Answer:
Therefore the characteristics to be found are:
* the focal length must be large and the focal length of the eyepiece must be small
* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light
* The system must be configured to the far sight tip,
Explanation:
The length of the telescope is
L = f_ocular + f_objetive
the magnification of the telescope is
m = - f_objective / f_ocular
These are the two equations that describe the behavior of the telescope. Therefore the characteristics to be found are:
* the focal length must be large and the focal length of the eyepiece must be small
* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light
* The system must be configured to the far sight tip,
A horizontal board of negligible thickness and area 1.0 m2 hangs from a spring scale that reads 50 N when a 4.0 m/s wind moves above the board. The air below the board is stationary. What does the scale read when the wind stops? The density of air is 1.25 kg/m3 .
Calculate the kinetic energy of a mass 2kg moving with a velocity of 0.1m/s
ANSWER-:
1/2 mv²
K.E = 1/2 mv²
K.E = 0.01 J.
Hence, the kinetic energy of a body is 0.01 Joule.
!! HOPE ITS HELP U !!
A projectile is shot from the edge of a cliff 80 m above ground level with an initial speed of 60 m/sec at an angle of 30° with the horizontal. Determine the time taken by the projectile to hit the ground below.
Answer:
8 seconds
Explanation:
Answer:
Explanation:
Going up
Time taken to reach maximum height= usin∅/g
=3 secs
Maximum height= H+[(usin∅)²/2g]
=80+[(60sin30)²/20]
=125 meters
Coming Down
Maximum height= ½gt²
125= ½(10)(t²)
t=5 secs
Activity
Spend time observing or thinking about events that involve matter and energy. Which events can you explain? Which
events can't you explain? Now answer the following questions.
Part A
Describe three events that you cannot explain.
Answer:Eating. Your muscles in your arms and mouth use energy to feed itself. Then your body digest the food which also takes energy.
Sleep. When your tired, you don’t have much energy. It is said that you use more energy while your sleeping. But how do you become energized if you were using even more energy than before?
Answer:
Different kinds of balls bounce to different heights when dropped on the same floor.
Sugar dissolves faster in hot milk than in cold milk.
Plants grow more slowly when they are not near a window.
Explanation:
The number of daylight hours, D, in the city of Worcester, Massachusetts, where x is the number of days after January 1 (), may be calculated by the function: What is the period of this function? N/A What is the amplitude of this function? 12 What is the horizontal shift? What is the phase shift? What is the vertical shift? How many hours of sunlight will there be on February 21st of any year?
Answer:
a. 365; b. 3; c. 78; d. 1.343 rad; e. 12; f. 10.66
Explanation:
Assume that the function is
[tex]D(x) = 3 \sin \left (\dfrac{2\pi}{365}(x - 78) \right ) + 12[/tex]
The general formula for a sinusoidal function is
y = A sin(B(x - C))+ D
|A| = amplitude
B = frequency
2π/B = period, P
C = horizontal shift (phase shift)
D = vertical shift
By comparing the two formulas, we find
|A| = 3
B = 2π/365
C = 78
D = 12
a. Period
P = 2π/B = 2π/(2π/365) = 2π × 365/2π = 365
The period is 365.
b. Amplitude
|A| = 3
The amplitude is 3.
c. Horizontal shift
C= 78
The horizontal shift is 78.
d. Phase shift (φ)
Ths phase shift is the horizontal shift expressed in radians.
φ = C × 2π/365 = 78 × 2π/365 ≈ 1.343
The phase shift is 1.343 rad.
e. Vertical shift
D = 12
The vertical shift is 12.
f. Hours of sunlight on Feb 21
Feb 21 is the 52nd day of the year, so x = 51 (the number of days after Jan 1),
[tex]\begin{array}{rcl}D(x) &=& 3 \sin \left (\dfrac{2\pi}{365}(x - 78) \right ) + 12\\\\&=& 3 \sin (0.01721(51 - 78) ) + 12\\&=& 3\sin(-0.4648) + 12\\&=& 3(-0.4482) + 12\\\&=& -1.345 + 12\\& = & \textbf{10.66 h}\\\end{array}[/tex]
There will be 10.66 h of sunlight on Feb 21 of any given year.
The figure below shows the graph of the function from 0 ≤ x ≤ 365.
Charge of uniform density (0.30 nC/m2) is distributed over the xy plane, and charge of uniform density (−0.40 nC/m2) is distributed over the yz plane. What is the magnitude of the resulting electric field at any point not in either of the two charged planes?
Answer: E = 39.54 N/C
Explanation: Electric field can be determined using surface charge density:
[tex]E = \frac{\sigma}{2\epsilon_{0}}[/tex]
where:
σ is surface charge density
[tex]\epsilon_{0}[/tex] is permitivitty of free space ([tex]\epsilon_{0} = 8.85.10^{-12}[/tex][tex]C^{2}/N.m^{2}[/tex])
Calculating resulting electric field:
[tex]E=E_{1} - E_{2}[/tex]
[tex]E = \frac{\sigma_{1}-\sigma_{2}}{2\epsilon_{0}}[/tex]
[tex]E = \frac{[0.3-(-0.4)].10^{-9}}{2.8.85.10^{-12}}[/tex]
[tex]E=0.03954.10^{3}[/tex]
E = 39.54
The resulting Electric Field at any point is 39.54N/C.
The magnitude of the resulting electric field at any point should be 28.2 N/C.
Calculation of the magnitude:Since the Charge of uniform density (0.30 nC/m2) should be allocated over the xy plane, and charge of uniform density (−0.40 nC/m2)should be allocated over the yz plane.
So,
E1
= σ1/2ε0
= 0.30e-9/(2*8.85e-12)
= 16.949 N/C
So, direction of E1 is +z
Now
E2 = σ2/2ε0
= 0.40e-9/(2*8.85e-12)
= 22.6 N/C
So, direction of E2 is -x
Now
E = √(E1*E1+E2*E2)
= √(16.949*16.949+22.6*22.6)
= 28.2 N/C
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Monochromatic light of wavelength, λ is traveling in air. The light then strikes a thin film having an index of refraction n1 that is coating a material having an index of refraction n2. If n1 is larger than n2, what minimum film thickness will result in minimum reflection of this light?
Answer:
tmin= lambda/2
Explanation:
See attached file pls
Intelligent beings in a distant galaxy send a signal to earth in the form of an electromagnetic wave. The frequency of the signal observed on earth is 2.2% greater than the frequency emitted by the source in the distant galaxy. What is the speed vrel of the galaxy relative to the earth
Answer:
Vrel= 0.75c
Explanation:
See attached file
what is liquid pressure? and tell me its si unit please
The SI unit of pressure is the pascal: 1Pa=1N/m2 1 Pa = 1 N/m 2 . Pressure due to the weight of a liquid of constant density is given by p=ρgh p = ρ g h , where p is the pressure, h is the depth of the liquid, ρ is the density of the liquid, and g is the acceleration due to gravity.
A thick wire with a radius of 4.0 mm carries a uniform electric current of 1.0 A, distributed uniformly over its cross-section. At what distance from the axis of the wire, and greater than the radius of the wire, is the magnetic field strength equal to that at a distance 2.0 mm from the axis. distance
Answer:
8 mm
Explanation:
From the information given:
The Ampere circuital law can be used to estimate the magnetic field strength at two points when the distance is less than the radius and when the distance is greater than the radius.
when the distance is less than the radius ; we have:
[tex]B_1 = \dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2}[/tex]
when the distance is greater than the radius; we have:
[tex]B_2 = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]
Equating both equations together ; we have :
[tex]\dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2} = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]
[tex]\dfrac{1}{R}= \dfrac{r}{d^2}[/tex]
[tex]R= \dfrac{d^2}{r}[/tex]
where; d = radius of the wire and r = distance;
[tex]R =\dfrac{4^2}{2}[/tex]
[tex]R =\dfrac{16}{2}[/tex]
R = 8 mm
What is the angle between a wire carrying an 8.40 A current and the 1.20 T field it is in, if 50.0 cm of the wire experiences a magnetic force of 2.55 N? ° (b) What is the force (in N) on the wire if it is rotated to make an angle of 90° with the field? N
Answer:
A. 30.38°
B 5.04N
Explanation:
Using
F= ILBsin theta
2 .55N= 8.4Ax 0.5mx 1.2T x sintheta
Theta = 30.38°
B. If theta is 90°
Then
F= 8.4Ax 0.5mx 1.2x sin 90°
F= 5.04N
"A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if"
Answer:
A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if
the dispersion is great
Three capacitors C1 = 10.7 µF, C2 = 23.0 µF, and C3 = 29.3 µF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Answer:
E = 1336.71875 J
Explanation:
We are given;. Capacitance of Capacitor 1; C1 = 10.7 µF
Capacitor 2; C2 = 23.0 µF
Capacitor 3; C3 = 29.3 µF
Supply voltage;V = 125 V
Formula for capacitance in series is;
Capacitors in series circuit: C(eq) = 1/C(1) +1/C(2) +1/C(3) .......
Thus, equivalent capacitance is;
C(eq) = (1/10.7) + (1/23) + (1/29.3) = 0.1711 µF = 0.1711 × 10^(6) F
Now, the formula for maximum energy stored is;
E = ½ × C(eq) × V²
E = ½ × 0.1711 × 10^(-6) × 125²
E = 1336.71875 J
g As observed on earth, a certain type of bacteria is known to double in number every 24 hours. Two cultures of these bacteria are prepared, each consisting initially of one bacterium. One culture is left on earth and the other placed on a rocket that travels at a speed of 0.893c relative to the earth. At a time when the earthbound culture has grown to 256 bacteria, how many bacteria are in the culture on the rocket, according to an earth-based observer
Answer:
86.4 hrs
Explanation:
The amount of bacteria is initially 1
It doubles every 24 hrs.
After first 24 hrs, the amount = 2
After next 24 hrs = 4
After next 24 hrs = 8
After next 24 hrs = 16
After next 24 hrs = 32
After next 24 hrs = 64
After next 24 hrs = 128
After next 24 hrs = 256
Total time taken to reach 256 = 24 x 8 = 192 hrs
For the bacteria culture on the rocket that travels at a speed of 0.893c relative to the earth, this time is contracted by the relationship
t = t'(1 - ¥^2)^0.5
Where t is the contracted time =?
t' is the time on earth
¥ = v/c
Where v is the speed of the rocket
c is the speed of light
since v = 0.893c
¥ = 0.893
Substituting, we have
t = 192 x (1 - 0.893^2)^0.5
t = 192 x 0.2025^0.5
t = 192 x 0.45 = 86.4 hrs
550 J of heat is added to the gas in an isothermal process. As the gas expands, pushing against the piston, how much work does it do
Answer:
The work done by the system is 550 J
Explanation:
Given;
heat added to the system, Q = 550 J
Apply the first law of thermodynamics;
ΔU = Q - W
Where;
ΔU is change in internal energy
Q is the heat added to the system
W is the work done by the system
During an isothermal process, the temperature of the system is constant for the entire process. During this process, the change in the internal energy is zero.
0 = Q - W
W = Q
W = 550 J
Therefore, the work done by the system is 550 J
The speed of a horse is 134 meters per second how long does it takes to travel a distance of 19,311?
Answer:
just need some focus
Explanation:
tan 13. The speed of a horse is 134 meters per second how long does it take to travel a distance of 19,311m? M+ tud V 134 14. You are walking down the block and see your neighbor's pitbull 30 meters away, out of the fence, starring at you. Suddenly, he starts running towards you at 20m/s. How long will it be before you're the pitbull's lunch? V 15. A pendulum has a frequency of 2 Hz what is it's period. T = 1/2 16. You have just finished a 1600 mile trip, and it took you 16 hours. What was your average speed V = Ad At 17. You are flying from New York, NY to SanFrancisco California, a distance of 2582 miles, it takes 6.33hrs to complete the flight what was your average speed? give answer in m/s. V = Ad = At 3 of 6
A cylindrical container with a cross-sectional area of 66.2 cm2 holds a fluid of density 856 kg/m3 . At the bottom of the container the pressure is 119 kPa . Assume Pat = 101 kPa
A) What is the depth of the fuild?
B) Find the pressure at the bottom of the container after an additional 2.35×10−3 m3 of this fluid is added to the container. Assume that no fluid spills out of the container.
Answer:
A. h = 2.15 m
B. Pb' = 122 KPa
Explanation:
The computation is shown below:
a) Let us assume the depth be h
As we know that
[tex]Pb - Pat = d \times g \times h \\\\ ( 119 - 101) \times 10^3 = 856 \times 9.8 \times h[/tex]
After solving this,
h = 2.15 m
Therefore the depth of the fluid is 2.15 m
b)
Given that
height of the extra fluid is
[tex]h' = \frac{2.35 \times 10^{-3}}{ area} \\\\ h' = \frac{2.35 \times 10^{-3}} { 66.2 \times 10^{-4}}[/tex]
h' = 0.355 m
Now let us assume the pressure at the bottom is Pb'
so, the equation would be
[tex]Pb' - Pat = d \times g \times (h + h')\\\\Pb' = 856 \times 9.8 \times ( 2.15 + 0.355) + 101000[/tex]
Pb' = 122 KPa
(A) The depth of the fluid is 2.14 m.
(B) The new pressure at the bottom of container is 121972 Pa.
Given data:
The cross-sectional area of the container is, [tex]A =66.2 \;\rm cm^{2}=66.2 \times 10^{-4} \;\rm m^{2}[/tex].
The density of fluid is, [tex]\rho = 856 \;\rm kg/m^{3}[/tex].
The container pressure at bottom is, [tex]P=119 \;\rm kPa=119 \times 10^{3} \;\rm Pa[/tex].
The atmospheric pressure is, [tex]P_{at}=101 \;\rm kPa=101 \times 10^{3}\;\rm Pa[/tex].
(A)
The given problem is based on the net pressure on the container, which is equal to the difference between the pressure at the bottom and the atmospheric pressure. Then the expression is,
[tex]P_{net} = P-P_{at}\\\\\rho \times g \times h= P-P_{at}[/tex]
Here, h is the depth of fluid.
Solving as,
[tex]856\times 9.8 \times h= (119-101) \times 10^{3}\\\\h=\dfrac{ (119-101) \times 10^{3}}{856\times 9.8}\\\\h= 2.14 \;\rm m[/tex]
Thus, the depth of the fluid is 2.14 m.
(B)
For an additional volume of [tex]2.35 \times 10^{-3} \;\rm m^{3}[/tex] to the liquid, the new depth is,
[tex]V=A \times h'\\\\h'=\dfrac{2.35 \times 10^{-3}}{66.2 \times 10^{-4}}\\\\h'=0.36 \;\rm m[/tex]
Now, calculate the new pressure at the bottom of the container as,
[tex]P'-P_{at}= \rho \times g \times (h+h')\\\\\P'-(101 \times 10^{3})= 856 \times 9.8 \times (2.14+0.36)\\\\P'=121972 \;\rm Pa[/tex]
Thus, we can conclude that the new pressure at the bottom of container is 121972 Pa.
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within which type of system is the total mass conserved but not the total energy
In a closed system the mass is conserved, but the energy is not conserved.
To find the answer, we have to study about different systems in thermodynamics.
What is thermodynamic system?A system, which can be expressed in terms of thermodynamic coordinates is called Thermodynamic system.Open system: System can exchange both energy and matter, thus, both energy and matter is not conserved here.Closed system can exchange energy with its surroundings (as heat or work), but not matter.Isolated system: A system that is open to the environment can interchange energy and matter, but a system that is insulated from it cannot.Thus, we can conclude that, in closed system the mass is conserved, but the energy is not conserved.
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What is the speed of light (in m/s) in air? (Enter your answer to at least four significant figures. Assume the speed of light in a vacuum is 2.997 ✕ 108 m/s.) m/s What is the speed of light (in m/s) in polystyrene? m/s
Answer:
The speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.
Explanation:
To find the speed of light in air and in polystyrene we need to use the following equation:
[tex] c_{m} = \frac{c}{n} [/tex]
Where:
[tex]c_{m}[/tex]: is the speed of light in the medium
n: is the refractive index of the medium
In air:
[tex]c_{a} = \frac{c}{n_{a}} = \frac{2.997 \cdot 10^{8} m/s}{1.0003} = 2.996 \cdot 10^{8} m/s[/tex]
In polystyrene:
[tex]c_{p} = \frac{c}{n_{p}} = \frac{2.997 \cdot 10^{8} m/s}{1.6} = 1.873 \cdot 10^{8} m/s[/tex]
Therefore, the speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.
I hope it helps you!
local unit of measurement of length : confined to a particular place cise Choose the best answer from the given alterna MKS system stands for ........ i. mass, kilogram and second ii. metre, kilogram and second iii. metre, kilometre and second iv. metre, kilogram and standard
MKS stands for metre , kilogram and second
Suppose we want to calculate the moment of inertia of a 56.5 kg skater, relative to a vertical axis through their center of mass.
Required:
a. First calculate the moment of inertia (in kg-m^2) when the skater has their arms pulled inward by assuming they are cylinder of radius 0.11 m.
b. Now calculate the moment of inertia of the skater (in kg-m^2) with their arms extended by assuming that each arm is 5% of the mass of their body. Assume the body is a cylinder of the same size, and the arms are 0.875 m long rods extending straight out from their body being rotated at the ends.
Answer:
a. 0.342 kg-m² b. 2.0728 kg-m²
Explanation:
a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m
I = 1/2MR²
= 1/2 × 56.5 kg × (0.11 m)²
= 0.342 kgm²
So the moment of inertia of the skater is
b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)
I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m
I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²
= 0.1802 kg-m² + 0.6852 kg-m²
= 0.8654 kg-m²
The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².
So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²
a) The moment of inertia as the skater pulled his/her arm inward by assuming he/she is a cylinder is 0.3418kg-m².
b) If the body of the skater is assumed to be a cylinder of the same size, and the arms are rods extending straight out from his/her body being rotated at the ends, the moment of inertia is 1.7495kg-m².
Given the data in the question;
Mass of skater; [tex]M = 56.5kg[/tex]
a)
When the skater has his arms pulled inward by assuming they are cylinder of radius; [tex]R = 0.11 m[/tex]
Moment of inertia; [tex]I = \ ?[/tex]
From Parallel axis theorem; Moment of Inertia for a cylindrical body is expressed:
[tex]I = \frac{1}{2}MR^2[/tex]
Where M is the mass and R is the radius
We substitute our given values into the equation
[tex]I = \frac{1}{2}\ *\ 56.5kg\ *\ (0.11m)^2\\\\I = \frac{1}{2}\ *\ 56.5kg\ *\ 0.0121m^2\\\\I = 0.3418kg.m^2[/tex]
Therefore, the moment of inertia as the skater pulled his/her arm inward by assuming he/she is a cylinder is 0.3418kg-m²
b)
With the skater's arms extended by assuming that each arm is 5% of the mass of their body
Mass of each arm; [tex]M_a = \frac{5}{100} * M = \frac{5}{100} * 56.5kg = 2.825kg[/tex]
Remaining mass; [tex]M_b = M - 2M_a = 56.5kg - 2(2.825kg) = 50.85kg[/tex]
Assume the body is a cylinder of the same size and the arms are 0.875 m long rods extending straight out from their body being rotated at the ends.
Length of arm; [tex]L = 0.875 m[/tex]
From Parallel axis theorem; Moment of Inertia about vertical axis is expressed as:
[tex]I = \frac{1}{2}M_bR^2 + \frac{2}{3}M_aL^2[/tex]
We substitute in our values
[tex]I = \frac{1}{2}*50.85kg*(0.11m)^2 + \frac{2}{3}*2.825kg*(0.875m)^2\\\\I = [\frac{1}{2}*50.85kg * 0.0121m^2] + [\frac{2}{3}*2.825kg*0.765625m^2]\\\\I = 0.3076kg.m^2 + 1.4419kg.m^2\\\\I = 1.7495kg.m^2[/tex]
Therefore, if the body of the skater is assumed to be a cylinder of the same size, and the arms are rods extending straight out from his/her body being rotated at the ends, the moment of inertia is 1.7495kg-m².
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If the terminals of a battery with zero internal resistance are connected across two identical resistors in series, the total power delivered by the battery is 8.00 W. If the same battery is connected across the same resistors in parallel, what is the total power delivered by the battery
Answer:
24W
Explanation:
The series connection has a resistance of 2R
The parallel connection has a resistance of R/2 .. the resistance has decreased by a factor 4
Assuming the battery still provides the same pd .. the current increases by a factor of 4 .. increasing the power output by a factor of 4 also (P = V x A)
Power output = 4 x 8W .. .. So P = 24 W
What type of energy conversion occurs when you place your feet near the fireplace and they become warm?
O Radiant to thermal
o Thermal to mechanical
O Mechanical to chemical
O Nuclear to thermal
Answer:
It is the first one RADIENT TO THERMAL
Explanation:
The heat emitted from the campfires is an an example of radiant energy and thermal energy is refers to the energy contained within a system that is responsible for its tempreture with in this case is the campfires and heat energy being reflected upon your feet.
Answer:
A
Explanation:
A fan is turned off, and its angular speed decreases from 10.0 rad/s to 6.3 rad/s in 5.0 s. What is the magnitude of the angular acceleration of the fan?
A) 0.37 rad/s2
B) 11.6 rad/s2
C) 0.74 rad/s2
D) 0.86 rad/s2
E) 1.16 rad/s2
Answer:
chk photo
Explanation:
A 17.0 g bullet traveling horizontally at 785 m/s passes through a tank containing 13.5 kg of water and emerges with a speed of 534 m/s.
What is the maximum temperature increase that the water could have as a result of this event? (in degrees)
Answer:
The maximum temperature increase is [tex]\Delta T = 0.0497 \ ^oC[/tex]
Explanation:
From the question we are told that
The mass of the bullet is [tex]m = 17.0 \ g =0.017 \ kg[/tex]
The speed is [tex]v_1 = 785 \ m/s[/tex]
The mass of the water is [tex]m_w = 13.5 \ kg[/tex]
The velocity it emerged with is [tex]v_2 = 534 \ m/s[/tex]
Generally due to the fact that energy can nether be created nor destroyed but transferred from one form to another then
the change in kinetic energy of the bullet = the heat gained by the water
So
The change in kinetic energy of the water is
[tex]\Delta KE = \frac{1}{2} m (v_1^2 - v_2 ^2 )[/tex]
substituting values
[tex]\Delta KE =0.5 * 0.017 * (( 785)^2 - (534) ^2 )[/tex]
[tex]\Delta KE = 2814.1 \ J[/tex]
Now the heat gained by the water is
[tex]Q = m_w* c_w * \Delta T[/tex]
Here [tex]c_w[/tex] is the specific heat of water which has a value [tex]c_w = 4190 J/kg \cdot K[/tex]
So since [tex]\Delta KE = Q[/tex]
we have that
[tex]2814.1 = 13.5 * 4190 * \Delta T[/tex]
[tex]\Delta T = 0.0497 \ ^oC[/tex]
The elastic portion of the downward-sloping straight-line demand curve lies:_______
a. at the intersection with the supply curve.
b. anywhere to the right of the current market price.
c. above the point of unit elasticity.
d. below the point where total revenue is maximized.
Answer:
c. above the point of unit elasticity.
Explanation:
The elastic portion of the downward-sloping straight-line demand curve lies above the point of unit elasticity. Supply and demand are fundamental concept in economics. The demand curve shows how much of a good people will want at a different prices. The demands curves illustrates the intuition why people purchase a good for a lower price. For the demand curve, the price is always shown on the vertical axis and the demand curve is shown on the horizontal axis. Thus , the quantity demanded increases as the price gets lower. However, the price elasticity of the demand curve varies along the demand curve. This is because there is a key distinction between the gradient and the elasticity. The gradient which is the slope of the line is always the same in the demand curve but elasticity of the demand changes in the percentage of the quantity demand. Therefore, elasticity will vary along the downward-sloping straight - line demand curve. So, in a downward-sloping straight-line demand curve, the elastic portion is usually above the point of unit elasticity
Describe the change in motion and kinetic energy of the particles as thermal energy is removed from a liquid. Which change of state might happen?
Please I will give brainliest!!!!
Answer:
Change in state from liquid to solid.
Explanation:
Decrease in thermal energy will decrease energy and particles will slow down. Change in state from liquid to solid.
A child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the horizontal. The ball leaves her hand 1.0 m above the ground and the fence is 2.0 m high. The ball just clears the fence while still traveling upwards and experiences no significant air resistance. How far is the child from the fence?
Answer:
the child is 1.581 m far from the fence
Explanation:
The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.
From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:
[tex]x - x_o = u_xt[/tex]
[tex]\mathtt{x = u_xt \ \ \ since (x_o = 0)}[/tex] ---- (1)
the equation of the motion y is :
[tex]\mathtt{y - y_o =u_yt - 0.5 gt^2}[/tex]
[tex]\mathtt{y = u_yt-4.9t^2 \ \ \ since (y_o =0)}[/tex]
[tex]\mathtt{ 1= (u \ sin 40^0)t -4.9 \ t^2 }[/tex]
[tex]\mathtt{1 = 8 sin 40^0 t - 4.9 t^2}[/tex]
[tex]\mathtt{1 = 5.14t - 4.9t^2}[/tex]
[tex]\mathtt{4.9t^2 - 5.14t +1 = 0}[/tex]
By using the quadratic formula, we have;
[tex]\mathtt{ \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}} }[/tex]
where;
a = 4.9, b = -5.14 c = 1
[tex]= \mathtt{ \dfrac{ -(-5.14) \pm \sqrt{(-5.14)^2 - 4(4.9)(1)}}{2(4.9)}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14 \pm \sqrt{26.4196 -19.6}}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14 \pm \sqrt{6.8196}}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14+ \sqrt{6.8196}}{9.8} \ \ OR \ \ \dfrac{ 5.14- \sqrt{6.8196}}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14+ 2.6114}{9.8} \ \ OR \ \ \dfrac{ 5.14- 2.6114}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 7.7514}{9.8} \ \ OR \ \ \dfrac{ 2.5286}{9.8}} }[/tex]
[tex]= \mathbf{ 0.791 \ \ OR \ \ 0.258} }[/tex]
In as much as the ball is traveling upward, then we consider t= 0.258sec.
From equation (1)
[tex]\mathtt{x = u_x(0.258)}[/tex]
[tex]\mathtt{x = ucos 40^0 (0.258)}[/tex]
[tex]\mathtt{x = 8 \ cos 40^0 (0.258)}[/tex]
[tex]\mathbf{x = 1.581 \ m}[/tex]
Thus, the child is 1.581 m far from the fence
The highest mountain on mars is olympus mons, rising 22000 meters above the martian surface. If we were to throw an object horizontaly off the mountain top, how long would it take to reach the surface? (Ignore atmospheric drag forces and use gMars=3.72m/s^2
a. 2.4 minutes
b. 0.79 minutes
c. 1.8 minutes
d. 3.0 minutes
Answer:
t = 1.81 min , the correct answer is c
Explanation:
This is a missile throwing exercise
The object is thrown horizontally, so its vertical speed is zero (voy = 0), let's use the equation
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
the final height is y = 0 and the initial height is y₀ = 22000 m
0 = y₀ + 0 - ½ g t²
t = √y 2y₀ / g
let's calculate
t = √(2 22000 / 3.72)
t = 108.76 s
let's reduce to minutes
t = 108.76 s (1 min / 60 s)
t = 1.81 min
The correct answer is c
If a bus travels 50 km in 10 hours, how fast was the
bus travelling?
Answer:
5 kilometers per hour
Explanation:
Speed = distance / time
Distance: 50km
Time: 10 hours
Speed = 50/10 = 5kph
Answer:
5kmph
Explanation:
if the bus traveled 50 km in 10 hours, we have to divide 50 by 10 to see how fast it traveled per hour.
50/10 = 5
therefore, the bus was traveling 5 km per hour
hope this helps :)