1. Three students are using sports equipment to model the relative motion of
Earth, the moon and the sun.
A. Identify which ball would be used to model the Sun, Earth and the Moo
B. Describe the relative (in relation to each other) motions of these bodies

Answer:

The postulates of the Kinetic Molecular Theory, KMT, are;

(1) In an ideal gas the molecules are in constant motion

(2) The collisions between molecules of gases are perfectly elastic

(3) The volume occupied by the molecules are negligible

(4) The temperature of the gas is directly proportional to its kinetic energy

(5) The intermolecular forces in the gas are negligible

According to the KMT, gaseous water vapor molecules are in constant motion and move at a speed that depends on their temperature. The intermolecular forces between the molecules are negligible and when they collide with the cold glass, they lose temperature to the glass, thereby reducing their temperature, kinetic energy and therefore, their speed is reduced.

The increasingly temperature of the water vapor coming in contact with the cold glass gives rise to reduced speed of the cooled gas molecules, thereby causing them to move closer together after having elastic collisions and to cluster with tiny particles in the air, to form tiny droplets

The rapid cooling on the cold glass surface causes the droplets to form rapidly on the cold glass surface which makes them visible as condensed water on the surface of the cold glass of water

Explanation:

Answer:

Explanation:

To find the horizontal component, the x component specifically, use the formula:

[tex]V_x=Fcos\theta[/tex] and for the vertical component, the y component, use the formula:

[tex]V_y=Fsin\theta[/tex]

where F is the magnitude of the force and theta is the angle in degrees.

For the x-component:

[tex]V_x=6cos250[/tex] so

[tex]V_x=-2.1[/tex] and depending upon whether this is a displacement vector or a velocity vector, the label would be meters/feet or m/s, respectively.

For the y-component:

[tex]V_y=6sin250[/tex] so

[tex]V_y=-5.6[/tex]

Answer:

One standard kilogram is define as the mass of platinium iridium cylinder having equal diameter and height kept at the particular condition of international bureo of weigh and measure sevre near paris.

Answer:

* Point charge outside the radius of the sphere r> R, the force in the two systems is the same

* Point charge inside the sphere  r <R,  therefore the force in the system with the insulating sphere is greater

Explanation:

To answer this question let's use the relation

          F = q E

with q being the point charge and E the electric field created by the sphere.

If we use Gauss's law

The electric field flux is proportional to the wax charge within the surface.

Let's analyze our situation.

* Point charge outside the radius of the sphere

          r> R

where R is the radius of the sphere and r the distance from the center of the sphere to the point charge

in this case the waxed charge for the insulating and conducting sphere is the same, therefore the force in the two systems is the same

* Point charge inside the sphere

           r <R

conductive sphere.

     As the charges are mobile, they are located on the surface of the sphere and there is no waxed charge within a Gaussian surface that passes through the point charge, therefore the electric field is zero and consequently the force

             F = 0

insulating sphere

      Charges cannot move therefore there is a fraction of charge within a surface that passes through the point charge, consequently the electric field is different from zero

            Fe> 0

for this second position the force on the conducting sphere is zero

therefore the force in the system with the insulating sphere is greater

Answer:

The answer is "4.97 m".

Explanation:

[tex]u = 1.23\ \frac{m}{s}\\\\[/tex]

[tex]H= 1.45 \ m\\\\[/tex]

[tex]\mu = 0.231\\\\[/tex]

The law of conservation tells us that heat energy at the top with kinetic energy at the top equals kinetic energy at the base.

[tex]mgh+\frac{1}{2}mu^2=\frac{1}{2}mv^2\\\\2gh +u^2 =v^2\\\\v=\sqrt{u^2+2gh}[/tex]

[tex]v=\sqrt{(1.23\ \frac{m}{s})^2+2(9.81 \frac{m}{s^2}) +(1.45\ m)[/tex]

   [tex]=\sqrt{1.5129+19.62 +1.45}\\\\=\sqrt{22.5829}\\\\=4.75\ \frac{m}{s}[/tex]

Friction force is given by the formula

[tex]f=-\mu mg \\\\ma= -\mu mg\\\\a=-\mu g\\\\[/tex]

[tex]= -(0.231) \ (9.81\ \frac{m}{s^2})\\\\=-2.26611 \ \frac{m}{s^2}[/tex]

Now by using an equation of motion as

[tex]v^2-u^2= 2as[/tex]

From the above the distance traveled is

[tex]S=\frac{v^2-u^2}{2a}[/tex]

[tex]S=\frac{(0)^2-(4.75\ \frac{m}{s})^2}{2(-2.26611\ \frac{m}{s^2})}\\\\[/tex]

   [tex]=\frac{-(4.75\ \frac{m}{s})^2}{2(-2.26611\ \frac{m}{s^2})}\\\\=\frac{-22.5625}{-4.53222}\\\\=4.97[/tex]

In other words, the distance from the bottom of the chute to the point where the toy zebra comes to rest is [tex]s = 4.97\ m[/tex]

Answer:

F = 32.28 N

Explanation:

For this exercise we must use the rotational equilibrium relation

          Σ τ = 0

In the initial configuration it is in equilibrium, for which all the torque and forces are compensated. By the time the payment lands on the bar, we assume that the counter-clockwise turns are positive.

          W_bird  L / 2 - F_left 0.595 - F_right 0.595 = 0

we assume that the magnitude of the forces applied by the hands is the same

          F_left = F_right = F

          W_bird L / 2 - 2 F 0.595 = 0

          F = [tex]\frac{m_{bird} \ g L} { 4 \ 0.595}[/tex]

we calculate

         F = 0.560 9.8 14.0 /2.38

         F = 32.28 N

Answer:

Option A.

Explanation:

We define the half time T as the time such that an initial quantity A reduces to its half.

So we can model the quantity as a function of time like:

P(t) = A*e^(-k*t)

Then for the half time, T, we will have:

P(T) = A/2 = A*e^(-k*T)

solving for k, we get:

A/2 = A*e^(-k*T)

1/2 = e^(-k*T)

ln(1/2) = ln( e^(-k*T)) = -k*T

-ln(1/2)/T = k

Here we know that the half time is T = 26.5s

if we input that in the above equation, we get:

-ln(1/2)/26.5s = k = 0.0262 s^-1

Then the correct option is A

Answer:

88 m ^2

Explanation:

Answer:

The orbital velocity an aircraft orbiting around a heavenly body is found as follows;

At the orbital velocity, [tex]F_G[/tex] = [tex]F_C[/tex]

Where;

[tex]F_G[/tex] = The gravitational force = [tex]\dfrac{G \cdot M \cdot m}{R_E^2}[/tex]

[tex]F_C[/tex] = The centripetal force = [tex]\dfrac{m \cdot v_0^2}{R_E}[/tex]

Therefore

[tex]v_0 = \sqrt{\dfrac{G \cdot M}{R_E} }[/tex]

Therefore, at the orbital velocity of the spacecraft, the centripetal force attracting the person away from the central region heavenly body is equal to the gravitational force pulling the person towards the center of the heavenly body (which was felt as her or his weight), and the person feels weightless while inside the orbiting spacecraft

Explanation:

Answer:

F = 9/5 C + 32     conversion from C to F

F = 9/5 * F/5 + 32

25 F = 9 F + 800

16 F = 800

F = 50  

Check:

C = 5/9 ( F - 32)  = 5/9 (50 - 32) = 10  as requested

Q = c m change in temp

Q = 1 cal/gm-deg C * 500 gm * 45 deg C = 22,500 calories

50 Fahrenheit heat required.

Heat transfer is a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy  between physical systems.

F = 9/5 C + 32     conversion from C to F

F = 9/5 * F/5 + 32

25 F = 9 F + 800

16 F = 800

F = 50

The answer is 50 Fahrenheit.

Learn more about heat transfer

https://brainly.com/question15086449

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Answer:

MgO relative atomic mass is 40

Explanation:

Mg=24

O=16

Answer:

2 m/s^2

Explanation:

a = ∆v / ∆t

a = (50 - 30) / 10

a = 20 / 10

a = 2 m/s^2

acceleration = u-v/t

50-30/10

=2m/s

Answer:

D. 2.8 hp

Explanation:

1 hp=746 watts

2088.8÷746= 2.8

Answer:

The atomic number of phosphorus is 15.

Explanation:

It’s found after Si(Silicon) and before S(sulphur)

Answer:

1) The kinetic energy of the object after it drops for 2.0 seconds is approximately 384.9 Joules

2) The potential energy of the object after it drops for 2.0 seconds is approximately 204 J

Explanation:

1) The given mass of the object, m = 2.0 kg

The height from which the object is dropped, h = 30 m

The kinetic energy of the object after it drops for 2.0 seconds = Required

Kinetic energy, K.E. = (1/2)·m·v²

Where;

v = The velocity of the object

The kinematic equation for finding the velocity of the object is presented as follows;

v = u + g·t

Where;

u = The initial velocity of the object = 0

g = The acceleration due to gravity of the object ≈ 9.81 m/s²

t = The time of motion of the object = 2.0 seconds

∴ The velocity after 2 seconds, v ≈ 0 + 9.81 m/s² × 2 s = 19.62 m/s

The kinetic energy, K.E. after 2 seconds as the object drops is given as follows;

[tex]K.E._{(after \ two \ seconds)}[/tex] = (1/2) × 2.0 kg × (19.62 m/s)² = 384.9444 J ≈ 384.9 J

2) The total energy, M.E. of the object at the top, h = 30 m, u = 0, is given as follows;

The total mechanical energy, M.E. = P.E. + K.E.

M.E. = m·g·h + (1/2)·m·u²

∴ M.E. = 2.0 kg × 9.81 m/s² × 30 m + 0 = 588.6 J

M.E. = 588.6 J

Given that the total mechanical energy, M.E., is constant, we have;

At 2.0 seconds, M.E. = 588.6 J , K.E. ≈ 384.9 J, P.E. = M.E. - K.E.

∴ P.E. = 588.9 J - 384.9 J ≈ 204 J

The potential energy after it drops 2.0 seconds, P.E. ≈ 204 J

Answer:

[tex] \sf \: given \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \bf \: mass \: \: \: m \: = 7.00 \times {10}^{ - 4} \: kg \\ \\ \bf \: E=mc^2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = > E=7.00 \times {10}^{ - 4} \times ({3 \times {10}^{8} })^{2} \\ \\ = > \green{ \boxed{ E = 6.3 \times {10}^{13} \: J}}[/tex]

Answer:

13.759 % of the initial mechanical energy is lost as thermal energy.

Explanation:

By the First Law of Thermodynamics we know that increase in internal energy of the object ([tex]U[/tex]), in joules, is equal to the lost amount of the change in gravitational potential energy ([tex]U_{g}[/tex]), in joules:

[tex]\frac{x}{100} \cdot \Delta U_{g} = \Delta U[/tex] (1)

Where [tex]x[/tex] is the percentage of the energy loss, no unit.

By definition of the gravitational potential energy and internal energy, we expand this equation:

[tex]\frac{x\cdot m \cdot g \cdot h}{100} = m\cdot c\cdot \Delta T[/tex] (1b)

Where:

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

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

[tex]h[/tex] - Initial height of the object above the lunar ground, in meters.

[tex]c[/tex] - Specific heat of aluminium, in joules per degree Celsius-kilogram.

[tex]\Delta T[/tex] - Temperature increase due to collision, in degree Celsius.

If we know that [tex]m = 0.95\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]h = 75\,m[/tex], [tex]c = 920\,\frac{J}{kg\cdot ^{\circ}C}[/tex] and [tex]\Delta T = 0.11\,^{\circ}C[/tex], then the percentage of energy loss due to collision is:

[tex]x = \frac{100\cdot c\cdot \Delta T}{g\cdot h}[/tex]

[tex]x = \frac{100\cdot \left(920\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (0.11\,^{\circ}C)}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (75\,m)}[/tex]

[tex]x = 13.759\,\%[/tex]

13.759 % of the initial mechanical energy is lost as thermal energy.

Answer:

i. The pressure of due to the water, P, is given according to the following equation;

P = ρ·g·h

Where;

ρ = The density of the water (a constant) = 997 kg/m³

g = The acceleration due to gravity = 9.81 m/s²

h = The height of the water (minimum h = h₁, maximum h = h₂)

The pressure is directly proportional to the water height, and we have;

The pressure, P, will be maximum when the water height, h, is maximum or h = h₂, which is the level DC

ii. The thrust = The force acting on the body = Pressure × Area

The maximum areas exposed to the water are on side AB and DC

However, the pressure at level DC, which is the location of the maximum pressure, is larger than the pressure at level AB, therefore, the maximum thrust will be at the level DC

Explanation:

Answer:

nothing

Explanation:

bocouse of darkness

Answer:

Explanation:

Answer:

It was created during the French Revolution in 1799 and has enabled for the international exchange of scientific and technical information. Calculating with SI units is also a lot easier than using the English system.

Answer:

friction reduces the efficiency of machines, thus we must reduce the friction force that is acting upon it.

Answer:

Because it causes a lot of wear and tear in machine parts that move against each other. It erodes the surfaces and destroys their symmetries

Explanation:

Answer:

The separation between the two stones is 36 m, when the second stone is approximately 10.9 m below the top of the cliff

Explanation:

The given parameters are;

The height of the cliff from which the stones are dropped, h = 60 m

The time at which the second stone is dropped = 1.6 seconds after the first

The distance below the top of the cliff when the distance between the two stones is 36 m = Required

We have;

The kinematic equation of motion that can be used is s = u·t - (1/2)·g·t²

For the first stone, we have, s₁ = u·t₁ - (1/2)·g·t₁²

For the second stone, we get; s₂ = u·t₂ - (1/2)·g·t₂²

t₁ = t₂ + 1.6

g = The acceleration due to gravity ≈ 9.81 m/s²

s = The distance below the cliff top

The initial velocity of the stones, u = 0

Let t represent the time from which the second stone is dropped at which the distance between the two stones is 36 m, we have;

s₁ = u·(t + 1.6) + (1/2)·g·(t + 1.6)²

s₂ = u·t + (1/2)·g·t²

u = 0

∴ s₁ - s₂ = 36 =  (1/2)·g·(t + 1.6)² - (1/2)·g·t²

2 × 36/(g) = (t + 1.6)² - t²  = t² + 3.2·t + 2.56 - t² = 3.2·t + 2.56

2 × 36/(9.81) = 3.2·t + 2.56

t = (2 × 36/(9.81) - 2.56)/3.2 =  ≈ 1.49 s

t ≈ 1.49 s

s₂ = (1/2)·g·t²

∴ s₂ = (1/2) × 9.81 × 1.49² ≈ 10.9

The distance below the top of the cliff of the second stone when the the separation between the two stones is 36 m, s₂ ≈ 10.9 m.

Answer:

8392

Explanation:

d=s/t

Answer:

A) How the size of a tree affect the bird species that prefer to live in it

Explanation:

I took the quiz

Answer:

force=mass×acceleration

hence

acceleration is given by force÷mass

(500÷200)*5=12.5

Explanation:

electrical energy into heat energy

electrical , thermal

Answer:

each resistor draws 1/3 of an amp or 0.33333 amps

Explanation:

V = I * R

V = 10 volts

R = 30 ohms

10 = I * 30       Divide by 30

10/30 = I

I = 0.33333

Answer:

a) The velocity of the object as a function of time, v(t) is 2·b·t

b) The acceleration of the function of time, a(t) is 2·b

c) The time at which the object is at the flagpole is t = √(c/b)

Explanation:

The function that gives the position of the object north of the flagpole, x(t) is presented as follows;

x(t) = b·t² - c (b and c are constants)

a) The velocity of the object as a function of time, v(t), is derived as follows

v(t) = x'(t) = d(b·t² - c)/dt = 2·b·t

The velocity of the object as a function of time, v(t) = 2·b·t

b) The acceleration of the function of time, a(t) = v'(t) = d(2·b·t)/dt = 2·b

c) The time at which the object is at the flagpole is given by the x-intercept of the function, where x(t) = 0, as follows;

At the x-intercept, we have, x(t) = 0 and x(t) = b·t² - c

∴ 0 = b·t² - c, which gives

b·t² = c

t² = c/b

t = ±√(c/b), we reject the negative value to get;

The time at which the object is at the flagpole, t = √(c/b).

All three cars experience the same impulse.

Impulse is equal to change in momentum.

Each car starts with the same amount of momentum and ends up with zero, so the magnitudes of all three changes are equal.

Answer:

Explanation:

A plane mirror is the kind you look into when you look into a "regular" mirror. The image you see is right-side-up. These images are virtual. Real images are always upside down and are made by mirrors that are "parabolic" in shape. Virtual images are always right-side-up.