Answer:
a) W = 1.63 10⁻²⁸ J, b) W = 1.407 10⁻²⁷ J, c) W = 1.68 10⁻²⁸ J,
d) W = - 4.93 10⁻²⁸ J
Explanation:
a) In this problem we have an electron at the origin, work is requested to carry another electron from infinity to the point x₂ = 0, y₂ = 2.00m
If we use the law of conservation of energy, work is the change in energy of the system
W = ΔU = U_∞ -U
the potential energy for point charges is
U =k [tex]\sum \frac{q_i q_j}{r_{ij} }[/tex]
in this case we only have two particles
U = k [tex]\frac{q_1q_2}{r_{12} }[/tex]
the distance is
r₁₂ = [tex]\sqrt{(x_2-x_1)^2 + ( y_2-y_1)^2 }[/tex]
r₁₂ =[tex]\sqrt{ 0 + ( 2-0)^2}[/tex]Ra 0 + (2-0)
r₁₂ = √2= 1.4142 m
we substitute
W = k \sum \frac{q_i q_j}{r_{ij} }
let's calculate
W = [tex]\frac{ 9 \ 10^9 (1.6 \ 10^{-19})^2 }{1.4142}[/tex] 9 109 1.6 10-19 1.6 10-19 / 1.4142
W = 1.63 10⁻²⁸ J
b) the two electrons are fixed, what is the work to bring another electron to x₃ = 3.00 m y₃ = 0
in this case we have two fixed electrons
U = k [tex]( \frac{q_1q_3}{r_{13} } + \frac{q_2q_3}{r_{23} } )[/tex]
in this case all charges are electrons
q₁ = q₂ = q₃ = q
W = U = k q² [tex]( \frac{1}{r_{13} } + \frac{1}{r_{23} } )[/tex]
the distances are
r₁₃ = [tex]\sqrt{(3-0)^2 + 0}[/tex]RA (3.00 -0) 2 + 0
r₁₃ = 3
r₂₃ = [tex]\sqrt{ 3^2 + 2^2}[/tex]Ra (3 0) 2 + (2 0) 2
r₂₃ = √13
r₂₃ = 3.606 m
let's look for the job
W = U
let's calculate
W =[tex]{9 \ 10^3 ( 1.6 10^{-19})^2 }({\frac{1}{3} + \frac{1}{3.606} } )[/tex]
W = 1.407 10⁻²⁷ J
c) the three electrons are fixed, we bring the four electron to x₄ = 3.00m,
y₄ = 4.00 m
W = U = k [tex]( \frac{q_1q_4}{r_{14 }} + \frac{q_2q_4}{r_{24} } + \frac{q_3q_4}{r_{34} } )[/tex]
all charges are equal q₁ = q₂ = q₃ = q₄ = q
W = k q² [tex](\frac{1}{r_{14} } + \frac{1}{r_{24} } + \frac{1}{r_{34} } )[/tex]
let's look for the distances
r₁₄ = [tex]\sqrt{3^2 +4^2}[/tex]
r₁₄ = 5 m
r₂₄ = [tex]\sqrt{3^2 + ( 4-2)^2}[/tex]
r₂₄ = √13 = 3.606 m
r₃₄ = [tex]\sqrt{(3-3)^2 + (4-0)^2}[/tex]
r₃₄ = 4 m
we calculate
W = 9 10⁹ (1.6 10⁻¹⁹)² [tex]( \frac{1}{5} + \frac{1}{3.606} + \frac{1}{4} )[/tex]
W = 1.68 10⁻²⁸ J
d) we take the proton to the location x5 = 1m y5 = 1m
W = U = k [tex]( \frac{q_1q_5}{r_{15} } + \frac{q_2q_5}{r_{25} } + \frac{q_3q_5}{r_{35} } + \frac{q_4q_5}{r_{45} } )[/tex]
in this case the charges have the same values but charge 5 is positive and the others negative, so the products of the charges give a negative value
W = - k q² [tex]( \frac{1}{r_{15} } + \frac{1}{r_{25} } + \frac{1}{r_{35} } + \frac{1}{r_{45} } )[/tex]
we look for distances
r₁₅ = [tex]\sqrt{ 1^2 +1^2}[/tex]Ra (1-0) 2 + (1-0) 2
r₁₅ = √ 2 = 1.4142 m
r₂₅ = [tex]\sqrt{ (2-1)^2 +1^2}[/tex]
r₂₅ = √2 = 1.4142 m
r₃₅ = [tex]\sqrt{ ( 3-1)^2 +1^2}[/tex]
r₃₅ = √5 = 2.236 m
r₄₅ = [tex]\sqrt{ (3-1)^2 + (4-1)^2}[/tex]
r₄₅ = √13 = 3.606 m
we calculate
W = - 9 10⁹ (1.6 10⁻¹⁹)² [tex]( \frac{1}{1.4142} +\frac{1}{1.4142} + \frac{1}{2.236} + \frac{1}{3.606} )[/tex]
W = - 4.93 10⁻²⁸ J
A high-voltage direct-current generating station delivers 10 MW of power at 250 kV to a city, as depicted in Fig. P2.12. The city is represented by resistance RL and each of the two wires of the transmission line between the generating station and the city is represented by resistance RTL. The distance between the two locations is 2000 km and the transmission lines are made of 10 cm diameter copper wire. Determine (a) how much power is consumed by the transmission line and (b) 12 V I0 _
Answer:
The answer is below
Explanation:
The resistivity of copper is ρ = 1.72 * 10⁻⁸ Ωm, diameter d = 10 cm = 0.1 m
The resistance (R) of transmission line is given as:
Rtl = ρL / A; where ρ = resistivity of copper = 1.72 * 10⁻⁸ Ωm, L = length of transmission line = 2000 km = 2000000 m, A is the area of the wire = πd²/4 = π(0.1)²/4
[tex]R_{tl}=\frac{\rho L}{A}=\frac{1.72*10^{-8}*2000000}{\pi*0.1^2/4}=4.4 \ ohm[/tex]
Power = [tex]\frac{V_L^2}{R_L}[/tex]
Power = 10 MW = 10 * 10⁶ W
[tex]10*10^6=\frac{(250*10^3)^2}{R_L} \\\\R_L=\frac{(250*10^3)^2}{10*10^6} \\\\R_L=6250\ ohm[/tex]
[tex]I_L=\frac{V_L}{R_L} \\\\I_L=\frac{250*10^3}{6250} =40\ A[/tex]
a) Since there are two tranmission lines, the power consumed by the lines is:
[tex]P_{TL}=2*I_L^2*R_{TL}=2*40^2*4.4=14080\ W[/tex]
b) The energy generated by the source = 10 * 10⁶ W + 14080 W = 10014080 W
Fraction used = 10 * 10⁶ / 10014080 * 100% = 99.86%
Friction is necessary when you are on a bike to stay
Answer:
yes friction is needed hope this helps might of been to long tho
2. One tin for weight control is to:
Eat alone
Eat slowly
Answer:
Eat slowly
Explanation:
If you eat slower, you'll chew your food better, which leads to better digestion. Digestion actually starts in the mouth, so the more work you do up there, the less you'll have to do in your stomach. This can help lead to fewer digestive problems. Less stress.
Please help I don’t get this give me answers please
Answer:
c
Explanation:
A vibrating object produces periodic waves with a wavelength of 53 cm and a frequency of 15 Hz. How fast do these waves move away from the object?
Answer:
v = 7.95 m/s
Explanation:
Given that,
Wavelength of a wave, [tex]\lambda=53\ cm=0.53\ m[/tex]
Frequency of a wave, f = 15 Hz
We need to find the speed of the wave. The speed of a wave is given by :
[tex]v=f\lambda\\\\v=15\ Hz\times 0.53\ m\\\\v=7.95\ m/s[/tex]
So, the wave move with a speed of 7.95 m/s.
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Answer:
b
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this because if force (f) are resultant cos then the point is proportional to the direction of c at greatest possible forces
Determine the voltage Vab for the first circuit and also determine the voltages Vab and Vcd for the second circuit
Vab= E = 20V
because I = 0 and the voltage drop across the resistances R1 and R2 is also 0.
Second circuit:
Vab = 10V (no voltage drop across R1)
Vcd= E2-E1 = 20V
Series connection of voltage sources. But the sources are connected to the contrary and voltage drop across R1 or R2 is 0 V.
An ideal gas in a 50.0 L tank has a
pressure of 2.45 atm at 22.5°C.
How many moles of gas are in
the tank?
Answer:
5.05225 moles
Explanation:
The computation of the number of moles of gas in the tank is shown below:
Given that
Volume = V = 50 L = 50.0 × 10^-3m^3
Pressure = P = 2.45 atm = 2.45 × 101325
Temperature = T = 22.5°C = (22.5 + 273)k = 295.5 K
As we know thta the value of gas constant R is 8.314 J/mol.K
Now
PV = nRT
n = PV ÷ RT
= ((2.45 × 101325) (50.0 × 10^-3)) ÷ ((8.314) (295.5))
= 5.05225 moles
This is the build up of substance such as pesticides in an organism and occurs when an organism absorb a substance at a rate faster than that at which the substance is lost
Answer:
which the substance is lost by catabolism and excretion.
Explanation:
One species of eucalyptus tree, Eucalyptus regnans, grow to heights similar to those attained by California redwoods. Suppose a bird sitting on top of one specimen of eucalyptus tree drops a nut that is 1.7 ounces. If the speed of the falling nut at the moment it is 50.3 m above the ground is 42.7 m/s, how tall is the tree
Answer:
The tree is 143.325 meters tall
Explanation:
The given parameters of the eucalyptus tree are;
The mass of the eucalyptus tree nut = 1.7 ounces
The speed of the nut at 50.3 m above the ground, v = 42.7 m/s
The equation for free fall is given as follows;
v² = 2·g·h
Where;
v = The velocity after falling through a height, h
g = The acceleration due to gravity = 9.8 m/s²
h = The height through which the seed has already fallen
Therefore, we have;
h = v²/(2·g) = (42.7 m/s)²/(2 × 9.8 m/s²) = 93.025 m
The height through which the seed has already fallen, h = 93.025 m
The height of the tree = h + The height of the seed above ground at the moment it was falling at 42.7 m/s
The height of the tree = 93.025 m + 50.3 m = 143.325 m
The height of the tree = 143.325 m.
The height of the eucalyptus tree is approximately 111.9 meters.
To determine the height of the tree, we can use the equations of motion. The initial velocity of the nut, u, is 0 m/s (since it is dropped), the acceleration due to gravity, a, is approximately 9.8 m/s², and the final velocity, v, is 42.7 m/s. We need to find the height, h. Using the equation v² = u² + 2a(h - u), we can rearrange it to solve for h: h = (v² - u²) / (2a) Plugging in the values, we get: h = (42.7² - 0²) / (2 * 9.8) = 111.9 meters Therefore, the height of the eucalyptus tree is approximately 111.9 meters.
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The electric field from two charges in the plane of the paper is represented by the dashed lines and arrows below.
Select a response for each statement below. (Use 'North' towards top of page, and 'East' to the right)
The magnitude of the E-field at Ris .... than at M.
The force on a (+) test charge at P is zero.
The magnitude of the charge on the left is .... that on the right.
The force on a (+) test charge at L is directed ....
The force on a (-) test charge at J is directed
The force on a (-) test charge at N is directed ....
The sign of the charge on the right is negative.
Answer:
a) electric field at point P must be zero
b) harged must be positive
c) force ais in the direction of the electric field
d) force is in the opposite direction to the electric field
e) force is in the opposite direction to the field
Explanation:
After reading your exercise, it is unfortunate that the diagram did not come out, but we are going to answer the questions in general.
a) force on a charge (+) is zero
this implies that the electric field at point P must be zero
F = q E
b) the magnitude of the charge on the left is on the right
this indicates that the charged must be positive since the lines must exit the charge
c) force on load directed towards (direction not indicated)
since the charge is positive the force at point L is in the direction of the electric field at this point
d) force on test load (-) does not indicate direction
The force on a negative charge is in the opposite direction to the electric field at point J
e) Force on a test load (-) at point N
the force is in the opposite direction to the field at point N
An electron, tial well may be anywhere within the interval 2a. So the uncertainty in its position is Δx= 2a. There must be a corresponding uncertainty in the momentum of the electron and hence it must have a certain kinetic energy. Calculate this energy from the uncertainty relationship and compare it.
Answer:
[tex]K = \frac{h'}{8 m \ \Delta x^2}[/tex]K
Explanation:
The Heisenberg uncertainty principle is
Δx Δp ≥ h' / 2
h’ =[tex]\frac{h}{2\pi }[/tex]
The kinetic energy of a particle is
K = ½ m v²
p = mv
v = [tex]\frac{p}{m}[/tex]
substitute
K = [tex]\frac{1}{2} \frac{p^2}{m}[/tex]
from the uncertainty principle,
Δp = [tex]\frac{h'}{2 \ \Delta x}[/tex]
we substitute
K = [tex]\frac{1}{2m} ( \frac{h'}{2 \ \Delta x})^2[/tex]
[tex]K = \frac{h'}{8 m \ \Delta x^2}[/tex]
A question to think about on units: Suppose we wanted to exchange scientific information with a newly discovered species of intelligent life living on a planet orbiting the star Alpha Centauri. And suppose that our new friends have managed to become fluent in our language, but have not yet had the opportunity to visit the Earth. Which of the following statements would they understand?
a. The mass of the electron is 9.10938188 x 10^-31.
b. The speed of light is 2.99792458 x 10^8 meters/second.
c. The ratio of the proton's mass to the electron's mass is 1836.153. Explain your reasoning
Answer:
b. The speed of light is 2.99792458 x 10^8 meters/second.
Explanation:
Speed of light is a universal constant and its value is same throughout the universe . So alien living near Alpha Centauri will quickly understand about it . But other statements are not universal . Mass of electron can vary as per relativistic formula of Einstein . Similarly , mass of proton can also vary according to relativistic concept . It depends upon the velocity of particle . So, the ratio of mass of proton and mass of electron will also vary from one star to another .
• How much work is
required to lift a 2kg
object 2m high?
Answer You need to consider that the gravity on earth is 9.8 m/s/s. This means any object you let go on the earths surface will gain 9.8 m/s of speed every second. You need to apply a force on the object in the opposite direction to avoid this acceleration. If you are pushing something up at a constant speed, you are just resisting earths acceleration. The more massive and object is, the greater force is needed to accelerate it. The equation is Force = mass*acceleration. So for a 2kg object in a 9.8 m/s/s gravity you need 2kg*9.8m/s/s = 19.6 Newtons to counteract gravity. Work or energy = force * distance. So to push with 19.6 N over a distance of 2 meters = 19.6 N*2 m = 39.2 Joules of energy. There is an equation that puts together those two equations I just used and it is E = mgh
The amount of Energy to lift an object is (mass) * (acceleration due to gravity) * (height)
:Hence, the Work done to life the mass of 2 kg to a height of 10 m is 196 J. Hope it helps❤️❤️❤️
Explanation:
Young David who slew Goliath experimented with slings before tackling the giant. He found that he could revolve a sling of length 0.600 m at the rate of 5.00 rev/s. If he increased the length to 0.900 m, he could revolve the sling only 3.00 times per second. (a) What is the speed of the stone for each rate of rotation
Answer:
Explanation:
For circular motion of stone the formula is
v = ω R where ω is angular velocity , v is linear velocity .
For first motion ,
R = length of sling = .6 m
ω = 2π n , n is no of revolution per second
ω = 2 x 3.14 x 5 = 31.4 rad /s
v = 31.4 x .6 = 18.84 m /s
For second motion ,
R = length of sling = .9 m
ω = 2π n , n is no of revolution per second
ω = 2 x 3.14 x 5 = 31.4 rad /s
v = 31.4 x .9 = 28.26 m /s
2.) The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his/her head, forcing them to move quickly away from the net. Suppose that you loft the ball with an initial speed of 15m/s at an angle of 50 degrees from the horizontal. At this moment your opponent is 10m from the ball. They begin to run away from you 0.3 seconds after the ball was launched hoping to reach the ball and hit it back to you at a height of 2.1m above where you hit it. What is the minimum average speed that your opponent must move so that he is in position to hit this ball
Answer:
The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s
Explanation:
The given parameters of the ball are;
The initial speed of the ball = 15 m/s
The direction in which the ball is launched = 50° above the horizontal
The location of the other tennis player when the ball is launched = 10 m from the ball
The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched
The height at which the ball is hit back = 2.1 m above the height from which the ball is launched
The vertical position, 'y', at time, 't', of a projectile motion is given as follows;
y = (u·sinθ)·t - 1/2·g·t²
When y = 2.1 m, we have;
2.1 = (15·sin(50°))·t - 1/2·9.8·t²
∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0
Solving with the aid of a graphing calculator function, we get;
t = 0.199776187257 s or t = 2.14525782198 s
Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds
The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;
x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m
The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m
Therefore, we have;
The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s
The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m
The minimum average speed the other player has to move with, [tex]v_s[/tex] = d/t₂
∴ [tex]v_s[/tex] = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s
The minimum average speed the opponent must move so that he is in position to hit the ball, [tex]v_s[/tex] ≈ 5.79 m/s.
An insulated, vertical piston-cylinder assembly contains 50 L of steam at 105 oC. The outside pressure is 101 kPa. The piston has a diameter of 20 cm and the combined mass of the piston and the load is 75 kg. The electrical heater and the paddle wheel are turned on and the piston rises slowly by 25 cm with a constant pressure. The total internal energy increases by 3.109 kJ.
Determine:
a. The pressure of air inside the cylinder during the process.
b. The boundary work performed by the gas.
c. The combined work transfer by the shaft and electricity.
Answer:
Explanation:
From the given information:
The pressure of the air during the process = [tex]P_{atm} + P_{due \ to \ wt \ of \ piston}[/tex]
[tex]= 101 \ kPa + \dfrac{75 \ kg \times 9.8 \ m/s^2 \times \dfrac{1 \ N }{1 \ kg.m/s^2} }{\dfrac{\pi}{4}(0.2 \ m)^2} ( \dfrac{1 \ N }{m^2} \times \dfrac{1 \ kPa}{1000 \ n/m^2})[/tex]
The pressure of the air during the process = 124.42 kPa
The boundary work = P × ΔW
The boundary work = 124.42 kPa × (π/4) × (0.2 m)² × 0.25 m × (1 kJ/1 kPa.m³)
The boundary work = 0.977 kJ
The combined work transfer = [tex]W_{boundary} + \Delta U[/tex]
The combined work transfer = 0.977 + 3.109 kJ
The combined work transfer = 4.086 kJ
At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power generation potential of a wind turbine with 60-m-diameter blades at that location. Take the air density to be 1.25 kg/m3. Cengel, Yunus; Cengel, Yunus. Thermodynamics: An Engineering Approach (p. 98). McGraw-Hill Higher Education. Kindle Edition.
Answer:
1767Kw
Explanation:
Velocity of wind = 10 m/s
diameter of the blades= 60m
ρ= air density = 1.25 kg/m3
Acceleration due to gravity= 9.81 m/s^2
Mechanical energy of the wind can be calculated using the expression below
Energy= (e*m)
= ρ V A e............eqn(1)
Where A= area
ρ= air density
e= wind energy per unit mass of air
e= (v^2)/2..........eqn(2)
If we substitute the values into eqn (2) we have
e= [(10)^2]/2
=50J/Kg
But Area=A= (πd^2)/4
Area= ( π× 60^2)/4
Area=2827.8m^2
If we input substitute the values into eqn (1) we have
Energy= 1.25 ×10 × 50×2827.8
=1767145.7W
We can convert to kilo watt
=1767145.7W/ 1000
= 1767Kw
Hence, the mechanical energy of air per unit mass and the power generation potential of a wind turbine is 1767Kw
What is electronegativity
A proton accelerates from rest in a uniform electric field of 664 N/C. At some later time, its speed is 1.46 106 m/s. (a) Find the magnitude of the acceleration of the proton. m/s2 (b) How long does it take the proton to reach this speed? µs (c) How far has it moved in that interval? m (d) What is its kinetic energy at the later time? J
Answer:
Explanation:
In an electric field E force on charge q
F = Eq , acceleration a = Eq / m
a = 664 x 1.6 x 10⁻¹⁹ / 1.67 x 10⁻²⁷
= 636.16 x 10⁸ m /s²
b )
initial velocity u = 0
final velocity v = 1.46 x 10⁶ m/s
v = u + at
1.46 x 10⁶ = 0 + 636.16 x 10⁸ x t
t = 2.29 x 10⁻⁵ s
c )
s = ut + 1/2 a t²
= 0 + .5 x 636.16 x 10⁸ x ( 2.29 x 10⁻⁵ )²
= 1668 x 10⁻²
= 16.68 m
d )
Kinetic energy = 1/2 m v²
= .5 x 1.67 x 10⁻²⁷ x ( 1.46 x 10⁶ )²
= 1.78 x 10⁻¹⁵ J .
1
What kind of adaptation is a long neck on a tortoise? *
(10 Points)
O
A. Structural
B. Behavioral
a
C. Functional
a
D. Physiological
Answer:
The answer is ......... structural adaptation
Explanation:
because structural adaptations is a physical thinng in their body so its A please give me brainliest
Energy Transformation and Conservation
Explain how different forms of energy are related.
Answer:
Energy transformation is when energy changes from one form to another – like in a hydroelectric dam that transforms the kinetic energy of water into electrical energy. While energy can be transferred or transformed, the total amount of energy does not change – this is called energy conservation.
Explanation:
I love you
Please answer my question :-)
Answer:
A- Astronomical body
C- Galaxy
D- Comet
B- Moon
Hope this helps you! Have a great day!
Answer:
1. A
2. C
3. D
4. B
Explanation:
PLZZZZ HELPPPPPPPPPppppp
what happens when a wave passes through a medium ?
Answer:
When waves travel from one medium to another the frequency never changes. As waves travel into the denser medium, they slow down and wavelength decreases. Part of the wave travels faster for longer causing the wave to turn. The wave is slower but the wavelength is shorter meaning frequency remains the same.
Explanation:
Formula One racers speed up much more quickly than normal passenger vehicles, and they also can stop in a much shorter distance. A Formula One racer traveling at 90m/s can stop in a distance of 110m. What is the magnitude of the car's acceleration as it slows during braking?
Answer:
The magnitude of the car's acceleration as it slows during braking is 36.81 m/s²
Explanation:
From the question, the given values are as follows:
Initial velocity, u = 90 m/s
final velocity, v = 0 m/s
distance, s = 110 m
acceleration, a = ?
Using the equation of motion, v² = u² + 2as
(90)² + 2 * 110 * a = 0
8100 + 220a = 0
220a = -8100
a = -8100/220
a = -36.81 m/s²
The value for acceleration is negative showing that car is decelerating to a stop. The magnitude of the car's acceleration as it slows during braking is therefore 36.81 m/s²
If a satellite is orbiting the Earth in elliptical motion, then it will move _______________ (slowest, fastest) when its closest to the Earth. While moving towards the Earth (along the path from D to A) there is a component of force in the __________________ (same, opposite) direction as the motion; this causes the satellite to ___________________ (slow down, speed up). While moving away from the Earth (along the path from A to D) there is a component of force in the _________________ (same, opposite) direction as the motion; this causes the satellite to ___________________ (slow
Answer:fastest,same,slow down,opposite,slow
Explanation:
A satellite move fastest when its closest to the Earth. The other correct options are same direction, speed up, opposite direction and slow.
Velocity of a satellite around the planet.If a satellite is orbiting the Earth in elliptical motion, then it will move fastest when its closest to the Earth (based on Kepler's, law).
While moving towards the Earth (along the path from D to A) there is a component of force in the same direction as the motion; this causes the satellite to speed up.
While moving away from the Earth (along the path from A to D) there is a component of force in the opposite direction as the motion; this causes the satellite to slow.
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Bartender slides a beer mug at 1.1 m/s towards a customer at the end of the bar which is 1.8 m tall. The customer makes a grab for the mug and misses and mug sails at the end of the bar. a) How far away from the end of the bar does the mug hit the floor
Answer:
Δx = 0.7 m
Explanation:
Once the mug is moving in the horizontal direction, it keeps moving at the same speed of 1.1 m/s, due to no other force acts on it in this direction.Since the horizontal and vertical movements are independent each other (due to they are mutually perpendicular), in the vertical direction, the initial speed is just zero.In the vertical direction, the mug is accelerated by the force of gravity at all times, with a constant value of 9.8 m/s2, aimed downward.So, we can use the following kinematic equation in order to get the time passed from the instant that the mug left the bar, until it hit the floor, as follows:[tex]\Delta y = \frac{1}{2} * g* t^{2} = (1)[/tex]where Δy = 0-1.8m = -1.8m, g= -9.8m/s2.Replacing these values in (1) and solving for t, we get:[tex]t = \sqrt{\frac{2*1.8m}{ 9.8m/s2} } = 0.6 s (2)[/tex]
Now, since the mug obviously finishes its horizontal trip at this same time (hitting ground), we can find the horizontal distance traveled, just applying the definition of average speed, as follows:[tex]\Delta x = v_{o} * t = 1.1 m/s* 0.6 s = 0.7 m (3)[/tex]
Which statement describes what will most likely occur when warm air cools and the temperature drops to the
point?
A.air will contain more water vapor. B. Dew will form on leaves C.clouds will disappear. D. Water vapor in the air will evaporate
Answer:
The person who did the comment is correct, it B - Dew will form on leaves
Explanation:
Answer:
B
Explanation:
Valeriie.07 tap in g
Which of the following hydrocarbons are SATURATED hydrocarbons?
I. alkanes II. alkenes III. alkynes IV. cycloalkanes
A. I and IV
B. II and III
C. I and III
D. II and IV
Answer:
i think c
Explanation:
