Answer:
t_{out} = [tex]\frac{v_s - v_r}{v_s+v_r}[/tex] t_{in}, t_{out} = [tex]\frac{D}{v_s +v_r}[/tex]
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is
[tex]v_{sg 1} = v_{sr} + v_{rg}[/tex]
The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
[tex]v_{sg1}[/tex] = D / [tex]t_{out}[/tex]
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point
[tex]v_{sg 2} = v_{sr} - v_{rg}[/tex]
[tex]v_{sg 2}[/tex] = D / [tex]t_{in}[/tex]
D = v_{sg 2} t_{in}
with the distance is the same we can equalize
[tex]v_{sg1} t_{out} = v_{sg2} t_{in}[/tex]
t_{out} = t_{in}
t_{out} = [tex]\frac{v_s - v_r}{v_s+v_r}[/tex] t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D / [tex]v_{sg2}[/tex]
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} = [tex]\frac{D}{v_s +v_r}[/tex]
At what tempreture will the of and oC
be the same
Answer:
-40 degreesTo find the temperature when both are equal, we use an old algebra trick and just set �F = �C and solve one of the equations. So the temperature when both the Celsius and Fahrenheit scales are the same is -40 degrees.
Explanation:
Hope it is helpful.....
Answer:
To find the temperature when both are equal, we use an old algebra trick and just set �F = �C and solve one of the equations. So the temperature when both the Celsius and Fahrenheit scales are the same is -40 degrees.
what is a vector quantity?
Answer:
A quantity that has magnitude and direction. It's usually represented by an arrow whose direction is the same direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude
PROBLEM 5 (Problem 4-145 in 7th edition) Consider a well-insulated horizontal rigid cylinder that is divided into two compartments by a piston that is free to move but does not allow either gas to leak into the other side. Initially, one side of the piston contains 1 m3 of N2 gas at 500 kPa and 120oC while the other side contains 1 m3 of He gas at 500 kPa and 40oC. Assume the piston is made of 8 kg of copper initially at the average temperature of the two gases on both sides. Now thermal equilibrium is established in the cylinder as a result of heat transfer through the piston. Using constant specific heats at room temperature, determine the final equilibrium temperature in the cylinder. What would your answer be if the piston were not free to move
Answer:
The answer is "[tex]\bold{83.8^{\circ} \ C}[/tex]".
Explanation:
Formula for calculating the mass in He:
[tex]\to m = \frac{PV}{RT}\\[/tex]
[tex]= \frac{500 \times 1}{ 2.0769 \times (40 + 273)}\\\\ = \frac{500 }{ 2.0769 \times 313}\\\\ = \frac{500 }{ 650.0697}\\\\= 0.76914 \ Kg[/tex]
Formula for calculating the mass in [tex]N_2[/tex]:
[tex]\to m = \frac{PV}{RT}\\[/tex]
[tex]= \frac{500 \times 1}{ 0.2968 \times (120+ 273)}\\\\ = \frac{500 }{ 0.2968 \times 393}\\\\ = \frac{500 }{ 116.6424}\\\\= 4.2866\ Kg[/tex]
by using the temperature balancing the equation:
[tex]T' = \frac{mcT (He) + mcT ( N_2 )}{ mc (He) + mc ( N_2)}[/tex]
[tex]= \frac{0.76914 \times 3.1156 \times 313 + 4.2866 \times 0.743 \times393}{ 0.76914 \times 3.1156 + 4.2866 \times 0.743} \\\\ = 357 \ \ K \approx 83.8^{\circ} \ C[/tex]
As mentioned in the text, the tangent line to a smooth curve r(t) = ƒ(t)i + g(t)j + h(t)k at t = t0 is the line that passes through the point (ƒ(t0), g(t0), h(t0)) parallel to v(t0), the curve’s velocity vector at t0. In Exercises 23–26, find parametric equations for the line that is tangent to the given curve at the given parameter value t = t0.
Answer:
[tex]x = t[/tex]
[tex]y = \frac{1}{3}t[/tex]
[tex]z =t[/tex]
Explanation:
Given
[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex] at [tex]t = 0[/tex]
Point: [tex](f(t0), g(t0), h(t0))[/tex]
[tex]r(t) = ln\ t_i + \frac{t-1}{t+2}j + t\ ln\ tk[/tex], [tex]t0 = 1[/tex] -- Missing Information
Required
Determine the parametric equations
[tex]r(t) = ln\ ti + \frac{t-1}{t+2}j + t\ ln\ tk[/tex]
Differentiate with respect to t
[tex]r'(t) = \frac{1}{t}i +\frac{3}{(t+2)^2}j + (ln\ t + 1)k[/tex]
Let t = 1 (i.e [tex]t0 = 1[/tex])
[tex]r'(1) = \frac{1}{1}i +\frac{3}{(1+2)^2}j + (ln\ 1 + 1)k[/tex]
[tex]r'(1) = i +\frac{3}{3^2}j + (0 + 1)k[/tex]
[tex]r'(1) = i +\frac{3}{9}j + (1)k[/tex]
[tex]r'(1) = i +\frac{1}{3}j + (1)k[/tex]
[tex]r'(1) = i +\frac{1}{3}j + k[/tex]
To solve for x, y and z, we make use of:
[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex]
This implies that:
[tex]r'(1)t = xi + yj + zk[/tex]
So, we have:
[tex]xi + yj + zk = (i +\frac{1}{3}j + k)t[/tex]
[tex]xi + yj + zk = it +\frac{1}{3}jt + kt[/tex]
By comparison:
[tex]xi = it[/tex]
Divide by i
[tex]x = t[/tex]
[tex]yj = \frac{1}{3}jt[/tex]
Divide by j
[tex]y = \frac{1}{3}t[/tex]
[tex]zk = kt[/tex]
Divide by k
[tex]z = t[/tex]
Hence, the parametric equations are:
[tex]x = t[/tex]
[tex]y = \frac{1}{3}t[/tex]
[tex]z =t[/tex]
a car acceleration from rest to 90km/h in 10 seconds. what is its acceleration in meter per second square?
Answer:
2.5 m/s^2
Explanation:
First, convert 90 km/hr into m/s:
90/3.6 = 25 m/s
vf = final velocity = 25 m/s
vi = initial velocity = 0 m/s
t = time = 10 seconds
a = acceleration, unknown
Then, find a using the following equation:
(vf - vi)/t = a
(25 m/s)/10 s = 2.5 m/s^2
a = 2.5 m/s^2
Hope this helps!! :)
How do you classify flammable liquid to gas and solid?
Answer:
A flammable liquid enjoys the attention of at least three different federal agencies: the DOT in matters of its transportation, OSHA as it might affect workplace safety, and the EPA concerning its cradle-to-grave management.A flammable liquid enjoys the attention of at least three different federal agencies: the DOT in matters of its transportation, OSHA as it might affect workplace safety, and the EPA concerning its cradle-to-grave management.Add to that a gaggle of state, regional, and local authorities; and we feel compelled to begin this blog entry with our favorite caveat: get expert advice before deciding what to do with that rusting drum of stale gasoline out back.Explanation:
I hope it help you;)
Two strings with linear densities of 5 g/m are stretched over pulleys, adjusted to have vibrating lengths of 0.50 m, and attached to hanging blocks. The block attached to string 1 has a mass of 20 kg and the block attached to string 2 has a mass of M. Listeners hear a beat frequency of 2 Hz when string 1 is excited at its fundamental frequency and string 2 is excited at its third harmonic. What is one possible value for mass M
Answer:
2.18 kg
Explanation:
The frequency of a wave in a stretched string f = n/2L√(T/μ) where n = harmonic number, L = length of string, T = tension = mg where m = mass of object on string and g = acceleration due to gravity = 9.8 m/s² and μ = linear density of string.
For string 1, its fundamental frequency f is when n = 1. So,
f = 1/2L√(T/μ) = 1/2L√(mg/μ)
Now for string 1, L = 0.50 m, m = 20 kg and μ = 5 g/m = 0.005 kg/m
substituting the values of the variables into f, we have
f = 1/2L√(mg/μ)
f = 1/2 × 0.50 m√(20 kg × 9.8 m/s²/0.005 kg/m)
f = 1/1 m√(196 kgm/s²/0.005 kg/m)
f = 1/1 m√(39200 m²/s²)
f = 1/1 m × 197.99 m/s
f = 197.99 /s
f = 197.99 Hz
f ≅ 198 Hz
For string 2, at its third harmonic frequency f' is when n = 3. So,
f' = 3/2L√(T/μ) = 3/2L√(mg/μ)
Now for string 2, L = 0.50 m, m = M kg and μ = 5 g/m = 0.005 kg/m
substituting the values of the variables into f, we have
f' = 3/2L√(Mg/μ)
f' = 3/2 × 0.50 m√(M × 9.8 m/s²/0.005 kg/m)
f' = 3/1 m√(M1960 m²/s²kg)
f' = 3/1 m√M√(1960 m²/s²kg)
f' = 3/1 m √M × 44.27 m/s√kg
f' = 132.81√M/s√kg
f' = 132.81√M Hz/√kg
Since the frequency of the beat heard is 2 Hz,
f - f' = 2 Hz
So, 198 Hz - 132.81√M Hz/√kg = 2 Hz
132.81√M Hz/√kg = 198 Hz - 2 Hz
132.81√M Hz/√kg = 196 Hz
√M Hz/√kg = 196 Hz/138.81 Hz
√M/√kg = 1.476
squaring both sides,
[√M/√kg] = (1.476)²
M/kg = 2.178
M = 2.178 kg
M ≅ 2.18 kg
What is the acceleration of a .3 kg mass when there is a net force of 25.9 N on it?
Answer:
86.33m/s^2
Explanation:
Acceleration = Force/Mass
= 25.9/0.3
= 86.33
An object initially traveling in a straight line with
a speed of 5.0 meters per second is accelerated
at 2.0 meters per second squared for 4.0 seconds.
The total distance traveled by the object in the
4.0 seconds is
Answer:
We conclude that the total distance traveled by the object in the 4 seconds is 36 m.
Explanation:
Given
Initial velocity u = 5.0 m/sAcceleration a = 2.0 m/s²Time t = 4 sTo determine
The total distance traveled by the object in the 4.0 seconds is
Important Tip:
We can determine the total distance traveled by the object in the 4.0 seconds by using the equation of motion such as
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s = distanceu = initial velocitya = accelerationt = timesubstituting u = 5.0, a = 2, and t = 4 in the formula
[tex]s=ut+\frac{1}{2}at^2[/tex]
[tex]s=\left(5\right)\left(4\right)+\frac{1}{2}\left(2\right)\left(4\right)^2[/tex]
[tex]s=20+16[/tex]
[tex]s=36[/tex] m
Therefore, we conclude that the total distance traveled by the object in the 4 seconds is 36 m.
The total distance traveled by the object is 36 meters.
Given the following data:
Initial velocity = 5.0 m/s Acceleration = 2.0 [tex]m/s^2[/tex] Time = 4.0 seconds.To find the total distance traveled by the object, we we would use the second equation of motion.
Mathematically, the second equation of motion is given by the formula;
[tex]S = ut + \frac{1}{2}at^2[/tex]
Where:
S is the total distance traveled.
u is the initial velocity.
a is the acceleration.
t is the time measured in seconds.
Substituting the values into the formula, we have;
[tex]S = 5(4) + \frac{1}{2}(2)(4^2)\\\\S = 20 + 1(16)[/tex]
Total distance, S = 36 meters.
Therefore, the total distance traveled by the object is 36 meters.
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A mole of a monatomic ideal gas at point 1 (101 kPa, 5 L) is expanded adiabatically until the volume is doubled at point 2. Then it is cooled isochorically until the pressure is 20 kPa at point 3. The gas is now compressed isothermally until its volume is back to 5 L (point 4). Finally, the gas is heated isochorically to return to point 1.
a. Draw the four processes and label the points in the pV plane.
b. Calculate the work done going from 1 to 2.
c. Calculate the pressure and temperature at point 2.
d. Calculate the temperature at point 3.
e. Calculate the temperature and pressure and point 4.
f. Calculate the work done going from from 3 to 4.
g. Calculate the heat flow into the gas going from 3 to 4. g
Answer:
(a). Check attachment.
(b). 280.305 J.
(c). 31.81 kpa; 38.26K.
(d). 24.05K.
(e). 24.05k; 40kpa.
(f). -138.6J.
Explanation:
(a). Kindly check the attached picture for the diagram showing the four process.
1 - 2 = adiabatic expansion process.
2 - 3 = Isochoric process.
3 - 4 = isothermal process.
4 - 1 = isochoric process.
(b). Recall that the process from 1 to is an adiabatic expansion process.
NB: b = 5/3 for a monoatomic gas.
Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].
= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.
Thus, the workdone = 280.305 J.
(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.
T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.
(d). The process 2 - 3 is an Isochoric process, then;
T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.
(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.
The pressure can be determine as below;
P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.
(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J
Why are we seeing extremely old light from Canopus instead of light in real-time?
Answer:
Canopus is more than 300 light years away from earth. This means it takes the light we see more than 300 years to reach us.
For a given substance, the molecules
move fastest when the substance is
Answer:GAS
Explanation:
Four-wheel drive trucks do not stop better on icy
roads than a car. Is what law of motion (Newton's laws)
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%
How do you think that changing the mass of the pendulum bob will affect the period of the pendulum swing?
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
A flat screen tv uses 120 watts. How much energy is used up if it is left on for 15 min?
A.) 4j
B.) 15j
C.) 0.67j
D.) 108,000j
Answer:
d
Explanation:
A battery has an emf of ε = 15.0 V. THe terminal voltage of the battery is Vt = 11.6 V when it is delivering P = 20.0 W of power to an external load resistor R. (a) What is the value of R? (b) What is the internal resistance r of the battery?
AnswerHM???
Explanation:
I dONT KNOW
A block of mass m1 = 19.5 kg slides along a horizontal surface (with friction, μk = 0.35) a distance d = 2.6 m before striking a second block of mass m2 = 8.25 kg. The first block has an initial velocity of v = 6.5 m/s.
(a) Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s?
(b) How far does block two travel, d2 in meters, before coming to rest after the collision?
Answer:
19.5 m/s
87.8 m
Explanation:
The acceleration of block one is:
∑F = ma
-m₁gμ = m₁a
a = -gμ
a = -(9.8 m/s²) (0.22)
a = -2.16 m/s²
The velocity of block one just before the collision is:
v² = v₀² + 2aΔx
v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)
v = 7.63 m/s
Momentum is conserved, so the velocity of block two just after the collision is:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m₁u₁ = m₂v₂
(18.5 kg) (7.63 m/s) = (7.25 kg) v
v = 19.5 m/s
The acceleration of block two is also -2.16 m/s², so the distance is:
v² = v₀² + 2aΔx
(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx
Δx = 87.8 m
Explanation:
By using conservation of linear momentum and also by equating work done to kinetic energy, [tex]V_{2}[/tex] = 15.36 m/s and [tex]d_{2}[/tex] = 4.32 meters
Parameters given are :
[tex]m_{1}[/tex] = 19.5 kg
friction, μk = 0.35
distance d = 2.6 m
mass [tex]m_{2}[/tex] = 8.25 kg.
initial velocity of [tex]U_{1}[/tex] = 6.5 m/s.
a.) Since we assumed that the block one stops after it collides with block two, the final velocity for block one will be zero. That is, [tex]V_{1}[/tex] = 0 so its final momentum = 0
Let us also assume that block two was initially at rest. Therefore, it initial velocity and its momentum will be equal to zero.
The formula to use will be :
[tex]m_{1}U_{1} = m_{2}V_{2}[/tex]
Substitute all the parameters into the formula above
19.5 x 6.5 = 8.25[tex]V_{2}[/tex]
Make [tex]V_{2}[/tex] the subject of formula
[tex]V_{2}[/tex] = 126.75/8.25
[tex]V_{2}[/tex] = 15.36 m/s
b.) Let us first calculate the work done in by block one.
The K.E = [tex]1/2mU^{2}[/tex]
substitute its mass and velocity into the formula
K.E = 1/2 x 19.5 x [tex]6.5^{2}[/tex]
K.E = 411.94 Joule
The work done = Kinetic energy
But the resultant Force F = force f - friction
where Frictional force = 0.35 x 19.5 x 9.8
Frictional force = 66.89N
Work done will be the product of resultant Force F and the distance travelled
(F - 66.89) x 2.6 = 411.94
F - 66.89 = 411.94/2.6
F - 66.89 = 158.44
F = 225.3 N
The second block will experience the same force which is equal to 225.3N
Find the kinetic energy of the second block.
K.E = [tex]1/2mV^{2}[/tex]
K.E = 0.5 x 8.25 x 15.36^2
K.E = 973.2
Using The work done = Kinetic energy
225.3[tex]d_{2}[/tex] = 973.2
[tex]d_{2}[/tex] = 973.2/225.3
[tex]d_{2}[/tex] = 4.32 meters
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what is the mathematical formula associated with newton's 2nd law of motion?
Answer:
F= m x a
Explanation:
Force (f) = mass (m) x acceleration (a)
is a step in the scientific method. The step that follows this step involves forming
Answer: read this hope this helped
Explanation: A hypothesis is a possible explanation for a set of observations or an answer to a scientific question. ... The next step in the scientific method is to test the hypothesis by designing an experiment. This includes creating a list of materials and a procedure— a step-by-step explanation of how to conduct the experiment.
A crate rests on a flatbed truck which is initially traveling at 13.6 m/s on a level road. The driver applies the brakes and the truck is brought to a halt in a distance of 38.1 m. If the deceleration of the truck is constant, what is the minimum coefficient of friction between the crate and the truck that is required to keep the crate from sliding
Answer:
0.248
Explanation:
Initial speed u = 13.6
Final speed v = 0
Distance s = 38.1
We have umg = ma
We make u subject of the formula
u = a/g
V² = u² + 2as
a = v²-u²/2s
We substitute the values into the above
a = 0-(13.6)²/2*38.1
a = 184.96/76.2
a = 2.427m/sec
Remember that
u = a/g
u = 2.427/9.8
= 0.2476
This is approximately 0.248
This is the minimum coefficient of friction required to keep the crate from sliding.
A 4.0-g bead carries a charge of 20 μC. The bead is accelerated from rest through a potential difference V, and afterward the bead is moving at 3.0 m/s. What is the magnitude of the potential difference V? *
1 point
900 V
400 V
200 V
400 kV
Answer:
The magnitude of the potential difference is 900 V.
Explanation:
Given;
mass of the bead, m = 4.0 g = 0.004 kg
charge of the bead, Q = 20 μC = 20 x 10⁻⁶ C
final velocity of the bead, v = 3 m/s
What is the magnitude of the potential difference V?
Apply the principle of conservation of energy;
The electric potential energy at the beginning is equal to kinetic energy of the bead at the end of the journey.
qV = ¹/₂mv²
[tex]V = \frac{mv^2}{2q} \\\\V = \frac{0.004 \ \times \ (3)^2}{2(20 \times 10^{-6})}\\\\V = 900 \ V[/tex]
Therefore, the magnitude of the potential difference is 900 V.
The magnitude of the potential difference (V) is equal to 900 Volts.
Given the following data:
Mass of bead = 4.0 g to kg = 0.004 kgCharge of bead = 20 μC = [tex]20 \times 10^{-6} \;C[/tex]Final velocity of bead = 3 m/sTo determine the magnitude of the potential difference (V):
How to calculate the potential difference (V).We would apply the law of conservation of energy, which states that the electric potential energy possessed by the bead at the beginning is equal to the kinetic energy possessed by the bead at the end of the journey:
[tex]qV = \frac{1}{2} mv^2\\\\V = \frac{\frac{1}{2} mv^2}{q} \\\\V = \frac{\frac{1}{2} \times 0.004 \times 3.0^2}{20 \times 10^{-6}} \\\\V = \frac{ 0.002 \times 9}{20 \times 10^{-6}}[/tex]
V = 900 V.
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please help me I'm begging you Define and give examples of elements and compounds the structure of atoms (electrons, neutrons, and protons)
please answer correctly
Answer:
616000 J.
Explanation:
Bi. Determination of the work done.
Force (F) = 220 N
Distance (s) = 2800 m
Workdone (Wd) =?
Workdone is simply defined as the product of force and the distance moved in the direction of the force. Mathematically, is can be expressed as:
Workdone = Force × distance
Wd = F × s
With the above formula, we can obtain the Workdone as follow:
Force (F) = 220 N
Distance (s) = 2800 m
Workdone (Wd) =?
Wd = F × s
Wd = 220 × 2800
Wd = 616000 J.
Any change in the cross section of the vocal tract shifts the individual formant frequencies, the direction of the shift depending on just where the change in area falls along the standing wave. Constriction of the vocal tract at a place where the standing wave of a formant exhibits minimum-amplitude pressure oscillations generally causes the formant to drop in frequency; expansion of the tract at those same places raises the frequency. Three other major tools for changing the shape of the tract in such a way that the frequency of a particular formant is shifted in a particular direction are the jaw, the body of the tongue and the tip of the tongue. Moving the various articulatory organs in different ways changes the frequencies of the two lowest formants over a considerable range [18].
One way to increase formant frequency is to ________ the vocal tract at a place where the standing wave of a formant frequency exhibits minimum-amplitude pressure oscillations.
a. Stretch
b. Vibrate
c. Contract
d. Expand
Answer:
The correct answer is option D.
Explanation:
It is stated in the question that constriction of the vocal tract at a place where the standing wave of a formant exhibits minimum-amplitude pressure oscillations generally causes the formant to drop in frequency so to increase formant frequency, the vocal should expand where the standing wave of a formant exhibits minimum-amplitude pressure oscillations. The answer is D.
I hope this helps.
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.
Learn more about motion of satellite here: https://brainly.com/question/25721729
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. [30%] We first showed that The electric field for a point charge radiating in 3-dimensions has a distance dependence of 1/r 2 (see Equation 1). In Problem 1 you showed that the electric field for a point charge radiating in 2-dimensions has a distance dependence of 1/r . Consider again the 2-dimensional case described in Problem 1. What distance dependence do you expect for the electric potential
Answer:
Answer is explained in the explanation section below.
Explanation:
Note: This question is incomplete and lacks necessary data to solve. As it mentioned the reference of problem number 1, which is missing in this question. However, I have found that question on the internet and will be solving the question accordingly.
Solution:
The relation between electric field and the electric potential is:
E = [tex]\frac{dV}{dr}[/tex]
So, making dV the subject, we have:
dV = E x dr
Integrating the above equation, we get.
V = [tex]\int\limits^_ {} \,[/tex]E x dr Equation 1
Now, in 2-D
E is inversely proportional to the radius r.
E ∝ 1/r
So, we can write: replacing E ∝ 1/r in the equation 1
V ∝ [tex]\int\limits^_ {} \,[/tex][tex]\frac{1}{r}[/tex] x dr
Which implies that,
V ∝ log (r)
Hence, distance dependence expected for the electric potential = ln (r)
PLEASE HELP ASAP! WILL GIVE BRAINLIEST TO CORRECT ANSWER! HELP!! HELP!!
The diagram shows the structure of an animal cell.
The image of an animal cell is shown with some organelles labeled numerically from 1 to 6. The outer double layer boundary of the cell is labeled 1. A stacked disc like structure is labeled 2. A broad rod shaped structure with an irregular shape inside it is labeled 3. The entire plain section that forms the background of the cell and is within the outer boundary is labeled 4. A small circular shape within the large circular shape is labeled 5. The large central circular shape is labeled 6.
Which number label represents the cell membrane?
1
2
4
6
(this is middle school science)
Answer:
1. cell membrane
2. golgi body
3. mitochondrion
4. cytoplasm
5. nucleolus
6. nucleus
Explanation:
The correct answer to this question is Option A; 6.
Why?
In a plant cell, the nucleus surrounds the nucleolous, which would be number 5. Therefore, number 6 would be your correct answer.
~Thank you~
Which example describes a nonrenewable resource?
Answer: Examples of nonrenewable resources include crude oil, natural gas, coal, and uranium. These are all resources that are processed into products that can be used commercially. For example, the fossil fuel industry extracts crude oil from the ground and converts it to gasoline.