Answer:
The angular momentum is [tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]
Explanation:
From the question we are told that
The mass of the woman is [tex]m = 50 \ kg[/tex]
The angular speed of the rim is [tex]w = 0.80 \ rev/s = 0.8 * [\frac{2 \pi}{1} ] = 5.024 \ rad \cdot s^{-1}[/tex]
The mass of the disk is [tex]m_d = 110 \ kg[/tex]
The radius of the disk is [tex]r_d = 4.0 \ m[/tex]
The moment of inertia of the disk is mathematically represented as
[tex]I_D = \frac{1}{2} m_d r^2_d[/tex]
substituting values
[tex]I_D = \frac{1}{2} * 110 * 4^2[/tex]
[tex]I_D = 880 \ kg \cdot m^2[/tex]
The moment of inertia of the woman is
[tex]I_w = m * r_d^2[/tex]
substituting values
[tex]I_w = 50 * 4^2[/tex]
[tex]I_w =800\ kg[/tex]
The moment of inertia of the system (the woman + the large disk ) is
[tex]I_t = I_w + I_D[/tex]
substituting values
[tex]I_t = 880 +800[/tex]
[tex]I_t =1680 \ kg \cdot m^2[/tex]
The angular momentum of the system is
[tex]L = I_t w[/tex]
substituting values
[tex]L = 1680 * 5.024[/tex]
[tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]
How can socialism
impact populations?
Answer:
it represents a fundamental difference. (more info below)
Explanation:
Production is incessantly developing and expanding in socialist countries, and employment is guaranteed for the entire productive population. Consequently, the relative overpopulation problem has been eliminated. This represents the fundamental difference between socialism's demographic law and capitalism's law.
hope this helped!
A scuba diver and her gear displace a volume of 68.5 L and have a total mass of 71.8 kg . Part A What is the buoyant force on the diver in sea water? FB = nothing N Request Answer Part B Will the diver sink or float?
Answer:
A) Fb = 671.3 N
B) The diver will sink.
Explanation:
A)
The buoyant force applied on an object by a fluid is given by the following formula:
Fb = Vρg
where,
Fb = Buoyant Force = ?
V = Volume of the water displaced by the object = 68.5 L = 0.0685 m³
ρ = Density of Water = 1000 kg/m³
g = 9.8 m/s²
Therefore,
Fb = (0.0685 m³)(1000 kg/m³)(9.8 m/s²)
Fb = 671.3 N
B)
Now, in order to find out whether the diver sinks or float, we need to find weight of the diver with gear.
W = mg = (71.8 kg)(9.8 m/s²)
W = 703.64 N
Since, W > Fb. Therefore, the downward force of weight will make the diver sink.
The diver will sink.
A diver shines light up to the surface of a flat glass-bottomed boat at an angle of 30° relative to the normal. If the index of refraction of water and glass are 1.33 and 1.5, respectively, at what angle (in degrees) does the light leave the glass (relative to its normal)?
A. 26
B. 35
C. 42
D. 22
E. 48
Answer:
35Explanation:
According to snell's law which states that the ratio of the sin of incidence (i) to the angle of refraction(n) is a constant for a given pair of media.
sini/sinr = n
n is the constant = refractive index
Since the diver shines light up to the surface of a flat glass-bottomed boat, the refractive index n = nw/ng
nw is the refractive index of water and ng is that of glass
sini/sinr = nw/ng
given i = 30°, nw = 1.33, ng = 1.5, r = angle the light leave the glass
On substitution;
sin 30/sinr = 1.33/1.5
1.5sin30 = 1.33sinr
sinr = 1.5sin30/1.33
sinr = 0.75/1.33
sinr = 0.5639
r = arcsin0.5639
r ≈35°
angle the light leave the glass is 35°
A transverse wave is traveling through a canal. If the distance between two successive crests is 2.37 m and four crests of the wave pass a buoy along the direction of travel every 22.6 s, determine the following.
(a) frequency of the wave. Hz
(b) speed at which the wave is traveling through the canal. m/s
Answer:
(a) 0.0885 Hz
(b) 0.21 m/s
Explanation:
(a) Frequency: This can be defined as the number of cycle completed in one seconds.
From the question,
Note: 2 crest = one cycle,
If four crest = 22.6 s,
Then two crest = (22.6/2) s
= 11.3 s.
T = 11.3 s
But,
F = 1/T
F = 1/11.3
F = 0.0885 Hz.
(b)
Using,
V = λF...................... Equation 1
Where V = speed of wave, F = Frequency of wave, λ = wave length.
Given: F = 0.0885 Hz, λ = 2.37 m.
Substitute these values into equation 1
V = 2.37(0.0885)
V = 0.21 m/s.
Car A is traveling at twice the speed of car B. They both hit the brakes at the same time and decrease their velocities at the same rate. If car B travels a distance D before stopping, how far does car A travel before stopping?
A) 4D
B) 2D
C) D
D) D/2
E) D/4
Answer:
A) 4D
Explanation:
The distance traveled by the cars before coming to rest can be determined by 3rd equation of motion:
2as = Vf² - Vi²
s = (Vf² - Vi²)/2a
where,
s = distance traveled
Vf = Final Speed = 0 m/s
Vi = Initial Speed
a = deceleration rate
First, we consider Car B and we assign a subscript 2 for it:
Vf₂ = 0 m/s (As, car finally stops)
s₂ = D
a₂ = - a (due to deceleration)
D = (0² - Vi₂²) /(-2a)
D = Vi₂²/2a -------- equation (1)
Now, we consider Car A and we assign a subscript 1 for it:
Vf₁ = 0 m/s (As, car finally stops)
s₁ = ?
a₁ = - a (due to deceleration)
Vi₁ = 2 Vi₂ (Since, car A was initially traveling at twice speed of car B)
s₁ = (0² - Vi₁²) /(-2a)
s₁ = (2Vi₂)²/2a
s₁ = 4 (Vi₂²/2a)
using equation (1), we get:
s₁ = 4D
Therefore, the correct option is:
A) 4D
Constants Canada geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at speeds up to about 100 km/h. The one goose is flying at 100 km/h relative to the air but a 44 km/h wind is blowing from west to east.
1. At what angle relative to the north-south direction should this bird head so that it will be traveling directly southward relative to the ground?2. How long will it take the bird to cover a ground distance of 450 from north to south?
Answer:
a. 63.89° in the north-southward manner
b. 2.2 sec
Explanation:
The goose is flying at 100 km/h
Air from east to west is 44 km/h
angle relative to the north-south direction for the bird to travel south will be
cos∅ = 44/100 = 0.44
∅ = [tex]cos^{-1}[/tex]0.44 = 63.89° in the north-southward manner
Speed south relative to the ground will be v
Tan 63.89 = v/100
2.04 = v/100
v = 2.04 x 100 = 204 km/hr
to cover a distance of 450 m from north to south at this speed time will be
t = d/v = 450/204 = 2.2 sec
help yall 13 points!!
Answer:
Explanation:
12.)
A. Opposite poles attract
B. Same poles repel
13.)
IDK
50 points!! please help :((
Answer:
Loudness: decreases
Amplitude: decreases
Pitch: stays the same
Frequency: stays the same
Explanation:
1.
An oscilloscope measures how much the microphone is vibrating, or how much electricity it is sending. This means that a louder noise will register higher on the oscilloscope. Since the size of the waves at Y is lower than at X, the loudness of the sound has decreased.
2.
Similarly to loudness, amplitude measures how far the crests of the waves are from the nodes. Since Y is closer to the center line than X, it has a lower amplitude.
3 and 4.
The pitch and frequency, for our purposes, are essentially the same thing here. They are dependent on how close together the waves on the oscilloscope are, or how quickly the microphone is vibrated. Since this stays the same throughout the entire sound, they both stay the same.
Hope this helps!
(a) What is the cost of heating a hot tub containing 1440 kg of water from 10.0°C to 40.0°C, assuming 75.0% efficiency to take heat loss to surroundings into account? The cost of electricity is 9.00¢/(kW · h) and the specific heat for water is 4184 J/(kg · °C). $ 67 Incorrect: Your answer is incorrect. How much heat is needed to raise the temperature of m kg of a substance? How many joules are in 1 kWh? (b) What current was used by the 220 V AC electric heater, if this took 3.45 h? 88.2 Correct: Your answer is correct. A
Answer:
a) [tex]E = 6.024\,USD[/tex], For m kilograms, it is 4184m J., 3600000 joules, b) [tex]i = 88.200\,A[/tex]
Explanation:
a) The amount of heat needed to warm water is given by the following expression:
[tex]Q_{needed} = m_{w}\cdot c_{w}\cdot (T_{f}-T_{i})[/tex]
Where:
[tex]m_{w}[/tex] - Mass of water, measured in kilograms.
[tex]c_{w}[/tex] - Specific heat of water, measured in [tex]\frac{J}{kg\cdot ^{\circ}C}[/tex].
[tex]T_{f}[/tex], [tex]T_{i}[/tex] - Initial and final temperatures, measured in [tex]^{\circ}C[/tex].
Then,
[tex]Q_{needed} = (1440\,kg)\cdot \left(4184\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (40^{\circ}C - 10^{\circ}C)[/tex]
[tex]Q_{needed} = 180748800\,J[/tex]
The energy needed in kilowatt-hours is:
[tex]Q_{needed} = 180748800\,J\times \left(\frac{1}{3600000}\,\frac{kWh}{J} \right)[/tex]
[tex]Q_{needed} = 50.208\,kWh[/tex]
The electric energy required to heat up the water is:
[tex]E = \frac{50.208\,kWh}{0.75}[/tex]
[tex]E = 66.944\,kWh[/tex]
Lastly, the cost of heating a hot tub is: (USD - US dollars)
[tex]E = (66.944\,kWh)\cdot \left(0.09\,\frac{USD}{kWh} \right)[/tex]
[tex]E = 6.024\,USD[/tex]
The heat needed to raise the temperature a degree of a kilogram of water is 4184 J. For m kilograms, it is 4184m J. Besides, a kilowatt-hour is equal to 3600000 joules.
b) The current required for the electric heater is:
[tex]i = \frac{Q_{needed}}{\eta \cdot \Delta V \cdot \Delta t}[/tex]
[tex]i = \frac{180748800\,J}{0.75\cdot (220\,V)\cdot (3.45\,h)\cdot \left(3600\,\frac{s}{h} \right)}[/tex]
[tex]i = 88.200\,A[/tex]
Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 16 m/s at an angle 32 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.
1) What is the horizontal component of the ball’s velocity when it leaves Julie's hand?
2) What is the vertical component of the ball’s velocity when it leaves Julie's hand?
3) What is the maximum height the ball goes above the ground?
4) What is the distance between the two girls?
5) How high above the ground will the ball be when it gets to Julie? (note, the ball may go over Julie's head.)
Answer:
Explanation:
1. [tex]V_{x}[/tex] = [tex]V_{0}[/tex] * cos[tex]\alpha[/tex] ⇒ 16*cos32 ≈ 13.6 m/s (13.56)
2. [tex]V_{y}[/tex] = [tex]V_{0}[/tex] * sin[tex]\alpha[/tex] ⇒ 16* sin32 ≈ 9.4 m/s
3. [tex]y_{max}[/tex] = [tex]\frac{v_{0}^2*sin^2\alpha}{2g}[/tex]= [tex]\frac{16^2*sin^232}{2*9.8}[/tex] (the g (gravity) depends on the country but i'll take the average g which is 9.2m/s^2)
[tex]y_{max}[/tex] ≈ 3.6677+1.5 ≈ 5.2m
4. [tex]x_{max}[/tex] = [tex]\frac{v_{0}^2*sin(2\alpha)}{g}[/tex]=[tex]\frac{16^2*sin(2*32)}{9.8}[/tex] ≈ 23.5m (23.47)
5. -
answer 4 could be wrong, not certain about that one and i don't know 5
Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. A particle with charge q is placed at the center of the cavity. The net charge on the inner surface of the conducting shell is
Answer: in this question, the only charge in the cavity is Q. Inside the conducting spherical shell, the electric field is zero.
While outside the shell, the electric field is given by: k(q + Q)/r²
Where;
K= is a constant which is given as, 8.99 x 10^9 N m² / C².
Q= source charge which creates the electric field
q= is the test charge which is used to measure the strength of the electric field at a given location.
r= is the radius
Explanation: Inside the conducting spherical shell, the electric field is zero since the Electric field vanishes everywhere inside the volume of a good conductor.
A jet plane is flying at a constant altitude. At time t1=0t 1=0, it has components of velocity vx=90m/s,vy=110m/sv x = 90m/s,v y=110m/s. At time t2=30.0st 2=30.0s, the components are vx=−170m/s,vy=40m/sv x =−170m/s,v y=40m/s.
(a) Sketch the velocity vectors at t1and t2.
How do these two vectors differ? For this time interval calculate
(b) the components of the average acceleration, and
(c) the magnitude and direction of the average acceleration.
The average acceleration [tex]\vec a_{\rm ave}[/tex] over some time interval [tex][t_1,t_2][/tex] is equal to the ratio of the change in velocity [tex]\vec v_2-\vec v_1[/tex] over the duration of the interval [tex]t_2-t_1[/tex], or
[tex]\vec a_{\rm ave}=\dfrac{\Delta\vec v}{\Delta t}=\dfrac{\vec v_2-\vec v_1}{t_2-t_1}[/tex]
which can be split into the [tex]x[/tex] and [tex]y[/tex] components as
[tex]a_{\rm{ave},x}=\dfrac{v_{2,x}-v_{1,x}}{t_2-t_1}=\dfrac{-170\frac{\rm m}{\rm s}-90\frac{\rm m}{\rm s}}{30.0\,\mathrm s-0}\approx-8.67\dfrac{\rm m}{\mathrm s^2}[/tex]
[tex]a_{\rm{ave},y}=\dfrac{v_{2,y}-v_{1,y}}{t_2-t_1}=\dfrac{40\frac{\rm m}{\rm s}-110\frac{\rm m}{\rm s}}{30.0\,\mathrm s-0}\approx-2.33\dfrac{\rm m}{\mathrm s^2}[/tex]
The magnitude of this average acceleration is
[tex]\left\|\vec a_{\rm ave}\right\|=\sqrt{{a_{\rm{ave},x}}^2+{a_{\rm{ave},y}}^2}\approx8.98\dfrac{\rm m}{\mathrm s^2}[/tex]
and its direction is [tex]\theta[/tex] such that
[tex]\tan\theta=\dfrac{a_{\rm{ave},y}}{a_{\rm{ave},x}}\implies\theta\approx-164.9^\circ[/tex]
which corresponds to a direction of about 15.1º South of West.
describe Piaget's four stages of cognitive development. Include the major hallmarks of each stage.
Answer:
Explanation:
Sensorimotor Infants "think" by acting on the world with their eyes, ears, hands, and mouth.
Preoperational. Development of language and make-believe play takes place.
Concrete Operational children think in a logical, organized fashion only when dealing with concrete information they can perceive directly.
Formal Operational. Adolescences can also evaluate the logic of verbal statements without referring to real-world circumstances.
Sensorimotor, preoperational, concrete operational, and formal operational are Piaget's four phases of cognitive development.
What is cognitive development?The way youngsters think, investigate, and figure things out is referred to as cognitive development.
Piaget defined four stages of cognitive development:
1. Sensorimotor. From birth through the age of 18-24 months.
2. Preoperational.From infancy (18-24 months) until toddlerhood (age 7)
3. Operational concrete. 7 to 11 years old
4. Formal operational. From adolescence to adulthood
Hence, sensorimotor, preoperational, concrete operational, and formal operational are Piaget's four phases of cognitive development.
To learn more about the cognitive development refer to:
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When the distance between a point source of light and a light meter is reduced from 6.0m to 2.0 m, the intensity of illumination at the meter will be the original value multiplied by _____.
Answer:
Explanation:
Let the point source have power P .
At distance r , the intensity I
I = P / 4πr² . If intensity at 6 m and 2 m be I₁ and I₂
I₁ = P / 4π x 6²
I₂ = P / 4π x 2²
I₁ / I₂ = 2² / 6²
= 1 / 9
I₂ = 9 I₁
Intensity will be 9 times that at 6 m .
0.92 kg of R-134a fills a 0.14-m^3 weighted piston–cylinder device at a temperature of –26.4°C. The container is now heated until the temperature is 100°C. Determine the final volume of R-134a.
Answer:
The final volume of R-134a is 0.212m³Explanation:
Using one of the general gas equation to find the final volume of the R-134a.
According to pressure law; The volume of a given mas of gas is directly proportional to its temperature provided that the pressure remains constant.
VαT
V = kT
k = V/T
V1/T1 = V2/T2 = k
Given V1 = 0.14-m³ at T1 = –26.4°C = –26.4° + 273 = 246.6K
V2 = ? at T = 100°C = 100+273 = 373K
On substituting this values for T2;
0.14/246.6 = V2/373
373*0.14 = 246.6V2
V2 = 373*0.14 /246.6
V2 = 0.212m³
The final volume of R-134a is 0.212m³
Consider a weather balloon floating in the air. There are three forces acting on this balloon: the force of gravity is FG, the force from lift towards balloon is FL, and the force from the wind is labeled Fw. The orientation of these forces along with a coordinate system is given below:
Assume that || FG || = 20 N, || FL ||= 25 N, and || Fw ll = 15 N.
Required:
Find the magnitude of the resultant force acting on the weather balloon and round your answer to two decimal places.
To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the end of the bar, and string 2 is attached a distance 0.6 m from the other end. A monkey of mass 1.35 kg walks from one end of the bar to the other. Find the tension T1 in string 1 at the moment that the monkey is halfway between the ends of the bar.
Answer:
[tex]T_{1}[/tex] = 14.88 N
Explanation:
Let's begin by listing out the given variables:
M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,
g = 9.8 m/s²
At equilibrium, the sum of all external torque acting on an object equals zero
τ(net) = 0
Taking moment about [tex]T_{1}[/tex] we have:
(M + m) g * 0.5L - [tex]T_{2}[/tex](L - d) = 0
⇒ [tex]T_{2}[/tex] = [(M + m) g * 0.5L] ÷ (L - d)
[tex]T_{2}[/tex] = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)
[tex]T_{2}[/tex]= 59.535 ÷ 2.4
[tex]T_{2}[/tex] = 24.80625 N ≈ 24.81 N
Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N
Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N
Using sum of equilibrium in the vertical direction, we have:
[tex]T_{1}[/tex] + [tex]T_{2}[/tex] = W + w ------- Eqn 1
Substituting T2, W & w into the Eqn 1
[tex]T_{1}[/tex] + 24.81 = 26.46 + 13.23
[tex]T_{1}[/tex] = 14.88 N
A space ship traveling east flies directly over the head of an inertial observer who is at rest on the earth's surface. The speed of the space ship can be found from this relationship: . The navigator's on-board instruments indicate that the length of the space ship is 20 m. If the length of the ship is measured by the inertial earth-bound observer, what value will be obtained
Answer:
10 metres
Explanation:
So, we are given the following data or parameters or information which is going to assist us in solving this particular problem or Question efficiently.
=> The speed of the space ship can be found from this relationship: ✓(1 - [v^2/c^2] ) = 1/2.
=> The length of the space ship = 20 m.
=> Assumption = '' If the length of the ship is measured by the inertial earth-bound observer".
Thus, from the speed of the space ship can be found from this relationship we can determine the value;
✓(1 - [v^2/c^2] ) = 1/2.
V = 20 × 1/2 = 10 metres.
Note that we use the contraction formula to solve for V.
Assuming 100% efficient energy conversion, how much water stored behind a 50
centimeter high hydroelectric dam would be required to charge the battery?
Answer:
Explanation:
The power rating of the battery isn't provided. But let us assume that it is one of the common batteries with ratings of 12 V and 50 A.h
Potential energy possessed by water at that height = mgh
m = mass of the water = ρV
ρ = density of water = 1000 kg/m³
V = volume of water = ?
g = acceleration due to gravity = 9.8 m/s²
h = height of water = 50 cm = 0.5 m
Potential energy = ρVgh = 1000 × V × 9.8 × 0.5 = (4900V) J
Energy of the battery = qV
q = 50 A.h = 50 × 3600 = 180,000 C
V = 12 V
qV = 180,000 × 12 = 2,160,000 J
Energy = 2,160,000 J
At a 100% conversion rate, the energy of the water totally powers the battery
(4900V) = (2,160,000)
4900V = 2,160,000
V = (2,160,000/4900)
V = 440.82 m³
Hence, with our assumed power ratings for the battery (12 V and 50 A.h), 440.82 m³ of water at the given height of 50 cm would power the battery.
Incase the power ratings of the battery in the complete question is different, this solution provides you with how to obtain the correct answer, given any battery power rating.
Hope this Helps!!!
A beam of light is incident upon a flat piece of glass (n = 1.50) at an angle of incidence of 30.00. Part of the beam is transmitted and part is reflected. Determine the angle between the reflected and transmitted rays
Answer:
130.528779365 degrees
Explanation:
The angle of incidence is 30 degrees. From this, we can use Snell's Law to calculate the angle of refraction.
n1/n2 = sin(theta2)/sin(theta1)
let theta1 be 30 degrees, and n1 be the refractive index of air = 1
1/1.5 = sin(theta2)/sin(30deg)
solve:
sin(theta2) = 2/3 sin(30deg) = 1/3
theta2 = arcsin (1/3) = 19.4712206345 degrees
The angle of reflection will always be equal to the angle of incidence, in this case, 30 degrees.
Because these angles are measured relative to the normal, the angle formed between the two rays is the difference between the normal line (180 degrees) and the sum of the two angle measures.
Angle between = 180-30-19.4712206345 = 130.528779365 degrees
The angle between the reflected and transmitted rays 130.5287 degrees
What is the refraction of light?The angle of incidence is 30 degrees. From this, we can use Snell's Law to calculate the angle of refraction.
[tex]\dfrac{n_1}{n_2} = \dfrac{sin(\theta_2)}{sin(\theta_1)}[/tex]
let [tex]\theta_1[/tex] be 30 degrees, and n1 be the refractive index of air = 1
[tex]\dfrac{1}{1.5} = \dfrac{sin(\theta_2)}{sin(30)}[/tex]
solve:
[tex]sin(\theta_2) = \dfrac{2}{3} sin(30) = \dfrac{1}{3}[/tex]
[tex]\theta_2 = sin ^{-1}\dfrac{1}{3} = 19.4712 \ degrees[/tex]
The angle of reflection will always be equal to the angle of incidence, in this case, 30 degrees.
Because these angles are measured relative to the normal, the angle formed between the two rays is the difference between the normal line (180 degrees) and the sum of the two angle measures.
Angle between = 180-30-19.4712206345 = 130.528779365 degrees
Hence the angle between the reflected and transmitted rays 130.5287 degrees
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Our Sun shines bright with a luminosity of 3.828 x 1025 Watt. Her energies
responsible for many processes and the habitable temperatures on the earth that
make our life possible.
a) Calculate the amount of energy arriving on the Earth in a single day
b) To how many litres of heating oil (energy density 37.3 x 10^6 J/litre is the equivalent?
C) The Earth reflects 30% of this energy : Determine the temperature on Earth's sufact
d) what other factors should be considered to get an even more precisa temperature postiache
Note: The Earth's radius is 6370km; the Sun's sadius is 696 ×10^3km, I AU is 1.495 × 10^8km)
Answer:
a) E = 1.58 10²¹ J , b) Oil = 4,236 107 liter , e) T = 54.3 C
Explanation:
a) To calculate the energy that reaches Earth, let us combine that the power emitted by the Sun is distributed uniformly on a spherical surface
I = P / A
A = 4π r²
in this case the radius of the sphere is the distance from the Sun to Earth r = 1.5 10¹¹ m
I = P / A
I = P / 4π r²
let's calculate
I = 3,828 10²⁵/4 pi (1.5 10¹¹)²
I = 1.3539 10²W / m² = 135.4 W / m2
the energy that reaches the disk of the Earth is
E = I A
the area of a disc
A = π r²
E = I π r²
where r is the radius of the Earth 6.37 10⁶ m
E = 135.4 π(6.37 10⁶)
E = 1,726 10¹⁶ W
This is the energy per unit of time that reaches Earth
t = 1 dai (24h / 1day) (3600s / 1h) = 86400 s
E = 1,826 10¹⁶ 86400
E = 1.58 10²¹ J
b) for this part we can use a direct proportions rule
Oil = 1.58 10²¹ (1 / 37.3 10⁶)
Oil = 4,236 10⁷ liter
c) to silence the surface temperature of the Earth we use the Stefan-Bolztman Law
P = σ A e T⁴
T = [tex]\sqrt[4]{P/Ae}[/tex]
nos indicate the refect, therefore the amount of absorbencies
P_absorbed = 0.7 P
let's calculate
T = REA (0.7 1.58 1021 / [pi (6.37 106) 2 1)
T = RER (8,676 106)
T = 54.3 C
b) Among the other factors that must be taken into account is the greenhouse effect, due to the absorption of gases from the atmosphere
two blocks with masses 2 kg and 4 kg are pushed from rest by the same amount of fore for a distance of 100 m on a frictionless floor. the final kinetic energy of the 2 kg block after the 100 m distance is
Answer:
the kinetic energy of the 2 kg mass after the 100 m is equal to 1962 J
Explanation:
mass of block A = 2 kg
mass of block B = 4 kg
distance the blocks were pushed = 100 m
NB: Blocks were pushed the same distance at the same equal time period. And the ground is without friction.
Work done in moving the 2 kg mass along the 100 m distance is,
work = force x distance moved
force exerted by the 2 kg mass = 2 x 9.81 m/s^2(acceleration due to gravity)
force = 19.62 N
therefore,
work done = 19.62 x 100 = 1962 Joules of work.
According to energy conservation principles, the kinetic energy impacted of the 2 kg mass through this distance will be equal to the work done in moving the 2 kg mass through this distance.
Therefore, the kinetic energy of the 2 kg mass after the 100 m is equal to 1962 J
A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.65 kg and its length is 0.55 m. You may want to review (Pages 314 - 318) . Part A What is the rotational energy of the blade at its operating angular speed of 3510 rpm
Complete Question
A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.65 kg and its length is 0.55 m. You may want to review (Pages 314 - 318) .
Part A What is the rotational energy of the blade at its operating angular speed of 3510 rpm
Part B
If all of the rotational kinetic energy of the blade could be converted to gravitational potential energy, to what height would the blade rise?
Answer:
Part A
[tex]R = 1081 \ J[/tex]
Part B
[tex]h = 169.7 \ m[/tex]
Explanation:
From the question we are told that
The mass of the blade is [tex]m_b = 0.65 \ kg[/tex]
The length is [tex]l = 0.55 \ m[/tex]
The angular speed is [tex]w = 3510 rpm = 3510 * \frac{2 \pi }{60} = 367.6 \ rad/sec[/tex]
Generally the moment of inertia of the of this mower is mathematically evaluated as
[tex]I = \frac{m_b * l^2 }{12}[/tex]
substituting values
[tex]I = \frac{0.65 * 0.55^2 }{12}[/tex]
[tex]I = 0.016 \ kg m^2[/tex]
Generally the rotational kinetic energy of the bland is
[tex]R = \frac{1}{2} * I * w^2[/tex]
substituting values
[tex]R = \frac{1}{2} * 0.016 * 367.6^2[/tex]
[tex]R = 1081 \ J[/tex]
At point where the gravitational potential energy is equal to the rotational kinetic energy we have that
[tex]P = R = m_b * h * g[/tex]
Where P is the gravitational potential energy
substituting values
[tex]1081 = 0.65 * 9.8 * h[/tex]
=> [tex]h = 169.7 \ m[/tex]
A dart is inserted into a spring-loaded dart gun by pushing the spring in by a distance . For the next loading, the spring is compressed a distance . How much faster does the second dart leave the gun compared with the first
Complete question is;
A dart is inserted into a spring - loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is compressed a distance 2x. How much faster does the second dart leave the gun compared to the first?
Answer:
The second dart leaves the gun two times faster than the first one.
Explanation:
If we assume there was no energy loss during the spring - dart energy transfer, we can easily apply the principle of conservation of energy. So;
Potential energy = kinetic energy
Thus;
½kx² = ½mv²
Making velocity "v" the subject, we have;
v = √(kx²/m)
Since the initial distance is "x", thus initial launching velocity is;
v1 = √(kx²/m)
Since next distance is 2x, thus, second launch velocity is;
v2 = √(k(2x)²/m)
Expanding, we have;
v2 = √(4kx²/m)
v2 = 2√(kx²/m)
Comparing this to the one gotten for v1 earlier, we can see that it is double v1.
So, v2 = 2v1
Hence, The second dart leaves the gun two times faster than the first one.
b) A non-inductive load takes a current of 15 A at 125 V. An inductor is then connected
in series in order that the same current shall be supplied from 240 V, 50 Hz mains.
Ignore the resistance of the inductor and calculate:
i. the inductance of the inductor;
ii. the impedance of the circuit;
iii. the phase difference between the current and the applied voltage.
Assume the waveform to be sinusoidal.
Answer:
i. 43.5 mH ii. 16 Ω. In phasor form Z = (8.33 + j13.66) Ω iii 58.64°
Explanation:
i. The resistance , R of the non-inductive load R = 125 V/15 A = 8.33 Ω
The reactance X of the inductor is X = 2πfL where f = frequency = 50 Hz.
So, x = 2π(50)L = 100πL Ω = 314.16L Ω
Since the current is the same when the 240 V supply is applied, then
the impedance Z = √(R² + X²) = 240 V/15 A
√(R² + X²) = 16 Ω
8.33² + X² = 16²
69.3889 + X² = 256
X² = 256 - 69.3889
X² = 186.6111
X = √186.6111
X = 13.66 Ω
Since X = 314.16L = 13.66 Ω
L = 13.66/314.16
= 0.0435 H
= 43.5 mH
ii. Since the same current is supplied in both circuits, the impedance Z of the circuit is Z = 240 V/15 A = 16 Ω.
So in phasor form Z = (8.33 + j13.66) Ω
iii. The phase difference θ between the current and voltage is
θ = tan⁻¹X/R
= tan⁻¹(314.16L/R)
= tan⁻¹(314.16 × 0.0435 H/8.33 Ω)
= tan⁻¹(13.66/8.33)
= tan⁻¹(1.6406)
= 58.64°
Which of the following is often found in individuals who are active and eating a healthy diet?
Answer:
Increased blood circulation to the body.
Explanation:
plato/edmentum
What is the relationship between electric force and distance between charged objects and the amount of charge?
Explanation:
The relationship between electric force and distance between charged objects is given by the formula as follows :
[tex]F=\dfrac{kq_1q_2}{d^2}[/tex]
k is electrostatic constant and d is distance between charges
The electric force between charges is inversely proportional to the square of distance between them.
A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release, the object moves across the surface until it encounters a rough incline. The object moves UP the incline and stops a height of 1.5 m above the horizontal surface.
(a) How much work must be done to compress the spring initially?
(b) Compute the speed of the mass at the base of the incline.
(c) How much work was done by friction on the incline?
Answer with Explanation:
We are given that
Mass of spring,m=3 kg
Distance moved by object,d=0.6 m
Spring constant,k=210N/m
Height,h=1.5 m
a.Work done to compress the spring initially=[tex]\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J[/tex]
b.
By conservation law of energy
Initial energy of spring=Kinetic energy of object
[tex]37.8=\frac{1}{2}(3)v^2[/tex]
[tex]v^2=\frac{37.8\times 2}{3}[/tex]
[tex]v=\sqrt{\frac{37.8\times 2}{3}}[/tex]
v=5.02 m/s
c.Work done by friction on the incline,[tex]w_{friction}=P.E-spring \;energy[/tex]
[tex]W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J[/tex]
An automobile being tested on a straight road is 400 feet from its starting point when the stopwatch reads 8.0 seconds and is 550 feet from the starting point when the stopwatch reads 10.0 seconds.
A. What was the average velocity of the automobile during the interval from t = 10.0 seconds to t = 8.0 seconds
B. What was the average velocity of the automobile during the interval from t - Ostot - 10.0 s? (Assume that the stopwatch read t = 0 and started at the same time as the auto.)
C. If the automobile averages 100 ft/s from t - 10.0 stot - 20.0 s, what distance does it travel during this interval?
D. The automobile has a special speedometer calibrated in feet/s instead of in miles/hour. Att 85 the speedometer reads 65 ft/s; and at t = 10 s it reads 80 ft/s. What is the average acceleration during this interval?
Answer:
a) v = 75 ft / s , b) v = 55 ft / s , c) Δx = 1000 ft
Explanation:
We can solve this exercise with the expressions of kinematics
a) average speed is defined as the distance traveled in a given time interval
v = (x₂-x₁) / (t₂-t₁)
v = (550 - 400) / (10 -8)
v = 75 ft / s
b) we repeat the calculations for this interval
v = (550 - 0) / (10 -0)
v = 55 ft / s
c) we clear the distance from the average velocity equation
Δx = v (t₂ -t₁)
Δx = 100 (20-10)
Δx = 1000 ft
A student is given a small object that is hanging from a ring stand on a nylon thread. The student attempts to charge the object electrically in several ways. Based upon his results, he concludes the object is made of an insulating material. Which set of results must he have collected?
A. The object could be charged only by contact.
B. The object could be charged by either contact or induction.
C. The object could be charged by either contact or polarization.
D. The object could be charged only by polarization.
Answer:(a)
Explanation:
Student must have known that insulators can only be charged when they are rubbed against each other. In this process, one becomes electrically negative while other becomes electrically positive such that both have the same magnitude. The one which gains electrons becomes electrically negative due to the transfer of electrons while others lose the electron becomes positive due to the transfer of an electron to another body.
