time is not possible, so the answer is t1. At 3.12 seconds