The moment of inertia for a disk with mass of 500g and radius of 12cm rotated about its diameter is 7.2 × 10-³kgm².
How to calculate moment of inertia?Moment of inertia is a measure of a rigid body's resistance to change in its angular velocity around a given axis.
The formula for the moment of inertia is the “sum of the product of mass” of each particle with the “square of its distance from the axis of the rotation”. The formula of Moment of Inertia is expressed as follows:
I = mr²
Where;
I = moment of inertiam = mass r = radiusAccording to this question, a disk with mass of 500g and radius of 12 cm rotated about its diameter. The moment of inertia can be calculated as follows:
I = 0.5kg × 0.12²m
I = 7.2 × 10-³kgm²
Therefore, 7.2 × 10-³kgm² is the moment of inertia of the disk.
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a cylindrical space colony 13 km in diameter and 30 km long has been proposed as living quarters for future space explorers. such a habitat would have cities, land, and lakes on the inside surface and air and clouds in the center. all this would be held in place by the rotation of the cylinder about the long axis. how fast would such a cylinder have to rotate at its perimeter to produce a 1-g gravitational field at the walls of the cylinder?
0.00048 rad/s should be the angular velocity of the cylinder to produce a 1-g gravitational field at the walls of the cylinder.
The following problem requires understanding angular velocity and centripetal acceleration to obtain the result. The angular velocity of a body is defined by its virtue of velocity at a particular angle on an axis.
Centripetal acceleration is acceleration when the body is rotating and that acts towards the midpoint of the circular movement.
centripetal acceleration = rω² (derived from v²/r)
here
ω = angular velocity of the cylinder
r = radius of the space colony = 13km / 2 ⇒ 6.5km or 6500m
Substituting the values,
a = g ⇒ 9.8m/s²
ω = √9.8 / 6500 = 0.00048 rad/s
∴The cylinder should rotate at 0.00048 rad/s for producing a 1-g gravitational field.
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falling stars can only be observed by scanning the sky at night. given your knowledge of falling stars, what specific schedule of reinforcement maintains scanning the skies in search of falling stars?
The specific schedule of reinforcement maintains scanning the skies in search of falling stars is VI (Variable-interval schedule).
Which reinforcement schedule demands the end of a specific?Before reinforcement is given, a ratio plan demands the completion of a certain number of responses, whereas an interval schedule calls for the occurrence of at least one right response after a predetermined amount of time.
Opportunistic conditioning uses a schedule of reinforcement as a strategy. By setting up a gap between reinforcers and replies, the goal is to determine how and when a desired behavior occurs.
This is an illustration of a schedule with variable intervals, where the reinforcement is based on the passage of time but the intervals change in a random manner.
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A 5 kg block is pulled across the ground to the right by a tension force of 40 N with a frictional force of 8 N opposing the motion. What is the acceleration of the object?
The Acceleration of the object = 6.4 m/s²
Mass of block (m) = 5 kg
Action force on block, (F₁) = 40 N
Frictional force opposing the motion (F₂) = 8 N
Acceleration of the object (a) = ?
⇒ Net force = Action force on block - Opposing friction force
⇒ F = F₁ - F₂
⇒ F = 40 - 8
⇒ F = 32 N
Net force of the block (F) = 32 N
Mass of block (m) = 5 kg
F is the Force in N.
m is the Mass in kg.
a is the Acceleration in m/s².
F = ma
⇛ a = F/m
⇛ a = 32/5
⇛ a = 6.4 m/s²
a gyroscope flywheel of radius 2.83 cm is accelerated from rest at 14.2 rad/s2 until its angular speed is 2760 rev/min. (a) what is the tangential acceleration of a point on the rim of the flywheel during this spin-up process? (b) what is the radial acceleration of this point when the flywheel is spinning at full speed? (c) through what distance does a point on the rim move during the spin-up?
The tangential acceleration of the gyroscope flywheel is 40.186 x [tex]10^{-2[/tex] m/[tex]s^2[/tex], Radial acceleration of gyroscope flywheel is 2364.08 [tex]m/s^2[/tex] and During the spin up process the distance moved by a point on rim is 83.24 m.
What is meant by tangential acceleration?The centripetal force causes an acceleration known as radial acceleration, which is measured in radians per square second. Radial acceleration is directed toward the center.Tangential acceleration occurs when a body or object moves at a speed that is not uniform.a gyroscope flywheel of radius, R = 2.83 cm =2.83 x [tex]10^{-2}[/tex] m
[tex]\omega_0 = 0[/tex]
[tex]\omega = 2760 X \frac{2\pi}{60} rad/sec[/tex]
[tex]\alpha = 14.2 rad/ sec^2[/tex]
The tangential acceleration :
[tex]a_t = R\alpha[/tex] =2.83 x [tex]10^{-2}[/tex] m x 14.2 [tex]rad/s^2[/tex] =40.186 x [tex]10^{-2[/tex] m/[tex]s^2[/tex]
The radial Acceleration :
[tex]a_r = R\omega^2\\=(2.83 X 10^-2) (2760 X \frac{\pi}{30} )^2[/tex] =2364.08 [tex]m/s^2[/tex]
During the spin up process the distance moved by a point on rim
d = [tex]R\theta[/tex]
where [tex]\theta =\frac{ \omega^2 - \omega_0^{2}}{2\alpha}[/tex]
= [tex]\frac{(2760 X \pi/30)^2}{2 X 14.2}[/tex] =2941.42 rad
d = 2.83 x [tex]10^{-2}[/tex] x 2941.42 =83.24.m
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a car speeds up from 22 m/s to 26 m/s in 2 seconds. what is the average acceleration of the car
Answer:
2 m/s^2
Explanation:
Acceleration is change in velocity over change in time
(26-22) m/s / 2 s = 2 m/s^2
A spherical object has a certain density. If it is compressed under high pressure to half of its original diameter, its density will now be.
To calculate the density, known the volume and density formula.
What is density?
How tightly a material is packed together is determined by its density. The definition of it is the mass per unit volume. Density Formula: D = m/V, where D stands for density, m for mass, and V for volume, denotes the mass and volume of the object, respectively.
What is volume?
Any three-dimensional solid's volume is simply the amount of space it takes up. A cube, cuboid, cone, cylinder, or sphere can be one of these solids.
Let mass of the sphere be m and radius be r, then,
m= v × p ∴ v= volume
or
m= 4/3πr³× p⇒1
now new diameter, d¹= 1/4d and d=2r
so, d¹= 1/4 × 2r⇒ d¹=1/2r also,
d¹=2r¹ so, 2r¹ = 1/2r⇒ r¹= 1/4r
logs is same, there is no change in it then,
m=v¹×p¹
m=4/3πr¹^3ₓp¹
or m=4/3π(1/4r)^3×p¹⇒2
(1) is equal to (2) so,
4/3π(1/4r)^3×p¹=4/3πr^3×p
1/64r^3×p¹ = r^3×p
p¹=64p
Therefore denisty will be 64.
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Light passes straight through two vertical polarizers. The axes of the filters are aligned parallel to each other. How does the light that passes through the second filter compare to the light that passed through the first filter?.
The direction of the polarized light after passing the first and the second filter are the same, however the intensity of the light after passing the first and second polarizer are half and fourth of its original intensity, respectively.
Light polarization is a filter of electromagnetic waves such that it propagates into one transmission axis only.
When light passes through a polarizer, its intensity will decrease by half.
Suppose the intensity of unpolarized light is I₀, after passing the first polarizer, its intensity becomes:
I₁ = 1/2 . I₀
After passing the second polarizer, the intensity will further decrease by half, or:
I₂ = 1/2 . I₁
I₂ = 1/2 .1/2 . I₀ = 1/4. I₀
Since both polarizers are vertical filters, then the direction of the polarized light after passing the first and the second filter are the same.
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Calculate the average speed of a complete round trip in which the outgoing 300 km is covered at 97 km/h , followed by a 1.0- h lunch break, and the return 300 km is covered at 55 km/h .
The average speed of the complete round trip is 62.85 km/h.
The average speed is nothing but the total distance travelled per unit time.
or in other words,
Average speed = [tex]\frac{Total distance travelled}{total time taken}[/tex]
Now, Total distance travelled = 300km + 300 km = 600 km
and the time taken = t1 + t2 + t3
t1 = distance/speed
t1 = 300km/(97km/h)
t1 = 300/97 Hour = 3.0927 Hour
t2 = 1 Hour
and t3 = distance/speed
t3 = 300km/(55km/h)
t3 = 300/55 Hour = 5.4545 Hour
Total time = t1+t2+t3
T = 3.0927+1+5.4545 = 9.547 Hour
So, average speed = total distance / total time
Average speed = 600km/9.547 Hour
Average speed = 62.85 km/h
Hence the average speed is 62.85 km/h.
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A gas receives from an external thermal source an amount of heat equal to 1000 J. This energy, in addition to producing heating in the gas, causes its expansion, with the consequent performance of work equivalent to 600 J. What was the change in the internal energy of the gas? gas?
help someone help me
Hello..!
Subject: ThermodynamicsThe first law of thermodynamics relates work and transferred heat exchanged in a system through a new thermodynamic variable, internal energy. This energy is neither created nor destroyed, only transformed.
We can think of gas as a thermodynamic system, all because gases can work and absorb heat, and then they can turn all that into energy.
The formula for the change in energy is given by the first law of thermodynamics expressed as:
[tex] \: \: \: \: \: \: \: \: \: \: {\boxed{\boxed{ \sf\large \rm \Delta U = Q - W }}}[/tex]
Being:
ΔU = change in energyQ = added heatW = Work doneProblem:
A gas receives from an external thermal source an amount of heat equal to 1000 J. This energy, in addition to producing heating in the gas, causes its expansion, with consequent performance of work equivalent to 600 J. What was the change in the internal energy of the gas? gas?
Data:
ΔU = ¿? (Meet)Q = 1000 JW = 600 JNow adding the data in the formula to find the energy change:
[tex] \sf\large \rm \Delta U = 1000J - 600J[/tex]
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{\boxed{\boxed{\large \rm \Delta U = 400 J}}}[/tex]
[tex]\begin{gathered}\rule{7cm}{0.01mm}\\\texttt{Good studies! :D}\\\rule{7cm}{0.01mm}\end{gathered}[/tex]
two cars, car x and car y, are moving with the same velocity, and slam on the brakes coming to a stop. car x is more massive than car y. for which car is the amount of work done by friction in stopping it the highest?
The amount of work done by friction in stopping the car is given as
Wx > Wy.
There are two cars moving with the same velocity
car X is more massive than car Y
mass of car X = Mx
mass of car Y = My
here given, Mx > My
The distance travelled will be the same as the initial velocity, and it is independent of mass.
Let, μ be the coefficient of kinetic friction
work done by the frictional force is given as:
F = μMg
Fx = μMxg
Fy = μMyg
W = F.s
Wx = μMxg.s
Wy = μMyg.s
Wx > Wy
The amount of work done by friction in stopping the car is given as
Wx > Wy.
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The oxidation of copper(I) oxide, Cu 2 O ( s ) , to copper(II) oxide, CuO ( s ) , is an exothermic process. 2 Cu 2 O ( s ) + O 2 ( g ) ⟶ 4 CuO ( s ) The change in enthalpy upon reaction of 98. 93 g Cu 2 O ( s ) is − 100. 9 kJ. Calculate the work, w , and energy change, Δ U rxn , when 98. 93 g Cu 2 O ( s ) is oxidized at a constant pressure of 1. 00 bar and a constant temperature of 25 ∘ C. Note that Δ E rxn is sometimes used as the symbol for energy change instead of Δ U rxn
Copper II oxide is created through the oxidation of copper I oxide, cu2os cu2os. Cuo s An exothermic reaction is Cuo S. The formula is 2cu2o s)+ o2 G 4cuo S.
A simple definition of oxidation is what?Cu2O(s), a copper(I) oxide, is oxidized to copper(II) oxide, CuO, by exothermic processes (s). Cu2O(s), a copper(I) oxide, is oxidized to copper(II) oxide, CuO, by exothermic processes (s).
An oxidation reaction is a chemical process that takes place when an object comes into contact with oxygen or another oxidizing agent (OK-sih-DAY-shun). Oxidation is characterized by the brown hue of sliced apples and rust. In the past, the term "oxidation" was used to describe any chemical reaction that involved oxygen and an element. When magnesium metal and oxygen react, magnesium oxide is produced as a result, which serves as an illustration of this.
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two balls of mass 4kg and 2kg are moving with speed 10 m/s and 8m/s respectively with the ball of heavier mass behind the lighter ball
Answer:
Explanation:
And what to find in the problem?
when an astronomer rambles on and on about the luminosity of a star she is studying, she is talking about:
When an astronomer rambles on and on about the luminosity of a star she is studying, she is talking about the amount of energy the star gives off each second.
Luminosity and apparent brightness are about brightness, but from a different point of view. The difference between luminosity and apparent brightness is that luminosity tells us exactly how bright a star really is while apparent brightness only tells us its brightness seen from Earth.
L / Lsun = ( d / dsun )² = b / bsun
L = Luminosity of a star
Lsun = Luminosity of sun
d = Distance of star
dsun = Distance of sun
b = Brightness of star
bsun = Brightness of sun
Therefore, when an astronomer rambles on and on about the luminosity of a star she is studying, she is talking about the amount of energy the star gives off each second.
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5. In an emergency, the driver of a 1.3 x 10³-kg car slams on the brakes, causing the car to skid forward on the road. The coefficient of kinetic friction between the tires and the road is 0.97, and the car comes to a stop after travelling 27 m horizontally. Determine the work done by the force of friction during the skidding.
The work done by the force of friction during the skidding, given that the car comes to a stop after travelling 27 m horizontally is 333660.6 J
How do I determine the work done?We'll begin by obtaining the frictional force. This can be obtained as follow:
Mass (m) = 1.3 x 10³ KgAcceleration due to gravity (g) = 9.8 m/s²Normal reaction (N) = mg = 1.3 x 10³ × 9.8 = 12740 NCoefficient of kinetic friction (μK) = 0.97Frictional force (F) = ?F = μKN
F = 0.97 × 12740
Frictional force = 12357.8 N
Finally, we shall determine the work done by the force of friction. This is illustrated below:
Frictional force = 12357.8 NDistance (d) = 27 mWorkdone (Wd) =?Work done (Wd) = force (F) × distance (d)
Wd = 12357.8 × 27
Wd = 333660.6 J
Thus, we can conclude that the work done by the force of friction is 333660.6 J
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after the switch is closed for a little while, the current through r1 is 1.4 a. what is the voltage across the capacitor?
To calculate the voltage across the capacitor is (1.44)R.
What is speed?
The speed at which electrons move past a particular location in an electrical circuit is known as the "current." In the most basic terms, current = flow. The international unit for measuring current is the ampere, pronounced "amp" (AM-pir).
What is voltage?
Voltage is the "pressure" under which electricity is pushed. Higher voltages result in more electricity flowing to an electronic device. The amount of voltage is measured in units known as volts (V).
I=1.4A
V=?
V=(1.44)R
R⇒ resistance
Therefore, the voltage across the capacitor is (1.44)R.
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flux i - suppose just a north pole of a permanent magnet is placed inside a gaussian surface. what is a true statement about the magnetic flux through the gaussian surface in this situation?
The true statement about magnetic flux through the Gaussian surface when a north pole of a permanent magnet is placed inside a Gaussian surface in this situation is "The net flux is zero," option A.
What is the magnetic flux through any surface?Magnet flux through a surface, in physics, is a surface that is integral of a normal component with a magnetic field, B over the given surface. This means that if a section of a magnet is enclosed in a chosen Gaussian surface, the net magnetic flux moving through the surface will be zero.
This magnetic flux continues inside the magnet and it leaves the Gaussian surface where the magnet is cut.
The full question is:
flux i - suppose just a north pole of a permanent magnet is placed inside a Gaussian surface. what is a true statement about the magnetic flux through the Gaussian surface in this situation?
a. The net flux is zero
b. The net flux is positive
c. The net flux is negative
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if the ladder is just on the verge of slipping when the firefighter is 9.10 m from the bottom, what is the coefficient of static friction between ladder and ground?
Static friction between the ladder and the ground has a coefficient of
[tex]\mu_{s}[/tex] = 0.303.
A measure of static friction.The maximum static friction force (F) that can exist between two surfaces before movement starts is divided by the normal force (N) to determine the coefficient of static friction.
x₁ = 8
x₂ = a₁
x₃ = x₁ cos θ
x₃ = 8 cos61 = 3.87m
x₄ = x₂cosθ = 9.1cos 61
x₄ = Lsin 60 = 16sin 60
[tex]x_{s}[/tex] = 13.856m
w₁ = mg = 840
w₂ = mg = 490
N₁ = mg + mg = 490+840 =1330N
N₂ = f = [tex]\mu_{s}[/tex]N₁
Torque about point is zero
[tex]\sum[/tex]Z = 0
mgx₃ +mgx₁ = N₂[tex]x_{s}[/tex]
N₂ = 490×387+840×4.41/13856
N₂ = 404.268
[tex]\mu_{s}[/tex] = N₂/N₁ = 404.268/1330 = 0.303
[tex]\mu_{s}[/tex] = 0.303
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a skater spins with an angular speed of 12.0rad/s with his arms outstretched. he lowers his arms, decreasing his moment of inertia from 41kgm2 to 36kgm2. a. calculate his initial and final rotational kinetic energy. b. does his rotational kinetic energy increase or decrease? c. what causes this change in kinetic energy?
It has a 2952 J initial rotational energy.
492 kg-m2/s represents its moment of momentum.
13.67 rad/s of angular momentum is conserved.3364J
A) is the final rotational kinetic energy. The rotational energy of it at first is
KE=[tex]\frac{1}{2}\\[/tex]Iω²
=[tex]\frac{1}{2}\\[/tex]
=[tex]2952J[/tex]
B) Her angular momentum is
L=Iω
=(41)(12)
= 492 kg-m²/s
C) Angle momentum is kept constant. Therefore, even if he draws his hands in, nothing will change. She still has 492 kg-m2/s of angular momentum after pulling her arms back. Her moment of inertia has altered, though. L must therefore change in order for its angular velocity to remain constant.
L=Iω
ω=L/I
=492/36
= 13.67 rad/s
D) So her final rotational kinetic energy is
KE=[tex]\frac{1}{2}\\[/tex]Iω²
=[tex]\frac{1}{2}\\[/tex][tex](36) (13.67)[/tex]
=[tex]3364J[/tex]
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the top of a ladder slides down a vertical wall at a rate of 0.15 mys. at the moment when the bottom of the ladder is 3 m from the wall, it slides away from the wall at a rate of 0.2 mys. how long is the ladder?
The length of the ladder which is placed 3 meters from the wall is 5 meters
The rate of the change of ladder decreasing in vertical wall = - 0.15 m/s
The distance between the bottom of the ladder and the wall = 3 meter
The rate of change of the bottom of the ladder away from the wall = 0.2 m/s.
The length of the ladder can be found using the Pythagoras theorem,
x² + y² = L²
where x is the distance between the bottom of the ladder and the wall
y is the distance from the top of the ladder to the bottom of the wall.
L is the length of the ladder
Let us differentiate in terms of the rate of change in the above equation,
2x dx/dt + 2y dy/dt = 0
Now let us substitute the known values,
2(3)(0.2) + 2y(-0.15) = 0
1.2 - 0.3y = 0
0.3y = 1.2
y = 1.2 / 0.3
= 4
Then, the length of the ladder is
3² + 4² = L²
9 + 16 = L²
L = √25
L = 5
Therefore, the length of the ladder is 5 meters
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suppose we compare to something that simply slides with zero friction (but doesn't roll at all). should our rolling cylinder be released from the same height, or from a larger height, or from a smaller height if we would like our cylinder to (barely) completely make it through the circle?
The cylinder must be released from a greater height for it to complete the circle.
From the concept of work-energy theorem we know that for the sliding object, the beginning work done/ energy of the object is the potential energy due to gravity at a height.
P.E = m.g.h
Here h=height
At the top of the circle, total energy will include both Kinetic energy and potential energy
From conservation of energy, it’s inferred
m.g.h= (½ mv²) + (mg(2R))
Here R=radius of circular motion
Taking m common
m.g.h= m(v²/2 + 2gR)
Cancelling m
We get height for releasing the object as
h= (v²/2g) + (2R)
The similar formulas can be applied for the rolling object as well
but along with rotational energy
m.g.h = (½ mv²) + (½ Iω²) + (mg(2R))
Here I = moment of inertia of the rolling object
And angular velocity is denoted by ω
h = ½ (v²/g) + ½ (Iω²/mg)+ 2R
From the extra term in the above equation i.e. ½ (Iω²/mg) it’s deduced that the release height is more in case of a rolling object.
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At rest, a car's horn has a frequency of
395 Hz. Car A passes car B on the street
in the same direction. If car A is traveling
at 22.0 m/s and car B is traveling
at 19.5 m/s, what frequency does
car B hear when car A honks?
(Speed of sound = 343 m/s)
(Unit = Hz)
The frequency heard by car A is determined as 398.4 Hz.
What is the frequency heard by car A?
The frequency heard by car A is determined by applying the following equation.
f = fs(v - v₀) / (v - vs)
where;
v is the speed of sound = 343 m/sv₀ is the speed car B = 19.2 m/svs is the speed of car A = 22 m/sfs is the frequency of car A = 395 Hzf is the frequency of car B = ?f = 395(343 - 19.2) / (343 - 22)
f = 395(1.0087)
f = 398.4 Hz
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How long does it take a car to travel 330 m at a speed of 30.0 m/s
It will take 11 seconds for a car to travel 330 m at a speed of 30.0 m/s
By going through the formula of calculating the speed of the vehicle
that is, S=D/T
We get the equation T=D/S.........(i)
Where T stands for Time
D stands for Distance
S stands for Speed
Now, as per the given question
Distance(D)= 330 meters
Speed(S)= 30.0 meters per second
Time(T)= ?
Putting the values in equation (i),
T=330 meters/30 meters per second
T = 11 seconds.
Therefore, A car moving at a speed of 30.0 meters per second will traverse 330 meters in 11 seconds.
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if more massive stars have more hydrogen in their cores than less massive stars, why do they have shorter lifetimes?
Thermonuclear fusion occurs faster in massive stars so larger stars use up all their fuel in less time.
The more mass a star has the faster it will exhaust its fuel supply and the shorter its lifespan. The most massive stars could burn up and explode in a supernova after just a few million years of nuclear fusion. Massive stars are the largest, hottest, and brightest main-sequence stars and are blue, blue-white, or white in color.
Massive stars run out of hydrogen fuel very quickly and therefore have short lives. This is because the more massive the star the greater the fuel consumption. Even if a high-mass star has more fuel it uses it up very quickly so it does not live as long as a low-mass star. A main-sequence star's mass determines the fundamental properties of its luminosity surface temperature radius and lifetime.
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a ball is thrown straight up. it passes a 2.45 m high window 7.45 m off the ground on its path up and takes 1.45 s to go past the window. what was the ball's initial velocity (in m/s)?
The ball's initial velocity is 14 m/s
Since the equation of motion we referring to for calculating the initial velocity are :
s=ut+1/2at² ....1
v²= u²+2as .... 2
where u and v are initial and final velocity, t is the time taken, a is the acceleration due to gravity and s is the distance or we can also say the height .As we are given h₁ and h₂ which are 2.45 m and 7.45 m and the time is taken, t which is 1.45 s, and at last we are having a = -9.8ms^-2
using the 2 equations, we get
v₁= √(u²−2gh₁),
from equation 1, we get
h₂ =√(u₂−2gh₁t)−1/2gt₂
there the initial velocity will be:
u= √(((h₂+1/2gt²)/t)²+2gh1
=√(((7.35+1/2*9.80*(1.45)^2)/1.45)^2+2×9.80×2.45)
= 14
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Use the information from the graph to answer the
What is the total displacement of the object?
question.
m
Velocity vs. Time
Velocity (m/s)
40
30
20
10
0
Time (s)
0
5
10
15
20
25
Answer:
The total displacement 560 m
Explanation:
Given:
t₀ = 0 s
t = 25 s
V₀ = 10 m/s
V = 35 m/s
___________
D - ?
Acceleration:
a = (V - V₀) / (t - t₀) = (35 - 10) / (25 - 0) = 25 / 25 = 1.0 m/s²
The total displacement:
D = V₀·t + a·t² / 2
D = 10·25 + 1.0·25² / 2 ≈ 560 m
3) mary looked in her science book at a picture of the solar system. the planets were large and colorful but she knew it was not an accurate model. why?
mary looked in her science book at a picture of the solar system. the planets were large and colorful but she knew it was not an accurate model. The planets should have been much farther apart.
The gravitationally bound system of the Sun and the satellites in its orbit is referred to as the Solar System. The gravitational collapse of a massive interstellar molecular cloud gave it birth 4.6 billion years ago. The Sun is the system's primary mass, accounting for 99.86% of its total mass, with Jupiter making up the majority of the remaining mass. Mercury, Venus, Earth, and Mars are the four planets in the inner solar system, and they are all terrestrial planets with rocky and metallic cores. In comparison to the terrestrial planets, the four giant planets of the outer solar system are significantly bigger and more massive. The two biggest, Jupiter and Saturn, are gas giants made primarily of hydrogen and helium; the next two, Uranus and Neptune, are ice giants made primarily of highly volatile substances.
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if someone asked you whether their house could be heated with solar heat, what characteristics would you look for?
Solar house would have linear east-west floor designs with huge south-facing windows & roof overhangs to protect against the intense summer sun.
South-facing glass & thermal mass are the two main components of passive solar heat since they both collect, store, and disperse heat. The methods for putting such aspects into practice vary. A solar collector, thermal absorber, and distribution system make up the real living area.
This collection includes sizable circular or linear seating components that come with adjustable backrests and corner backrests.
The ideal windows for daylighting are those facing south. They provide all-day illumination and are less affected by heat and glare than windows with an east or west orientation.
The eaves of the house or the rafters on your roof are two names for this overhang. When this overhang is done, the bottom is referred to as the soffit, meaning literally translates to "something attached beneath".
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(a) the control in the experiment was 0 khz, or no sound. what information is gained by using the control with no sound that could not be obtained if no control were used?
In order to prevent deer from crossing the highway, the experiment involved placing sound-emitting equipment at the perimeter.
A procedure used to confirm or deny a hypothesis, as well as assess the effectiveness or likelihood of something that has never been tried before, is called an experiment. By showing what happens when a particular factor is manipulated, experiments shed light on cause-and-effect relationships. Experiments have a wide range of objectives and sizes but always rely on a repeatable process and a logical analysis of the outcomes. Additionally, there are experiment conducted in nature.
While teams of scientists may spend years conducting methodical research to advance their understanding of a phenomenon, a child may conduct simple experiment to understanding how things fall to the ground. In the science classroom, students benefit greatly from experiments and other hands-on activities. Experiments can improve test results and increase student engagement.
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kinetic energy problem an 82-kg stunt man falls off the top of a large building. after falling for 65 meters he has a speed of 22 m/s. assuming he was initially at rest, what was the average force of air resistance exerted on the stunt man during his fall?
The average force of air resistance exerted on the stunt man during his fall is 305.29 N
How do I determine the average force?We'll beging by obtaining the kinetic energy of the stunt man. This can be obtained as follow:
Mass (m) = 82 KgVelocity (v) = 22 m/sKinetic energy (KE) =?KE = ½mv²
KE = ½ × 82 × 22²
KE = 41 × 484
KE = 19844 J
Finally, we shall determine the average force of air resistance exerted on the stunt. This can be obtained as follow:
Distance = 65 metersEnergy = 19844 JAverage force =?Energy = work
Work (W) = force (F) × distance (d)
W = Fd
Thus,
E = W = Fd
19844 = F × 65
Divide both sides 65
F = 19844 / 65
F = 305.29 N
Thus, from the above calculation, we can conclude that the average force is 305.29 N
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The temperature, pressure, and precipitation conditions of the atmosphere for a specific place on a given day is called , whereas a longer term view of these same factors, typically taken over a period of many years, is called.
The temperature, pressure and precipitation condition of the atmosphere for a specific place on a given day is called weather, whereas are long term view of the same factors, typically taken over a period of many years, is called climate.
The basic difference between climate and weather is that climate exist for a very long period of time while weather can change in just a few hours.
Weather is just a way to way information about the temperature, pressure and precipitation condition of a atmosphere on a day today basis.
While climate is a kind weather information which provides information about the weather of a reason for a long period of time.
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