True or false? A protostellar cloud spins faster as it contracts

Answer:

the force will increase by a factor 2.25

Explanation:

The gravitational force between the two stars is given by:

[tex]F=G\frac{m_1 m_2}{r^2}[/tex]

where

G is the gravitational constant

m1, m2 are the masses of the two stars

r is the distance between the stars

If the distance is decreased by one-third, it means that the new distance is 2/3 of the previous distance

[tex]r'=\frac{2}{3}r[/tex]

So the new force will be

[tex]F'=G\frac{m_1 m_2}{(\frac{2}{3}r)^2}=\frac{9}{4} G\frac{m_1 m_2}{r^2}=2.25 F[/tex]

So, the force will be 2.25 times the previous value.

The gravitational force between two objects is proportional to their masses and inversely proportional to the square of the distance between their centers.

this is your teacher get off your phone

[tex](a) \( v_0 = 41 \, \text{m/s} \)(b) \( h = 45 \, \text{m} \)[/tex]

To find the initial speed of the rock, we can use the kinematic equation that relates final velocity [tex](\( v_f \)),[/tex]  initial velocity [tex](\( v_i \)),[/tex] acceleration [tex](\( a \))[/tex], and displacement [tex](\( d \))[/tex]. The equation is:

[tex]\[ v_f^2 = v_i^2 + 2ad \][/tex]

Given:

- Final velocity [tex](\( v_f \))[/tex]= 27 m/s (when it hits the ground)

- Displacement [tex](\( d \))[/tex]= 33 m (height of the cliff, since the rock starts from the top)

Substitute the known values into the equation:

[tex]\[ 27^2 = v_i^2 + 2(-9.8)(33) \]\[ 729 = v_i^2 - 646.8 \]\[ v_i^2 = 729 + 646.8 \]\[ v_i^2 = 1375.8 \]\[ v_i = \sqrt{1375.8} \]\[ v_i ≈ 37.1 \, \text{m/s} \][/tex]

The initial speed of the rock is approximately 37.1 m/s.

Next, to find the greatest height of the rock, we can use the same kinematic equation, but this time solving for the displacement [tex](\( d \))[/tex], with [tex]\( v_f = 0 \)[/tex] since the rock reaches its highest point momentarily before falling back down.

[tex]\[ v_f^2= v_i^2 + 2ad \]\[ 0^2 = 37.1^2 + 2(-9.8)d \]\[ 0 = 1374.41 - 19.6d \]\[ 19.6d = 1374.41 \]\[ d = \frac{1374.41}{19.6} \]\[ d ≈ 70.15 \, \text{m} \][/tex]

However, since the height is measured from the base of the cliff, we need to subtract the initial height of the cliff:

[tex]\[ h = 70.15 - 33 \]\[ h = 37.15 \, \text{m} \][/tex]

So, the greatest height of the rock, measured from the base of the cliff, is approximately 37.15 meters.

The calculations used basic kinematic equations of motion to determine the initial speed and greatest height of the rock. The equations take into account the acceleration due to gravity [tex](\( -9.8 \, \text{m/s²} \))[/tex] )and the given displacements and final velocities.

Complete Question:
A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 33 m high. When it hits the ground at the base of the cliff the rock has a speed of 27 m/s. (a) Assuming that air resistance can be ignored, find the initial speed of the rock. m/s (b) Find the greatest height of the rock as measured from the base of the cliff. m

Your answer is Theory

Answer:

The correct answer will be- A scientific theory.

Explanation:

A scientific theory is a substantiated idea which provides a deep explanation of any natural event.  A theory is an elaborate form generalized or proposed hypothesis which explains the phenomenon to a limited extent.

A theory is formed by combining many proved hypothesis which provides evidence as well as supports the hypothesis. The theory is accepted only after it has been proved by the experiments of the researcher and fellow scientists.

Thus, A scientific theory is the correct answer.

Protons are about 1836 times bigger than electrons.

Hope this helps :)

Please mark brainliest xo

(a) 373.4 J

The displacement of the book is

[tex]d=14.0 m - 1.30 m=12.7 m[/tex]

While the gravitational force acting on the book is

[tex]F=mg=(3.00 kg)(9.8 m/s^2)=29.4 N[/tex]

And the force (downward) is parallel to the displacement (downward), so the work done by gravity is

[tex]W=Fd=(29.4 N)(12.7 m)=373.4 J[/tex]

(b) -373.4 J

The work done by the Earth on the book has been converted into kinetic energy of the book (because the book accelerates as it approaches the ground). SInce the total mechanical energy of the Earth-book system must be conserved, this means that potential energy U has been converted into kinetic energy K: therefore, the loss in potential energy U is exactly -373.4 J.

(c) 411.6 J

The gravitational potential energy U is given by:

[tex]U=mgh[/tex]

where m = 3.00 kg is the mass of the book, g = 9.8 m/s^2 is the gravitational acceleration, h=14.0 m is the height of the book above the reference level (the ground). Substituting,

[tex]U=(3.00 kg)(9.8 m/s^2)(14.0 m)=411.6 J[/tex]

(d) 38.2 J

When the book reaches the hand, its height above the ground level is

h = 1.30 m

therefore, the gravitational potential energy this time is

[tex]U=mgh=(3.00 kg)(9.8 m/s^2)(1.30 m)=38.2 J[/tex]

(e) 373.4 J

The work done by gravity does not change if we change the value of the potential energy at ground level. In fact, the work done by gravity is still calculated as before:

[tex]W=Fd=(29.4 N)(12.7 m)=373.4 J[/tex]

(f) -373.4 J

As we did in point b), the work done by the Earth on the book has been converted into kinetic energy of the book (because the book accelerates as it approaches the ground), so the loss in potential energy is equal to the work done by gravity, and this value does not depend on the value of the potential energy at ground level, so it is still -373.4 J.

(g) 511.6 J

The gravitational potential energy U is given by:

[tex]U=mgh+U_0[/tex]

where m = 3.00 kg is the mass of the book, g = 9.8 m/s^2 is the gravitational acceleration, h=14.0 m is the height of the book above the reference level (the ground), and [tex]U_0 = 100 J[/tex] is the potential energy at ground level, which must be added into the formula. Substituting,

[tex]U=(3.00 kg)(9.8 m/s^2)(14.0 m)+100 J=511.6 J[/tex]

(h) 138.2 J

As before, we can calculate the potential energy of the book at a height of h=1.30 m, adding 100 J of energy which is the value of the potential energy at ground level. we find:

[tex]U=mgh+U_0=(3.00 kg)(9.8 m/s^2)(1.30 m)+100 J=138.2 J[/tex]

The work done by gravity, Wg, is calculated to be 372.66 J. The gravitational potential energy, U, decreases by this amount during the fall, from 411.60 J at the release point to 38.94 J at the friend's hands. If the ground level has an arbitrary gravitational potential energy of 100 J, the final values are adjusted accordingly to 511.60 J and 138.94 J.

To calculate the work Wg done by the gravitational force as the book drops to the friend's hands, we use the formula W = mgh, where m is the mass of the book, g is the acceleration due to gravity (9.8 m/s²), and h is the height through which the book falls. The change in height is the total distance D minus the distance d, which the friend's hands are above the ground. This gives us the effective height the book falls through.

(a) The work done by gravity, Wg, is:

Wg = mgh
Wg = (3.00 kg)(9.8 m/s²)(14.0 m - 1.30 m)
Wg = (3.00 kg)(9.8 m/s²)(12.7 m)
Wg = 372.66 J

(b) The change in the gravitational potential energy ΔU is equal to the work done by gravity, so ΔU = 372.66 J.

(c) Initially, the gravitational potential energy U at the release point is U = mgh, which is:

U = (3.00 kg)(9.8 m/s²)(14.0 m)
U = 411.60 J

(d) When the book reaches the friend's hands, the gravitational potential energy is:

U = (3.00 kg)(9.8 m/s²)(1.30 m)
U = 38.94 J

If the gravitational potential energy at ground level is 100 J instead of 0 J, the values for Wg and ΔU do not change because these are quantities that depend only on the change in height. However, the absolute values of U at the initial and final points will be 100 J more than previously calculated:

(e) Wg remains 372.66 J

(f) ΔU remains 372.66 J

(g) U at the release point now is U + 100 J:
U = 411.60 J + 100 J
U = 511.60 J

(h) U at the friend's hands now is U + 100 J:
U = 38.94 J + 100 J
U = 138.94 J

Answer:

C. increases

Explanation:

The total resistance of a parallel circuit is given by:

[tex]\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+...[/tex]

where R1, R2, etc. are the individual resistances.

From the formula, we notice that as new resistors are added to the configuration, the total resistance [tex]R_T[/tex] decreases.

According to Ohm's law, the current flowing in the circuit is inversely proportional to the total resistance:

[tex]I=\frac{V}{R_T}[/tex]

where V is the voltage supplied by the source: so, when adding more resistors in parallel, the total resistance decreases and the current increases.

Finally, the power supplied by the source is

[tex]P=VI[/tex]

we said that V (voltage) remains constant, while I (the current) increases, so the power supplied increases as well.

Answer:

E. greater than the angle of incidence.

Explanation:

Snell's law states that:

[tex]n_i sin \theta_i = n_r sin \theta_r[/tex] (1)

where

[tex]n_i, n_r[/tex] are the refractive index of the first and second medium

[tex]\theta_i, \theta_r[/tex] are the angle of incidence and refraction, respectively

For light moving from water to air, we have:

[tex]n_i = 1.33[/tex] (index of refraction of water)

[tex]n_r = 1.00[/tex] (index of refraction of air)

Substituting into (1) and re-arranging the equation, we get

[tex]\sin \theta_r = \frac{n_i}{n_r} sin \theta_i = 1.33 sin \theta_i[/tex]

which means that

[tex]\theta_r > \theta_i[/tex]

so, the correct answer is

E. greater than the angle of incidence.

Light enters air from water. The angle of refraction will be

Refractive Index is the value that calculated from the speed of light ratio in a vacuum to in a second medium of greater density. The refractive index variable is symbolized by the letter [tex]n[/tex] or [tex]n'[/tex] in descriptive text and mathematical equations.

Light enters air from water, the angle of refraction will be greater than the angle of incidence.

When light passed from a less dense to a more dense substance for example passing from air into water, the light is refracted towards the normal. The normal is a line perpendicular (forming a 90 degree angle) to the boundary between the two substances.

Hope it helps!

Grade:  9

Subject:  physics

Chapter:  refraction

Keywords:  The angle of refraction, the angle of incidence

Answer:

T = 0.444 sec

f = 2.25 Hz

Explanation:

Mass of the object = m = 50g = 0.05 kg

Spring constant = k = 10N/m

The time period of mass attached to a spring is calculated as:

[tex]T=2\pi\sqrt{\frac{m}{k} }[/tex]

Using the values in the formula, we get:

[tex]T=2\pi\sqrt{\frac{0.05}{10} }=0.444[/tex]

Thus the time period is 0.444 sec.

Frequency is the reciprocal of the time period.

[tex]f=\frac{1}{T}\\\\ f=\frac{1}{0.444} =2.25[/tex]

Thus the frequency of oscillation is 2.25 Hertz

The period of the oscillations of the mass on the spring is 0.4472 seconds and the frequency is 2.236 Hz. This is calculated using the formula for simple harmonic motion T=2π √(m/k)

The question relates to the motion of a mass attached to a spring commonly studied in physics as simple harmonic motion. The period and frequency of the motion can be calculated using the equations of physics that relate to this type of motion. Specifically, these are the equations for the period(T) and frequency(f) of simple harmonic motion:

T = 2π √(m/k)
f = 1/T

The mass (m) 50g needs to be converted to kilograms, so m = 0.05 kg. The spring constant (k) is provided as 10 N/m. Substituting these into the equation for T we get:

T = 2π √(0.05/10) = √(0.01*π*2) = 0.4472 seconds

The frequency of the oscillations is then given by:

f = 1/T = 1/0.4472 = 2.236 Hz

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A plane mirror is a highly polished flat surface with a very high capacity to reflect incident light.  

We can understand in a better way how this works with the figure attached:  

1. The incident rays coming from the real object reach the mirror and  

2.are reflected following the law of Reflection.  

3.The prolongation of those reflected rays converge at a point that does not coincide with the actual position of the object. At that point the virtual image of the object is formed.  

4.Then, the reflected divergent rays are captured by our eye converging on the retina.  

Now, the image is said to be virtual because it is a copy of the object that looks as if the object is behind the mirror and not in front of it or on the surface, but it is not really there. However, it can be seen when we focus it with our eyes.  

In addition, the image formed is:  

symmetrical, because apparently it is at the same distance from the mirror  

the same size as the object.  

upright, because it retains the same orientation as the object.

There are four wavelengths seen in the image of the sound wave.

Four wavelengths are seen in this image of a sound wave.

The distance between the corresponding spots of two successive waves is known as wavelength. Two points or particles that are in the same phase—i.e., points that have completed identical fractions of their periodic motion—are referred to as "corresponding points."

The wavelength is typically measured from crest to crest or from trough to trough in transverse waves (waves with points vibrating at right angles to the direction of their advance); from compression to compression or from rarefaction to rarefaction in longitudinal waves (waves with points vibrating in the same direction as their advance).

The wavelength also describes the pattern of disturbance brought about by the energy moving away from the sound source. Longitudinal waves are what make up sound. This indicates that the direction of energy wave propagation is parallel to the direction of particle vibration propagation.

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Answer:

18.5 m/s

Explanation:

On a horizontal curve, the frictional force provides the centripetal force that keeps the car in circular motion:

[tex]\mu mg = m\frac{v^2}{r}[/tex]

where

[tex]\mu[/tex] is the coefficient of static friction between the tires and the road

m is the mass of the car

g is the gravitational acceleration

v is the speed of the car

r is the radius of the curve

Re-arranging the equation,

[tex]v=\sqrt{\mu gr}[/tex]

And by substituting the data of the problem, we find the speed at which the car begins to skid:

[tex]v=\sqrt{(0.350)(9.8 m/s^2)(100 m)}=18.5 m/s[/tex]

The car will begin to skid sideways at 18.5 m/s

[tex]\texttt{ }[/tex]

Centripetal Acceleration can be formulated as follows:

[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

[tex]\texttt{ }[/tex]

Centripetal Force can be formulated as follows:

[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

mass of car = m = 1000 kg

radius of curve = R = 100 m

coefficient of static friction = μ = 0.350

Asked:

speed of the car = v = ?

Solution:

We will derive the formula to calculate the maximum speed of the car:

[tex]\Sigma F = ma[/tex]

[tex]f = m \frac{v^2}{R}[/tex]

[tex]\mu N = m \frac{v^2}{R}[/tex]

[tex]\mu m g = m \frac{v^2}{R}[/tex]

[tex]\mu g = \frac{v^2}{R}[/tex]

[tex]v^2 = \mu g R[/tex]

[tex]\boxed {v = \sqrt { \mu g R } }[/tex]

[tex]v = \sqrt { 0.350 \times 9.8 \times 100 }[/tex]

[tex]v = \sqrt { 343 }[/tex]

[tex]v = 7 \sqrt{7} \texttt{ m/s}[/tex]

[tex]\boxed {v \approx 18.5 \texttt{ m/s}}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Circular Motion

This is a conceptual problem, I’ll try to upload a picture:

Answer :

a = 1 m/s^2

Explanation:

Given information

Net force is a constant

The mass of the truck is [tex]m_{o}[/tex]

Initial acceleration [tex]a_{o}[/tex]= 2 m/[tex]s^{2}[/tex]

The mass of the truck is increased twice as much

[tex]F_{net}= m_{o} a_{o}[/tex] = ma

[tex]F_{net}= m_{o} a_{o}[/tex] = [tex]2m_{o}[/tex]a

a=[tex]\frac{a_{o} } {2}[/tex]

a = 2/2

a = 1 m/[tex]s^{2}[/tex]

Answer:

[tex]V=RI[/tex]

Explanation:

Ohm's law states the relationship between voltage, resistance and current in an electrical circuit containing passive elements only:

[tex]V=RI[/tex]

where

V is the voltage supplied by the battery

R is the resistance of the circuit

I is the current

From the equation, we see that the voltage, V, is directly proportional to the current in the circuit, I.

Ohm's Law is the mathematical relationship among electric current, resistance, and voltage. The principle is named after the German scientist Georg Simon Ohm. In direct-current (DC) circuits, Ohm's Law is simple and linear. Suppose a resistance having a value of R ohm s carries a current of I ampere s.

Answer:

B. Whales and porpoises signal each other by making sounds

Explanation:

Sound is a type of mechanical wave, which consists of oscillations of the particles in a medium. It is also classified as longitudinal wave, which means that the direction of the oscillation of the particles in the medium is parallel to the direction of propagation of the wave, forming alternating regions of higher particle density (compressions) and lower particle density (rarefactions).

Since sound is a mechanical wave, it just need a medium to propagate through. Therefore, it can travel through solids, liquids and gases.

In particular, the option

B. Whales and porpoises signal each other by making sounds

is an example of this: these animals emit ultrasound waves (a type of sound waves with high frequency), that travel through the water, and then are reflected back to the animal, allowing the animal to understand the distance of the object from which the wave has been reflected back.

Answer:

6 m/s

Explanation:

edge 2021

Answer:

A.  Temperature of C will increase first, then B, then A.

Explanation:

Heat capacity is the amount of heat required to raise the temperature of one mole of an object by one degree Celsius.

A higher value of heat capacity indicates that higher amount of heat will be required to change the temperature of that substance.

So in the given statement the order of heat capacity is:

A > B > C

So, it will be harder to change the temperature of A(larger amount of heat will be required) as compared to B and C. And between B and C it will be hard to change the temperature of B.

So, if equal amount of heat is supplied, substance C must undergo a temperature change first, then the substance B and substance A in the end.

Therefore, the correct option is:

A.  Temperature of C will increase first, then B, then A.

This has a two word answer: sun's heat. The faster moving molecules near the ocean's surface are provided with enough energy from the sun to escape the surface they are near.

The average angular acceleration of the fan is

[tex]\alpha_{\rm avg}=\dfrac{\Delta\omega}{\Delta t}=\dfrac{200\frac{\rm rev}{\rm min}-1000\frac{\rm rev}{\rm min}}{2\,\rm s}=-6.67\dfrac{\rm rev}{\mathrm s^2}[/tex]

The number of revolutions after time [tex]t[/tex] is given by

[tex]\theta=\omega_0t+\dfrac{\alpha}2t^2[/tex]

Acceleration is assumed to be constant, so [tex]\alpha=\alpha_{\rm avg}[/tex] and over the 2 second interval we have

[tex]\theta=\left(1000\dfrac{\rm rev}{\rm min}\right)(2\mathrm s)+\dfrac{\alpha_{\rm avg}}2(2\,\mathrm s)^2=20\,\mathrm{rev}[/tex]

so the answer is A.

No of revolution is defined as the no of loops taken from the starting to the end to the end from the same point started. The number of the revolution blade experiences will be 20.

Angular acceleration is defined as the pace of change of angular velocity with reference to time.

[tex]\rm{\alpha _{avg}= \frac{w_f-w_i}{t_f-t_i}}[/tex]

given ,

initial angular velocity = 1000 rev/min = 16.67 rev /sec

final angular velocity =  200 rev/min = 3.33 rev/sec

time taken = 2 second

[tex]\rm{\alpha _{avg}= \frac{w_f-w_i}{t_f-t_i}}\\\\\\\rm{\alpha _{avg}= \frac{3.33-16.67}{0.3}}\\\\\\\rm{\alpha _{avg}= -6.67 rev/ sec^2[/tex]

According to newtons second law of motion,

No of revolution after time t

[tex]\rm{\theta= \omega_it+\frac{1}{2} \alpha t^{2} }\\\\\\\rm{\theta= \ 16.67\times2+\frac{1}{2}(-6.67)\times{2}^{2} }\\\\\\\rm{\theta= \ 16.67\times2-\frac{1}{2} (6.67)\times{2}^{2} }\\\\\\\rm{\theta= \ 16.67\times2-\frac{1}{2} (6.67)\times{2}^{2} }\\\\\\\rm{\theta= \ 33.3- 13.3}\\\\\\\rm{\theta= \ 20rev }\\\\\\[/tex]

Therefore the number of revolutions the blade undergoes will be 20 revolutions.

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All four values are in 3 sig. fig.

(a)

[tex]Q = C\cdot V = 20.0\times 10^{-6} \times 850\;\text{V} = 1.70\times 10^{-2}\;\text{J}[/tex].

(b)

Sum of the final charge on the two capacitors should be the same as the sum of the initial charge. Voltage of the two capacitors should be the same. That is:

[tex]C_1\cdot V_\text{final} +C_2 \cdot V_\text{final} = C_1\cdot V_\text{initial}[/tex];

[tex](C_1+C_2)\cdot V_\text{final} = C_1\cdot V_\text{initial}[/tex];

[tex]\displaystyle V_\text{final} = \frac{C_1}{C_1+C_2}\cdot V_\text{initial}\\\phantom{V_\text{final}} = \frac{20.0\;\mu\text{F}}{20.0\;\mu\text{F} + 12.0\;\mu\text{F}} \times 850\;\text{V}\\\phantom{V_\text{final}} =531\;\text{V}[/tex].

(c)

[tex]\displaystyle E = \frac{1}{2}\cdot C\cdot V^{2}[/tex].

[tex]\displaystyle E_\text{final} = \frac{1}{2} (C_1 + C_2) \cdot {V_\text{final}}^{2} \\\phantom{E_\text{final}} = \frac{1}{2} \times (20.0\times 10^{-6} + 12.0\times 10^{-6}) \times 531.25\\\phantom{E_\text{final}} = 4.52\;\text{J}[/tex].

(d)

Initial energy of the system, which is the same as the initial energy in the [tex]20.0\;\mu\text{F}[/tex] capacitor:

[tex]\displaystyle E_\text{initial} = \frac{1}{2} \times 20.0\times 10^{-6} \times 850^{2} = 7.225\;\text{J}[/tex].

Change in energy:

[tex]\Delta E = 7.225\;\text{J} - 4.516\;\text{J} = 2.70\;\text{J}[/tex].

The properties of the capacitors can be calculated the answers are:

     a) Q₁ = 1.70 10-2 C

     b) V_f = 531 V

     c) U_f = 4.52 J

     d) ΔU = 2.71 J

Given parameters

To find

     a) The initial charge

     b) The potential difference of the system connected capacitors

     c) The final energy of the system

     d) the energy change when connecting the capacitors

A capacitor is a system formed by two separate parallel plates that serves to store electrical charge,

          Q = C V

Where Q is the stored charge, C the capacitance and V the potential difference

a) ask for the initial charge

         Q₀ = C₁ V₀

         Q₀ = 20.0 10⁻⁶  850

         Q₀ = 1.70 10⁻² C

b) The law of conservation of charge establishes that the electric charge cannot be created or destroyed, therefore the initial charge (Q₀) must be distributed between the two connected capacitors

           Q₀ =  [tex]Q_{1f} + Q_{2f}[/tex]

           C₁ V₀ = C₁ [tex]V_{1f}[/tex]  + C₂  [tex]V_{2f}[/tex]

the Power Difference final  between the two capacitors must be the same, parallel connection

           C₁ V₀ = (C₁ + C₂) [tex]V_f[/tex]

           [tex]V_f[/tex] = [tex]\frac{C_1}{C_1+C_2} \ V_o[/tex]

           V_f = [tex]\frac{20}{20+12} \ 850[/tex]

           V_f = 531.25 V

c) The stored energy capacitor is

          U = ½ C V²

The final energy system is

          U = ½ (C₁ + C₂) [tex]V_f^2[/tex]

          U = ½ (20 + 12) 10⁻⁶  531.25²

          U = 4.516 J

d) To calculate the energy change

         ΔU = U₀ - [tex]U_f[/tex]

let's look for the initial energy

         U₀ = ½ C₁ V₀²

         U₀ = ½ 20 10⁻⁶  850²

         U₀ = 7.225 J

whereby the energy change is

         ΔU = 7.225 - 4.516

         ΔU = 2.71 J

In conclusion using the properties of the capacitors we were able to calculate the answers are:

         a) Q₁ = 1.70 10-2 C

         b) V_f = 531 V

         c) U_f = 4.52 J

         d) ΔU = 2.71 J

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Final answer:

The nucleus of an atom, which contains positively charged protons and neutral neutrons, is held together by the strong nuclear force. This force overcomes the repulsive electromagnetic force between protons, allowing the nucleus to remain stable.

Explanation:

The nucleus of an atom is at the center and contains protons and neutrons, known together as nucleons. While the protons carry a positive charge, the neutrons are neutral. The presence of protons with like charges would typically cause them to repel each other due to electromagnetic force. However, protons and neutrons in the nucleus do not fly apart because they are bound together by the strong nuclear force. This is a much stronger force than the electromagnetic force that causes like charges to repel each other and it is the key to keeping the nucleus stable, despite the repulsion between positively charged protons.

Many nuclei contain roughly equal numbers of protons and neutrons, with these nucleons making up most of the atom's mass. The atomic nucleus is incredibly dense and occupies only a tiny portion of the atom's volume, suggesting its strong nuclear forces are short-range but potent within that small space.

It's the strong nuclear force that prevents the nucleus from disintegrating under the repulsive force experienced by the protons. Without this force, the positive protons would indeed repel each other and the atom would not be stable. The strong nuclear force ensures that atoms can exist and form the matter that constitutes the world around us.

Neither source is a renewable one.

The lack of an atmosphere means convection cannot happen on the moon. Therefore, there is no form of heat dissipation on regions in direct sunlight. In addition, the lack of an atmosphere means there is no greenhouse effect on the moon. This is why regions facing away from sunlight are very cold.  

The moon's low gravity prevents it from retaining an atmosphere, which leads to drastic temperature changes due to the lack of an insulating layer of gases. Moreover the moon's surface's porous nature allows it to cool more rapidly than solid rock, contributing to the temperature extremes.

The primary reason for the large temperature differences on the moon's surface, despite the absence of an atmosphere, is related to the moon's gravity, surface composition, and the radiation from the Sun.

The moon has about one-sixth Earth's surface gravity. This is too low to retain an atmosphere. Gaseous molecules can easily escape from the moon into space, leaving it without an atmosphere. This means that there is no layer of gases to absorb and redistribute the Sun's energy, leading to extreme temperature fluctuations.

The fact that the moon's surface is also predominantly made up of lunar soil (also known as lunar regolith), which is porous and cools more rapidly than solid rock, aids in these temperature extremes. During lunar daytime when the Sun is high in the sky, the temperature can rise above the boiling point of water. However during the long lunar night, the temperature drops dramatically to approximately 100 K (-173 °C).

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A= -2 due to is a desacceleration

Answer:

[tex]a = -0.89 m/s^2[/tex]

Explanation:

initial speed of the car is given as

[tex]v_i = 20 miles/hour[/tex]

here we know that

[tex]v_i = 20 \times \frac{1609}{3600}[/tex]

[tex]v_i = 8.94 m/s[/tex]

final speed of the car is given as

[tex]v_f = 10 mile/hour[/tex]

[tex]v_f = 10 \times \frac{1609}{3600}[/tex]

[tex]v_f = 4.47 m/s[/tex]

now we have

[tex]a = \frac{v_f - v_i}{\Delta t}[/tex]

[tex]a = \frac{4.47 - 8.94}{5}[/tex]

[tex]a = -0.89 m/s^2[/tex]

Answer:

kettles: holes left by glaciers.

cirques: three-sided valleys

erratics: large, out-of-place rocks bouldersleft by glaciers.

drumlins: egg-shaped hills

Explanation:

Final answer:

Kettles are holes left by glaciers, drumlins are egg-shaped hills, erratics are large, out-of-place boulders, and cirques are three-sided valleys carved out by glaciers.

Explanation:

A. Kettles are holes left by glaciers. They are formed when a block of ice becomes buried in glacial sediments and then melts, leaving behind a depression or hole.

B. Drumlins are egg-shaped hills, typically found in clusters. They are formed when glaciers deposit sediments in elongated mounds parallel to the flow of the ice.

C. Erratics are large, out-of-place boulders that are transported by glaciers and left behind when the ice melts. They can be found in areas different from the type of rock they are made of.

D. Cirques are three-sided valleys carved out by glaciers. They are typically found at the head of a mountain valley and have steep walls.

In order of decreasing wavelength:

Radio Waves

Microwaves

Infrared light

Red light

Violet light

Ultraviolet light

X-rays

Gamma rays

Explanation:

The electromagnetic spectrum is a classification of all the electromagnetic waves depending on their wavelength and their frequency. The waves with longest wavelength are radio waves, which have wavelenght that spans from a few mm up to several km. The waves with shortes wavelength are gamma rays, which have wavelength shorter than 10 picometers.

More or less at the centre of the electromagnetic spectrum light the visible part of the spectrum, which is usually classified into 7 colors: red, orange, yellow, green, blue, indigo and violet (decreasing wavelength). Red light is the color with longer wavelenght (approx. 750 nanometers), while violet light has the shortest wavelength (approx. 380 nanometers).

Electromagnetic radiation can take multiple forms. The electromagnetic radiation with the longest wavelength is of radio while the shortest is of gamma rays.

Electromagnetic radiation can be defined as the flow of energy at the universal speed of light through a material medium or free space.

Some examples are radio waves, gamma-ray, etc.

The ranking of the following kinds of electromagnetic radiation in order of decreasing wavelength is shown below,

The topmost is having a low frequency, long wavelength, and low quantum energy, while the radiation at the bottom is having a high frequency, short wavelength, and high quantum energy.

Hence, the electromagnetic radiation with the longest wavelength is of radio while the shortest is of gamma rays.

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I think the situation is modeled by the scenario in the attached image. Some specific values seem to be missing (like the height of door [tex]d[/tex])...

The door forms a right triangles that satisfies

[tex]\tan\theta=\dfrac ab\implies\sec^2\theta\dfrac{\mathrm d\theta}{\mathrm dt}=\dfrac{b\frac{\mathrm da}{\mathrm dt}-a\frac{\mathrm db}{\mathrm dt}}{b^2}[/tex]

We also have

[tex]\tan\theta=\dfrac ab\implies\cos\theta=\dfrac bd[/tex]

so if you happen to know the height of the door, you can solve for [tex]b[/tex] and [tex]a[/tex].

[tex]d[/tex] is fixed, so

[tex]a^2+b^2=d^2\implies2a\dfrac{\mathrm da}{\mathrm dt}+2b\dfrac{\mathrm db}{\mathrm dt}=0\implies\dfrac{\mathrm da}{\mathrm dt}=-\dfrac ba\dfrac{\mathrm db}{\mathrm dt}[/tex]

We can solve for the angular velocity [tex]\dfrac{\mathrm d\theta}{\mathrm dt}[/tex]:

[tex]\dfrac{\mathrm d\theta}{\mathrm dt}=\cos^2\theta\dfrac{b\left(-\frac ba\frac{\mathrm db}{\mathrm dt}\right)-a\frac{\mathrm db}{\mathrm dt}}{b^2}=-\dfrac1a\dfrac{\mathrm db}{\mathrm dt}[/tex]

At the point when [tex]\theta=40^\circ[/tex] and [tex]\dfrac{\mathrm db}{\mathrm dt}=1.8[/tex] ft/s, we get

[tex]\dfrac{\mathrm d\theta}{\mathrm dt}=-\dfrac{1.8}a\dfrac{\rm deg}{\rm s}=-\dfrac{1.8}{d\sin40^\circ}\dfrac{\rm deg}{\rm s}[/tex]

The angular velocity of the door is [tex]\rm \( 0.744 \, \text{rad/s} \)[/tex], and the velocity of end D is approximately [tex]\rm \( 3.6 \, \text{ft/s} \)[/tex] in the horizontal direction and [tex]\rm \( 2.6964 \, \text{ft/s} \)[/tex] in the upward direction.

To determine the angular velocity of the door and the velocity of end D, we'll need to use the concept of relative velocity.

Let:

- [tex]\rm \( v_b = 1.8 \, \text{ft/s} \)[/tex] (velocity of wheel B)

- [tex]\( \theta = 40^\circ \)[/tex] (angle between horizontal and the line connecting wheel A to wheel B)

We'll use the following relations for the velocity components:

- For wheel B: [tex]\rm \( v_b = v_{b_x} \hat{i} + v_{b_y} \hat{j} \)[/tex]

- For end D: [tex]\rm \( v_d = v_{d_x} \hat{i} + v_{d_y} \hat{j} \)[/tex]

Since the door's velocity at point B is vertical [tex]\rm (\( v_{b_x} = 0 \))[/tex] and directed upward [tex]\rm (\( v_{b_y} = 1.8 \, \text{ft/s} \))[/tex], we can calculate the angular velocity [tex]\rm (\( \omega \))[/tex] of the door using the relative velocity formula:

[tex]\rm \[ \omega = \frac{{v_b}}{{R}} \][/tex]

Where [tex]\rm \( R \)[/tex] is the distance from point B to the axis of rotation.

Next, to calculate the velocity of end D, we use the following relationship:

[tex]\rm \[ v_d = v_b + \omega \times \vec{r} \][/tex]

Where [tex]\rm \( \vec{r} \)[/tex] is the position vector from point B to point D.

Since [tex]\rm \( \theta = 40^\circ \)[/tex], we can express [tex]\rm \( \vec{r} \)[/tex] as:

[tex]\rm \[ \vec{r} = r \cos(\theta) \hat{i} + r \sin(\theta) \hat{j} \][/tex]

Substituting the values and calculating:

[tex]\rm \[ R = r \cos(\theta) \]\\\\rm \omega = \frac{{v_b}}{{r \cos(\theta)}} \][/tex]

Now, substituting [tex]\rm \( v_b = 1.8 \, \text{ft/s} \), \( \theta = 40^\circ \), and \( r \)[/tex] (distance from B to D) in feet, we can find [tex]\rm \( \omega \)[/tex] and then the velocity [tex]\rm \( v_d \)[/tex].

(Note: Make sure to convert [tex]\rm \( \theta \)[/tex] to radians before plugging it into the equations.)

Let's assume the distance from B to D [tex](\( r \))[/tex] is 5 feet.

[tex]\[ \theta = 40^\circ = \frac{{40}}{{180}} \pi \, \text{radians} \approx 0.698 \pi \, \text{radians} \]\\\ \\R = 5 \cos(0.698 \pi) \approx 2.42 \, \text{ft} \]\\\\ \omega = \frac{{1.8}}{{2.42}} \approx 0.744 \, \text{rad/s} \][/tex]

Now, using [tex]\rm \( \vec{r} = 5 \cos(0.698 \pi) \hat{i} + 5 \sin(0.698 \pi) \hat{j} \)[/tex]:

[tex]\rm \[ v_d = 1.8 \hat{j} + (0.744 \, \text{rad/s}) \times (5 \cos(0.698 \pi) \hat{i} + 5 \sin(0.698 \pi) \hat{j}) \]\\\\\\rm v_d = 1.8 \hat{j} + 0.744 \, \text{rad/s} \times (2.42 \hat{i} + 3.63 \hat{j}) \]\\\ \\\rm v_d = 1.8 \hat{j} + 1.8 \hat{i} + 2.6964 \hat{j} \approx (1.8 + 1.8) \hat{i} + (2.6964) \hat{j} \approx 3.6 \hat{i} + 2.6964 \hat{j} \, \text{ft/s} \][/tex]

So, the angular velocity of the door is [tex]\rm \( 0.744 \, \text{rad/s} \)[/tex], and the velocity of end D is approximately [tex]\rm \( 3.6 \, \text{ft/s} \)[/tex] in the horizontal direction and [tex]\rm \( 2.6964 \, \text{ft/s} \)[/tex] in the upward direction.

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a is the right answer

C reactant. Both 2C0 and 02 are reactants and 2C02 is the product

Answer:

All the 4 options

Explanation:

Electromagnetic induction occurs when the magnetic flux through a coil of wire is changing over time:

[tex]\epsilon = -\frac{\Delta \Phi}{\Delta t}[/tex]

where

[tex]\epsilon[/tex] is the emf induced in the coil

[tex]\Delta \Phi[/tex] is the variation of magnetic flux

[tex]\Delta t[/tex] is the variation of time

The presence of an emf in the coil will generated an induced current.

The magnetic flux through the coil is given by

[tex]\Phi = BA cos \theta[/tex]

where

B is the intensity of the magnetic field

A is the area of the coil

[tex]\theta[/tex] is the angle between the direction of the field and the axis of the coil

We see that any actions that changes one of these 3 variables will change the magnetic flux through the coil, so it will also induce a current.

The 4 options are:

1.Moving the loop outside of the magnetic field region. --> this will decrease the intensity of the magnetic field, B, therefore it will change the flux, and it will induce a current

2.Spin the loop such that its axis does not consistently line up with the magnetic field direction. --> this will change the angle between the direction of the coil's axis and the field B, so this will also change the flux, and therefore will induce a current

3.Change the magnitude of the magnetic field. --> this will change the magnitude of B, so this will also change the flux, and therefore will induce a current

4.Change the diameter of the loop --> this will change the area of the coil A, so this will also change the flux, and therefore will induce a current

Therefore, all 4 options are correct.

An induced emf in a loop of wire can result from moving the loop out of the magnetic field region, spinning the loop to change its orientation, or changing the magnitude of the magnetic field itself.

An induced electromotive force (emf) occurs in a loop of wire when the magnetic flux through the loop changes. This can happen in several ways, such as:

Changing the diameter of the loop does not induce an EMF unless the magnetic field or the loop's orientation relative to the field changes at the same time.

Answer:

b. Decreases

Explanation:

The total resistance of a series circuit is equal to the sum of the individual resistances:

[tex]R_T=R_1+R_2+...+R_n[/tex] (1)

Therefore, as we add more lamps, the total resistance increases (because we add more positive tems in the sum in eq.(1).

The current in a circuit is given by Ohm's law:

[tex]I=\frac{V}{R_T}[/tex]

where V is the voltage provided by the power source and [tex]R_T[/tex] is the total resistance. We notice that the current, I, is inversely proportional to the total resistance: therefore, when more lamps are added to the series circuit, the total resistance increases, and therefore the current in the circuit decreases.