Answer:
Explanation:
a.) To find Secrional liftSectional lift Coefficient [tex](c_{I})=\frac{W}{(0.5*Rho*V^{2})}[/tex]
where Rho = 1.225Kg/m³ (density of air at sea level)
[tex](c_{I})=\frac{1000}{(0.5*1.225*55^{2})}=0.5397[/tex]
b.) To obtain the sectional induced coefficient:
[tex]C_{D} = \frac{(C_{L})^2}{π(AR)}[/tex]
[tex]C_{D} = \frac{(0.5397)^2}{π(5)}=0.0185[/tex]
C.) For the effective angle of attack ∝0
∝0 = [tex]\frac{C_{l}}{m_{0}}\\=\frac{0.5397}{5.7}\\=0.0947rad[/tex]
For induced angle of attack ∝i:
∝i= [tex]\frac{-C_{D}}{C_{L}}\\=\frac{-0.0185}{0.5397}\\=-0.0343rad[/tex]
For absolute angle of attack ∝a:
∝a = ∝0 - ∝i = 0.0947 - (-0.0343) = 0.1290 rad
For geometric angle of attack ∝g:
∝g = ∝a + ∝L given ∝L = -2°≅-0.0349 rad
= 0.1290-0.0349 = 0.0941rad
Final answer:
The sectional lift coefficient is 0.3102, the sectional induced-drag coefficient is 0.0185, and the effective, induced, and geometric angles of attack are -2 degrees.
Explanation:
The sectional lift coefficient can be found using the formula: CL = W / (0.5 * density * V^2 * S), where CL is the sectional lift coefficient, W is the wing loading, density is the air density, V is the airspeed, and S is the wing area.
Substituting the given values, we get: CL = 1000 / (0.5 * 1.29 * (55^2) * (15/5)), which simplifies to CL = 0.3102
The sectional induced-drag coefficient can be found using the formula: CDi = CL^2 / (pi * AR * e), where CDi is the sectional induced-drag coefficient, CL is the sectional lift coefficient, AR is the aspect ratio, and e is the span efficiency factor. Substituting the given values, we get: CDi = 0.3102^2 / (pi * 5 * 0.8), which simplifies to CDi = 0.0185
The effective angle of attack can be found using the formula: alpha_eff = alpha_i + alpha_0, where alpha_eff is the effective angle of attack, alpha_i is the induced angle of attack, and alpha_0 is the geometric angle of attack.
Substituting the given values, we get: alpha_eff = -2 + 0, which simplifies to alpha_eff = -2 degrees.
A student needs to prepare 100.0 mL of vitamin C solution according to the directions in Part 1 of the experiment. S/he weighs out 0.0499 g vitamin C (molar mass = 176.124 g/mol) and dissolves it in water to give 100.0 mL solution. S/he pipets 10.00 mL of the solution into a beaker and adds 20 mL of water and 10 mL of buffer. (a) How many mg of vitamin C are in the beaker?
Answer:
4.99 mg of vitamin C are in the beaker.
Explanation:
Given that,
Weight of vitamin = 0.0499 g
Molar mass = 176.124 g/mol
Weight of water = 100.0 ml
We need to calculate the mg of vitamin C in the beaker
We dissolve 0.0499 g vitamin C in water to from 100.0 ml solution.
100 ml solution contain 49.9 mg vitamin C
Now, we take 10 ml of this vitamin C solution in breaker
Since, 100 ml solution =49.9 mg vitamin C
Therefore,
[tex]10\ ml\ solution =\dfrac{10\times49.9}{100}[/tex]
[tex]10\ ml\ solution =4.99\ mg[/tex]
Hence, 4.99 mg of vitamin C are in the beaker.
It is weigh-in time for the local under-85-kg rugby team. The bathroom scale used toassess eligibility can be described by Hooke’s law and is depressed 0.75 cm byitsmaximum load of 120 kg. (a) What is the spring’s effective spring constant? (b) Aplayer stands on the scales and depresses it by 0.48 cm. Is he eligible to play on thisunder-85-kg team?
To solve this problem it is necessary to apply the equations given in Newton's second law as well as Hooke's Law.
Since Newton's second law we have that force is
F = mg
Where,
m = mass
g = gravity
From the hook law, let us know that
F = -kx
Where
k = Spring constant
x = Displacement
Re-arrange to find k,
[tex]k = -\frac{F}{x}[/tex]
PART A ) We can replace the Newton definitions here, then
[tex]k = -\frac{mg}{x}[/tex]
Replacing with our values we have that
[tex]k = \frac{120*9.8}{-0.75*10^{-2}}[/tex]
[tex]k = 1.57*10^5N/m[/tex]
Therefore the required value of the spring constant is [tex]1.57*10^5N/m[/tex]
PART B) We can also equating both equation to find the mass, then
[tex]mg = -kx[/tex]
[tex]m = \frac{-kx}{g}[/tex]
Replacing with tour values we have
[tex]m = \frac{1.568*10^5(-9.48*10^{-2})}{9.8}[/tex]
[tex]m = 76.8Kg[/tex]
Therefore the mass of the player can be of 76.8Kg, then the player is eligible to play because the mass is less than 85Kg
Final answer:
The spring's effective force constant is 156,800 N/m. The player is eligible to play on the under-85-kg team.
Explanation:
Using Hooke's law, the effective force constant of the spring can be calculated by dividing the maximum load by the displacement. In this case, the spring is depressed by 0.75 cm under a maximum load of 120 kg.
To find the effective force constant, we can use the formula:
F = k * x
Where F is the force, k is the force constant, and x is the displacement.
So, the effective force constant of the spring is:
k = F / x = 120 kg * 9.8 N/kg / 0.0075 m = 156,800 N/m
For part (b), we can use the same formula to determine if the player is eligible to play on the under-85-kg team. Given that the player depresses the spring by 0.48 cm, we can plug in the values to calculate the force:
F = k * x = 156,800 N/m * 0.0048 m = 753.6 N
Since the force exerted by the player is less than 85 kg * 9.8 N/kg, the player is eligible to play on the under-85-kg team.
Current flowing in a circuit depends on two variables. Identify these variables and their relationship to current.
Question is Incomplete, Complete question is given below.
Current flowing in a circuit depends on two variables. Identify these variables and their relationship to current.
A) Current is proportionate to the conductance of the circuit and precisely proportional to the voltage applied across the circuit.
B) Current is conversely proportional to the electrical tension of the circuit and corresponds to the resistance across the circuit.
C) Current is inversely proportional to the resistance of the circuit and directly proportional to the voltage applied across the circuit.
D) Current is commensurate to the resistance of the circuit and directly proportional to the electric pressure applied across the circuit.
Answer:
C) Current is inversely proportional to the resistance of the circuit and directly proportional to the voltage applied across the circuit.
Explanation:
Now Ohms Law states that, "So long as a physical state of a conductor remains the potential difference applied to the conductor is directly proportional to current flowing through it."
I ∝ V
V=IR also I=V/R
where R is the Resistance
Hence, From above equation we can say that Current increases when there is increase in Voltage but Current decreases as the resistance decreases.
Hence,Current is inversely proportional to the resistance of the circuit and directly proportional to the voltage applied across the circuit.
A soccer ball has a mass of 0.40 kg. Initially, it is moving to the left at 20m/s, but then it is kicked and given a velocity of 45 degrees upward and to the right, with a magnitude of 30m/s. Find the impulse of the net force and the average net force, assuming a collision time of .010 seconds.
Answer:
Impulse = 18.53 kg m / s
Net force = 1853N
Explanation:
Mass of the ball, m = 0.4 kg
Initial velocity = 20 m/s horizontally left
Final velocity = 30 m/s at 45 degrees from horizontal
Time of collision = 0.010 s
Check attachment to see full explanation
The impulse and the average net force is 16.484 Ns and 1648.4 N respectively.
To calculate the impulse of the ball, we use the formula below.
Formula:
I = m(v-u).............. Equation 1Where:
I = impulse of the net forcem = mass of the ballu = initial velocity of the ballv = final velocity of the ballFrom the question,
Given:
m = 0.40 kgu = -20 m/s (to the left)v = 30cos45° (to the right) = 21.21 m/s.Substitute these values into equation 1
I = 0.4[21.21-(20)]I = 0.4(21.21+20)I = 0.4(41.21)I = 16.484 Ns.To calculate the average force, we use the formula below.
F = I/t............. Equation 2Where:
F = Average forcet = time.From the question,
Given:
I = 16.484 Nst = 0.01 sSubstitute these values into equation 1
F = 16.484/0.01F = 1648.4 NHence, The impulse and the average net force is 16.484 Ns and 1648.4 N respectively.
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A parallel-plate capacitor with circular plates of radius R is being charged by a battery, which provides a constant current. At what radius inside and outside the capacitor gap is the magnitude of the induced magnetic field equal to half of its maximum value? (Use any variable or symbol stated above as necessary.)
To solve this problem it is necessary to apply the concepts related to the magnetic field.
According to the information, the magnetic field INSIDE the plates is,
[tex]B=\frac{1}{2} \mu \epsilon_0 r[/tex]
Where,
[tex]\mu =[/tex]Permeability constant
[tex]\epsilon_0 =[/tex]Electromotive force
r = Radius
From this deduction we can verify that the distance is proportional to the field
[tex]B \propto r[/tex]
Then the distance relationship would be given by
[tex]\frac{r}{R} = \frac{B}{B_{max}}[/tex]
[tex]r =\frac{B}{B_{max}} R[/tex]
[tex]r = \frac{0.5B_{max}}{B_{max}}R[/tex]
[tex]r = 0.5R[/tex]
On the outside, however, it is defined by
[tex]B = \frac{\mu_0 i_d}{2\pi r}[/tex]
Here the magnetic field is inversely proportional to the distance, that is
[tex]B \not\propto r[/tex]
Then,
[tex]\frac{r}{R} = \frac{B_{max}{B}}[/tex]
[tex]r = \frac{B_{max}{B}}R[/tex]
[tex]r = \frac{B_{max}{0.5B_{max}}}R[/tex]
[tex]r = 2R[/tex]
A wheel is spinning about its axle such that the wheel rotates by 2π radians (1 rev) every 0.43 s. While the wheel spins, the axle is tilted through an angle of π/2 radians (90°) in a time of 0.30 s. The magnitude of the angular momentum changes by 50.00 kg m2/s when the axle is tilted. What is the magnitude of the torque applied to the wheel in order to tilt the axle by 90°?
Answer:
τ = 166.67 N m
Explanation:
The torque is related to the angular momentum
τ = dL / dt
Bold indicates vectors, where tau is the torque and L is the angular momentum
In this case we put the variations, It's like looking for middle values
τ = ΔL / Δt
Let's calculate
τ = 50.00 / 0.30
τ = 166.67 N m
Applied in the direction of rotation at the end of the wheel
Suppose that instead of a long straight wire, a shortstraight wire was used. The distance from the wire to thepoint that the magnetic field is measured would then be comparableto the length of the short wire. How would you expect themeasured value of the magnetic field to change if the field weresampled at the same distance from each of the two wires?
Answer:
Thus, if field were sampled at same distance, the field due to short wire is greater than field due to long wire.
Explanation:
The magnetic field, B of long straight wire can be obtained by applying ampere's law
[tex]B= \frac{\mu_0 I}{2\pi r}[/tex]
I is here current, and r's the distance from the wire to the field of measurement.
The magnetic field is obviously directly proportional to the current wire. From this expression.
As the resistance of the long cable is proportional to the cable length, the short cable becomes less resilient than the long cable, so going through the short cable (where filled with the same material) is a bigger amount of currents. If the field is measured at the same time, the field is therefore larger than the long wire because of the short wire.
Speed deamplifier. Block 1 of mass m1 slides along an x axis on a frictionless floor at speed 4.00 m/s. Then it undergoes a onedimensional elastic collision with stationary block 2 of mass m2 = 2.00m1. Next, block 2 undergoes a one-dimensional elastic collision with stationary block 3 of mass m3 = 4.00m1. (a) After all the collisions, what is the speed of block 3?
Answer:
29J
Explanation:
Please see attachment .
Imagine a car tire that contains 5.1 moles of air when at a gauge pressure of 2.1×10^5N/m2 (the pressure above atmospheric pressure) and a temperature of 27 degrees C. The temperature increases to 37 degrees C, the volume decreases to 0.8 times the original volume, and the gauge pressure decreases to 1.6×10^5N/m2.
How many moles of air are left in the tire?
To solve this problem it is necessary to use the concepts given through the ideal gas equation.
For this it is defined that
[tex]PV = nRT[/tex]
Where,
P = Pressure
V = Volume
R = Gas ideal constant
T = Temperature
n = number of moles.
In this problem we have two states in which the previous equation can be applied, so
[tex]1) P_1V_1 = n_1RT_1[/tex]
[tex]2) P_2V_2 = n_2RT_2[/tex]
From the first state we can calculate the Volume
[tex]V_1 = \frac{n_1RT_1}{P_1}[/tex]
Replacing
[tex]V_1 = \frac{5.1*8.314*300.15}{3.1*10^5}[/tex]
[tex]V_1 = 0.041m^3[/tex]
From the state two we can calculate now the number of the moles, considering that there is a change of 0.8 from Volume 1, then
[tex]n_2 = \frac{P_2(0.8*V_2)}{RT_2}[/tex]
[tex]n_2 = \frac{2.6*10^5(0.8*0.041)}{8.314*310.15}[/tex]
[tex]n_2 = 3.3moles[/tex]
Therefore there are 3.3moles of air left in the tire.
Final answer:
To find the number of moles of air left in the tire, we can use the ideal gas law. By using the given values for pressure, volume, and temperature, we can calculate the initial number of moles of air in the tire. We can then use the new volume and pressure values to find the number of moles of air left in the tire.
Explanation:
To find the number of moles of air left in the tire, we can use the ideal gas law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we need to convert the temperatures from Celsius to Kelvin. The initial temperature is 27 degrees C, so it is 300 K, and the final temperature is 37 degrees C, so it is 310 K.
Next, we use the ideal gas law to find the initial number of moles of air in the tire:
5.1 moles = (2.1 x 10^5 N/m^2) x V / ((8.314 J/mol*K) x 300 K)
Solving for V gives us V = (5.1 moles x 8.314 J/mol*K x 300 K) / (2.1 x 10^5 N/m^2) = 0.614 m^3.
Finally, we can use the new volume and pressure values to find the number of moles of air left in the tire:
n = (1.6 x 10^5 N/m^2) x (0.8 x 0.614 m^3) / (8.314 J/mol*K x 310 K) = 2.23 moles.
A 25 kg child at a playground runs with an initial speed of 2.5 m/s along a path tanget to the rim of a merry-go-round with a radius of 2.0 m and jumps on. The merry-go-round, which is initially at rest, has a moment of inertia of 500 kg m2. Find the angular velocity of the child and merry-go-round. Hint: the child and the merry-go-round form a system.
Answer:
the angular velocity of the child and merry-go-round= 0.21 rad/sec
Explanation:
mass m= 25 kg
initial speed v_i= 2.5 m/s
radius of merry go round= 2.0 m
MOI = 500 Kg-m^2
The angular momentum of the child is given by
L=mvr
initial angular momentum
Li=25×2.5×2= 125 kg m^2 s^{-1}
The final angular momentum considering the child and the merry-go-round form a system
[tex]L_f= (I_D+ I_C)\omega[/tex]
MOI of child I_C= 25×2^2= 25×4= 100 Kg-m^2
now plugging values in the above equation
[tex]L_f= (500+ 100)\omega[/tex]
L_f= 600ω
Now , we know that the angular momentum is conserved we can write
Li =Lf
125= 600ω
⇒ω=125/600 = 0.21 rad/sec
the angular velocity of the child and merry-go-round= 0.21 rad/sec
The angular velocity of the child and merry-go-round after the child jumps on is approximately[tex]\( 0.2083 \) rad/s.[/tex]
To find the angular velocity of the child and the merry-go-round, we need to apply the principle of conservation of angular momentum. The angular momentum of the system before the child jumps onto the merry-go-round must be equal to the angular momentum of the system after the child has jumped on.
Before the child jumps onto the merry-go-round, only the child has angular momentum, since the merry-go-round is initially at rest. The angular momentum [tex]\( L_{\text{initial}} \)[/tex]of the child can be calculated using the formula:
[tex]\[ L_{\text{initial}} = mvr \][/tex]
We can calculate [tex]\( L_{\text{initial}} \):[/tex]
[tex]\[ L_{\text{initial}} = (25 \text{ kg}) \times (2.5 \text{ m/s}) \times (2.0 \text{ m}) \][/tex]
[tex]\[ L_{\text{initial}} = 125 \text{ kg m}^2/\text{s} \][/tex]
After the child jumps onto the merry-go-round, the total angular momentum \( L_{\text{final}} \) of the system is the sum of the angular momenta of the child and the merry-go-round. Since the child is now rotating with the merry-go-round, their angular momenta can be combined using the formula for the angular momentum of a rotating body:
[tex]\[ L = I\omega \][/tex]
The moment of inertia of the merry-go-round is given as [tex]\( I_{\text{merry-go-round}} = 500 \)[/tex] kg m², and the moment of inertia of the child (treated as a point mass at the rim of the merry-go-round) is [tex]\( I_{\text{child}} = mr^2 \).[/tex]
[tex]\[ I_{\text{child}} = (25 \text{ kg}) \times (2.0 \text{ m})^2 \][/tex]
[tex]\[ I_{\text{child}} = (25 \text{ kg}) \times (4.0 \text{ m}^2) \][/tex]
[tex]\[ I_{\text{child}} = 100 \text{ kg m}^2 \][/tex]
The total moment of inertia [tex]\( I_{\text{total}} \)[/tex] after the child jumps on is:
[tex]\[ I_{\text{total}} = I_{\text{merry-go-round}} + I_{\text{child}} \][/tex]
[tex]\[ I_{\text{total}} = 500 \text{ kg m}^2 + 100 \text{ kg m}^2 \][/tex]
[tex]\[ I_{\text{total}} = 600 \text{ kg m}^2 \][/tex]
Since angular momentum is conserved:
[tex]\[ L_{\text{initial}} = L_{\text{final}} \][/tex]
[tex]\[ I_{\text{total}}\omega = L_{\text{initial}} \][/tex]
Solving for [tex]\( \omega \):[/tex]
[tex]\[ \omega = \frac{L_{\text{initial}}}{I_{\text{total}}} \][/tex]
[tex]\[ \omega = \frac{125 \text{ kg m}^2/\text{s}}{600 \text{ kg m}^2} \][/tex]
[tex]\[ \omega = \frac{125}{600} \text{ rad/s} \][/tex]
[tex]\[ \omega = 0.2083 \text{ rad/s} \][/tex]
Therefore, the angular velocity of the child and merry-go-round after the child jumps on is approximately[tex]\( 0.2083 \) rad/s.[/tex]
The answer is: [tex]0.2083 \text{ rad/s}.[/tex]
A metal wire 19 m long is cooled from 41 to -3°C. How much of a change in length will the wire experience if the coefficient of thermal expansion for this metal is 21 x 10-6 (°C)-1? If the change in length is negative (i.e., the wire shrinks), insert a minus sign before the numerical answer.
Answer:
ΔL = - 0.017556 m
Explanation:
Given
L₀ = 19 m
T₀ = 41ºC
T₁ = -3ºC
α = 21*10⁻⁶ °C⁻¹
ΔL = ?
We can use the equation
ΔL = L₀*α*ΔT
where
ΔT = T₁ - T₀ = -3ºC - 41ºC
⇒ ΔT = -44ºC
then
ΔL = (19 m)*(21*10⁻⁶ °C⁻¹)*(-44ºC)
⇒ ΔL = - 0.017556 m
Final answer:
The metal wire will experience a change in length of -0.017 m when cooled from 41 to -3 °C.
Explanation:
To calculate the change in length of the metal wire, we can use the equation for linear thermal expansion, AL = aLAT, where AL is the change in length, a is the coefficient of thermal expansion, L is the initial length of the wire, and AT is the change in temperature.
In this case, the initial length of the wire is 19 m, the coefficient of thermal expansion is 21 x 10^-6 (°C)^-1, and the change in temperature is 41 - (-3) = 44°C.
Using the equation, AL = (21 x 10^-6)(19)(44) = 0.017 mL, where mL represents meters. Since the change in length is negative (the wire cools and shrinks), the result is -0.017 m.
A 150 g ball and a 240 g ball are connected by a 33-cm-long, massless, rigid rod. The balls rotate about their center of mass at 150 rpm . Part A What is the speed of the 150 g ball? Express your answer to two significant figures and include the appropriate units
Answer:
v= 3.18 m/s
Explanation:
Given that
m= 150 g = 0.15 kg
M= 240 g = 0.24 kg
Angular speed ,ω = 150 rpm
The speed in rad/s
[tex]\omega =\dfrac{2\pi N}{60}[/tex]
[tex]\omega =\dfrac{2\pi \times 150}{60}[/tex]
ω = 15.7 rad/s
The distance of center of mass from 150 g
[tex]r=\dfrac{150\times 0+240\times 33}{150+240}\ cm[/tex]
r= 20.30 cm
The speed of the mass 150 g
v= ω r
v= 20.30 x 15.7 cm/s
v= 318.71 cm/s
v= 3.18 m/s
The speed of the 150 g ball rotating about its center mass is 3.2 m/s.
The given parameters;
mass of the first ball, m₁ = 150 gmass of the second ball, m₂ = 240 glength of the second ball, L₂ = 33 cmangular speed of the balls, ω = 150 rpmThe angular speed of the balls in radian per second;
[tex]\omega = 150 \ \frac{rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1 \min}{60 \ s } \\\\\omega = 15.71 \ rad/s[/tex]
The length of the second mass is calculated as follows;
[tex]L_2 =\frac{240(33)}{150 + 240} \\\\L_2 = 20.31 \ cm = 0.2031 \ m[/tex]
The speed of the 150 g ball is calculated as follows;
[tex]v = 0.2031 \times 15.71\\\\\ v = 3.2 \ m/s[/tex]
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A uniform, thin rod of length h and mass M is held vertically with its lower end resting on a frictionless horizontal surface. The rod is then released to fall freely. (a) What is the speed of its center of mass just before it hits the horizontal surface? (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.) vCM =___________.
To solve this exercise it is necessary to apply the concepts related to the conservation of rotational, kinetic and potential energy, as well as the concepts of moments of inertia in this type of bodies.
By definition we know that,
[tex]KE_f +U_f = KE_i + U_i[/tex]
Where,
KE = Kinetic energy
U = Potential energy
Let us start by defining that the center of mass of the body is located at a distance h / 2 from the bar and that the moment of inertia of a bar is defined by
[tex]I = \frac{1}{12}Mh^2[/tex]
Where M means the mass and h the height, then,
[tex]KE_f +U_f = KE_i + U_i[/tex]
[tex]\frac{1}{2}Mv^2+\frac{1}{2}I\omega +mgh_f = \frac{1}{2}Mv_f^2+Mgh (\frac{h}{2})[/tex]
There is not potential energy at the beginning and also there is not Kinetic energy at the end then
[tex]\frac{1}{2}Mv^2+\frac{1}{2}I\omega +0 = 0+Mgh (\frac{h}{2})[/tex]
Replacing inertia values,
[tex]\frac{1}{2}Mv^2+\frac{1}{2}(\frac{1}{12}Mh^2)(\frac{v}{h/2})^2 = Mg \frac{h}{2}[/tex]
Re-arrange for v, we have
[tex]v = \sqrt{\frac{3gh}{4}}[/tex]
Note that the value for the angular velocity ([tex]\omega[/tex])we replace with the equivalent in tangential velocity, which is [tex]\frac{v}{R}[/tex], where v is the velocity and R the radius, at this case h/2
Therefore the center of mass hat a velocity equal to[tex]v = \sqrt{\frac{3gh}{4}}[/tex]
The speed of the center of mass of a uniform, thin rod just before it hits the horizontal surface can be calculated using the equation vCM = √(2gh).
Explanation:The speed of the center of mass of a uniform, thin rod just before it hits the horizontal surface can be calculated using the equation:
vCM = √(2gh)
Where vCM is the speed of the center of mass, g is the acceleration due to gravity, and h is the height of the rod.
In this case, since the rod is released from rest, its initial velocity is zero, therefore, h is equal to the length of the rod, which is given in the problem.
By substituting the values into the equation, we can calculate the speed of the center of mass.
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: A laser beam is incident on two slits with separation d-0.038 mm. A screen is placed L-1.6 m from the slits. The wavelength of the laser light is A 4750 Å. and θ2 are the angles to the first and second bright fringes above the center of the screen. Otheex ee 25% Part (a) Express sin(91) in terms of d and λ. sin(θ 1) Add V Correct! 25% Part (b) Express sin(92) in terms of d and λ. sin(02) (2 A )yd Correct 25% Part (c) Express the distance between the two bright fringes on the screen, y, in terms of θ1, θ2 and L. Grade S Deductic Potentia 9 cotan(0) cotan(02) in(a)N 4 5 6 cos(0) | (.ilM-17 8 cos(α) cos(φ) Submiss Attempts per detailed sin(φ) tan(6) sin(0) tan(01) Hint I give up! Submit Feedback: deduction per feedback Hints: 0%, deduction per hint. Hints remaining: 2 25% Part (d) Solve for the numerical value of y ¡n meters.
To solve this physics problem involving Young's double-slit experiment, we apply the conditions for constructive interference to determine the angles θ1 and θ2 for the first and second bright fringes. We then use the small angle approximation to determine the distance y between those fringes on the screen. Lastly, we substitute the given values into the formula to compute the numerical value of y.
Explanation:The topic at hand is Young's double-slit experiment, a foundational experiment in wave optics. Given the slit separation d, the laser light's wavelength λ, and the screen's distance L, we need to find the distance between two bright fringes on the screen. The formulas required here stem from the conditions for constructive interference in double-slit set-ups.
Part a) sin(θ1) = λ/d, which is the condition for the light to interfere constructively at the first bright fringe from the center on the screen.
Part b) sin(θ2) = 2λ/d, which gives us the location of the second bright fringe, as it occurs when the path differential is twice the wavelength.
Part c) The distance y between the two bright fringes on the screen can be found using the small angle approximation, which allows us to express the angles θ1 and θ2 in terms of the fringe order m. Thus, y ≈ L*(θ2 - θ1) = L*(2λ/d - λ/d) = L*λ/d.
Part d) Substituting the given values into the formula from Part c, we find y = 1.6m * (4750Å / 0.038mm), which gives us a numerical value once we make sure to convert the units properly. In this case, we need to convert Å to mm by multiplying 4750Å by 10^-4 mm/Å.
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The question is about Young's double-slit experiment in physics. Both θ1 and θ2 are expressed in terms of d (the separation between the slits) and λ (the wavelength of the laser light). The distance between the fringes on the screen is calculated using these variables, resulting in a numerical value in meters.
Explanation:The questions refer to the phenomenon of light diffraction and interference, particularly as it relates to Young's double-slit experiment. In this context, we'll use the formulas defined for this experiment:
For parts (a) and (b): we know that the condition for constructive interference, resulting in bright fringes, for a double-slit setup is given by:
d sin θ = mλ
where 'd' is the slit separation, 'θ' is the angle to the fringe from the central maximum, 'm' is the order of the fringe and 'λ' is the wavelength of the light. For the first fringe (m=1), sin(θ1) = λ/d and for the second fringe (m=2), sin(θ2) = 2λ/d.
For part (c): Considering small angles where tan θ is approximately equal to sin θ, the separation of two bright fringes 'y' on the screen is given by:
y = L (θ2 - θ1) = L λ/d.
For part (d): Now, we just need to substitute the values given into the above formula. Substituting the given values of L = 1.6 m, λ = 4750 Å (which is equivalent to 4750 x 10^-10 m), and d = 0.038 mm (or 0.038 x 10^-3 m), you get:
y = L λ/d ≈ 0.02 m.
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The bliss you feel in the early stages of a relationship when you are infatuated is due to significantly elevated levels of dopamine and norepinephrine in your brain a favorable line-up of the moon and the stars the novelty of the situation a. & c.
The blissful feelings experienced in the early stages of a relationship, characterized by infatuation, are primarily attributed to options a. and c.
Significantly elevated levels of dopamine and norepinephrine in the brain contribute to the intense emotional highs associated with infatuation.
Dopamine is linked to pleasure and reward, while norepinephrine is associated with excitement and arousal, creating a euphoric state.
Additionally, the novelty of the situation plays a crucial role. New experiences, emotions, and interactions with a romantic partner stimulate the brain's reward system, enhancing feelings of happiness and exhilaration.
Therefore, the combination of heightened neurotransmitter levels and the novelty of the relationship fosters the intense and blissful emotions characteristic of infatuation during the early stages of a romantic relationship.
The probable question may be:
The blissful feelings experienced in the early stages of a relationship, characterized by infatuation, are primarily attributed to:
a. Significantly elevated levels of dopamine and norepinephrine in your brain.
b. A favorable line-up of the moon and the stars.
c. The novelty of the situation.
d. Options a. & c.
If a metal has a specific heat, then the metal could make an excellent material for cookware because
A. it can absorb much more heat before the temperature increases.
B. it maintains a constant temperature.
C. it takes less heat to raise the temperature of the metal.
D. it will not burn food.
Answer:
C) Because it takes less heat to raise the temperature of the metal
Explanation:
A) No. Metals have low specific heats compared to other solids thus they would absorb less heat in order to increase the temperature
B) No. Since they have lower specific heats they are more vulnerable to temperature changes due to variations in the heat rate
C) Yes. It takes less heat to raise the temperature of water due to their lower specific heats
D) No. Because the temperature can be chosen regulating the heat rate to the cookware. Also food burning is related with thermal conduction of cookware and from the cookware to food.
Answer:
C is correct
Explanation:
Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked.
This banking, with the aid of tire friction and very stable car configurations, allows the curves to be taken at very high speed.
Calculate the speed at which a 100 m radius curve banked at 65.0 should be driven if the road is frictionless.
Answer:
Speed, v = 45.84 m/s
Explanation:
Given that,
Radius of the curve, r = 100 m
Banking of the curve, [tex]\theta=65^{\circ}[/tex]
On the banking of curve, the speed of the vehicle is given by :
[tex]v=\sqrt{rg\ tan\theta}[/tex]
[tex]v=\sqrt{100\times 9.8\times \ tan(65)}[/tex]
v = 45.84 m/s
So, the speed at which a 100 m radius curve banked is 45.84 m/s. Hence, this is the required solution.
Final answer:
The ideal speed to navigate a 100 m radius curve banked at 65 degrees on a frictionless surface is calculated to be approximately 45.8 m/s or about 165 km/h using the principles of centripetal force.
Explanation:
When considering ideal speeds to take a steeply banked tight curve like the ones at the Daytona International Speedway, the scenario assumes a frictionless surface. The goal is to calculate the speed at which a vehicle should navigate a 100 m radius curve banked at 65 degrees. Utilizing the centripetal force equation, which in this case is provided by the component of the gravitational force, we can set this up as follows:
tan(θ) = v² / (rg)
Where θ is the angle of banking (65.0 degrees), v is the velocity, r is the radius of the curve (100 m), and g is the acceleration due to gravity (9.8 m/s²).
Rearranging the formula to solve for velocity, we get:
v = √(tan(θ) * r * g)
Plugging in the values:
v = √(tan(65.0 degrees) * 100 m * 9.8 m/s²)
v ≈ √(2.14 * 100 m * 9.8 m/s²) = √(2097.2 m²/s²)
v ≈ 45.8 m/s (which is about 165 km/h)
The amount of radiant power produced by the sun is approximately 3.9 ✕ 1026 W. Assuming the sun to be a perfect blackbody sphere with a radius of 6.96 ✕ 108 m, find its surface temperature (in kelvins).
Answer:
T=5797.8 K
Explanation:
Given that
Power P = 3.9 x 10²⁶ W
Radius ,r= 6.96 x 10⁸ m
We know that ,From Plank's law
P = σ A T⁴
σ = 5.67 x 10 ⁻⁸
A= Area ,T= Temperature ( in Kelvin)
[tex]T=\left (\dfrac{P}{\sigma A} \right )^{\dfrac{1}{4}}[/tex]
Now by putting the values
[tex]T=\left (\dfrac{P}{\sigma A}\right )^{\dfrac{1}{4}}[/tex]
[tex]T=\left (\dfrac{3.9\times 10^{26}}{5.67\times 10^{-8}\times 4\times \pi \times (6.96\times 10^8)^2} \right )^{\dfrac{1}{4}}[/tex]
T=5797.8 K
The temperature of surface will be 5797.8 K
The surface temperature of the sun can be calculated using the Stefan-Boltzmann law, which describes how the energy a perfect blackbody emits is related to its temperature. Knowing the radiant power and the surface area of the sun, we can use this law to find the surface temperature in kelvins.
Explanation:To calculate the surface temperature of the sun, we can make use of the Stefan-Boltzmann law ( radiant power, surface area of the object, and temperature in kelvins), a principle in Physics that describes how the total energy a perfect blackbody like the sun emits is related to its temperature. The law is represented mathematically as P = σ A T^4, where P is the radiant power, A is the surface area, T is the temperature, and σ is Stefan-Boltzmann constant (5.67 x 10^-8 J/(s.m².K^4)). Given that the sun's radiant power is approximately 3.9 x 10^26 W and its radius is 6.96 x 10^8 m (meaning its surface area is 4πr^2), we can rearrange the equation and solve for T to find the sun's surface temperature in kelvins: T = (P/ σA) ^ (1/4).
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russell westbrook has a mass of 84.8kg.When jumps to dunk on z 3.04 meter goal,what will be his G.P.E at the apex of his jump?
The gravitational potential energy is 2526 J
Explanation:
The gravitational potential energy (GPE) of an object is the energy possessed by the object due to its position in the gravitational field. Near the Earth's surface, it can be calculated as
[tex]GPE=mgh[/tex]
where
m is the mass of the body
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
h is the height of the object
In this problem, we have
m = 84.8 kg
h = 3.04 m
Substituting,
[tex]GPE=(84.8)(9.8)(3.04)=2526 J[/tex]
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A box is hanging from two strings. String #1 pulls up and left, making an angle of 450 with the horizontal on the left, and string #2 pulls up and to the right with a force of 180 N, making an angle of 50 with the horizontal on the right.
Flnd the mass of the box.
Hint: Apply ΣFx-0 and ΣFy 0
Answer:
The mass of the box is 25.6 kg.
Explanation:
Given that,
Angle = 45°
Force = 180 N
Angle = 50°
We need to calculate the force of left side
Using balance equation
[tex]\sum F_{x}=0[/tex]
[tex]-F_{1}\cos45+180\cos50=0[/tex]
[tex]F_{1}=\dfrac{180\cos50}{\cos45}[/tex]
[tex]F_{1}=163.6\ N[/tex]
We need to calculate the mass of the box
Using balance equation
[tex]\sum F_{y}=0[/tex]
[tex]F_{1}\sin45+180\sin50-mg=0[/tex]
[tex]m=\dfrac{1}{g}(F_{1}\sin45+180\sin50)[/tex]
Put the value into the formula
[tex]m=\dfrac{1}{9.8}\times(163.6\sin45+180\sin50)[/tex]
[tex]m=25.8\ kg[/tex]
Hence, The mass of the box is 25.6 kg.
Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere A has a radius 2 times that of sphere B. Let QA and QB be the charges on the two spheres, and let EA and EB be the electric-field magnitudes at the surfaces of the two spheres.What is the ratio QB/QA?What is the ratio EB/EA?
Answer:
Explanation:
Given
Radius of A is twice of B i.e.
[tex]R_A=2R_B[/tex]
Also Potential of both sphere is same
[tex]V_A=V_B[/tex]
[tex]V=\frac{kQ}{R}[/tex]
thus
[tex]k\frac{Q_A}{R_A}=K\frac{Q_B}{R_B}[/tex]
[tex]\frac{Q_A}{Q_B}=\frac{R_A}{R_B}[/tex]
[tex]\frac{Q_A}{Q_B}=\frac{2}{1}=2[/tex]
[tex]\frac{Q_B}{Q_A}=\frac{1}{2}[/tex]
(b)Ratio of [tex]\frac{E_B}{E_A}[/tex]
Electric Field is given by [tex]E=\frac{kQ}{R^2}[/tex]
thus [tex]E_A=\frac{kQ_A}{R_A^2}[/tex]----1
[tex]E_B=\frac{kQ_B}{R_B^2}[/tex]----2
Divide 2 by 1
[tex]\frac{E_B}{E_A}=\frac{Q_B}{R_B^2}\times \frac{R_A^2}{Q_A}[/tex]
[tex]\frac{E_B}{E_A}=\frac{1}{2}\times 4=2[/tex]
The ratio QB/QA for the charges on the spheres is 2, and the ratio EB/EA for the magnitudes of the electric field at the surfaces of the spheres is 1/2.
Explanation:The question revolves around the principles of electric fields and electric charges on two spherical objects. The spheres in question here are metal and have different sizes, but they hold the same electric potential at their surfaces. This situation is governed by the principles of physics, specifically electrostatics.
To solve this, we begin by noting that the electric potential (V) at the surface of a charged sphere is given by the equation V = KQ/r, where K is Coulomb's constant, Q is the charge on the sphere, and r is the radius of the sphere. Since the potential is the same for both spheres, we get that QA/2R = QB/R. Simplifying, we get the ratio QB/QA = 2.
Next, we calculate the electric field. The electric field (E) at the surface of a charged sphere is given by E = KQ/r^2. Therefore, EA = K×QA/(2R)^2 and EB = K×QB/R^2. Inserting the earlier found ratio of QB/QA = 2, the ratio EB/EA simplifies to 1/2.
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An astronaut is a short distance away from her space station without a tether rope. She has a large wrench. What should she do with the wrench to move toward the space station?
a. Throw it directly toward the space station.
b. Throw it directly away from the space station.
c. Throw it opposite to the direction of the station's orbit.
d. Throw it parallel to the direction of the station's orbit.
e. Throw it toward the station without letting go of it.
Answer: b. Throw it directly away from the space station.
Explanation:
According to Newton's third law of motion, when two bodies interact between them, appear equal forces and opposite senses in each of them.
To understand it better:
Each time a body or object exerts a force on a second body or object, it (the second body) will exert a force of equal magnitude but in the opposite direction on the first.
So, if the astronaut throws the wrench away from the space station (in the opposite direction of the space station), according to Newton's third law, she will be automatically moving towards the station and be safe.
Final answer:
To move toward the space station, the astronaut should throw the wrench directly away from it. By doing so, the astronaut imparts a momentum in the opposite direction, causing herself to move towards the space station.
Explanation:
The astronaut should throw the wrench directly away from the space station. When the astronaut throws the wrench in the opposite direction of the space station, the principle of conservation of momentum comes into play. According to this principle, the total momentum of a closed system remains constant. By throwing the wrench away from the space station, the astronaut imparts a momentum in the opposite direction, causing herself to move towards the space station. This is similar to how an ice skater moves faster when she throws her arms outward while spinning.
Heat in the amount of 100 kJ is transferred directly from a hot reservoir at 1320 K to a cold reservoir at 600 K. Calculate the entropy change of the two reservoirs and determine if the increase of entropy principle is satisfied. (Round the final answers to four decimal places.)
Answer:0.0909 kJ/K
Explanation:
Given
Temperature of hot Reservoir [tex]T_h=1320 K[/tex]
Temperature of cold Reservoir [tex]T_l=600 K[/tex]
Heat of 100 kJ is transferred form hot reservoir to cold reservoir
Hot Reservoir is Rejecting heat therefore [tex]Q_1=-100 kJ[/tex]
Heat is added to Reservoir therefore [tex]Q_2=100 kJ[/tex]
Entropy change for system
[tex]\Delta s=\frac{Q_1}{T_1}+\frac{Q_2}{T_2}[/tex]
[tex]\Delta s=\frac{-100}{1320}+\frac{100}{600}[/tex]
[tex]\Delta s=-0.0757+0.1666=0.0909 kJ/K[/tex]
As entropy change is Positive therefore entropy Principle is satisfied
A small mass m on a string is rotating without friction in a circle. The string is shortened by pulling it through the axis of rotation without any external torque. What happens to the angular velocity of the object?
a. It decreases.
b. It increases.
c. It remains the same.
A small mass m on a string is rotating without friction in a circle. When the string is shortened by pulling it through the axis of rotation without any external torque, angular velocity of the object increases.
What is angular momentum?The characteristic of any rotating object determined by the product of the moment of inertia and the angular velocity is known as angular momentum.
It is a characteristic of rotating bodies determined by the product of their moment of inertia and angular velocity. Since it is a vector quantity, the direction must also be taken into account in addition to the magnitude.
As no external torque applied; total angular momentum of the object remains conserved.
Angular momentum of an object of mass m moving with angular velocity ω in a circular path of radius r is given by: L = mω²r.
As the string is shortened by pulling it through the axis of rotation without any external torque, the angular momentum of the object remains same and its angular velocity increases.
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A small boat coasts at constant speed under a bridge. A heavy sack of sand is dropped from the bridge onto the boat. The speed of the boat
a. does not change.
b. increases.
c. Without knowing the mass of the boat and the sand, we can't tell.
d. decreases
Final answer:
The speed of the boat decreases when a heavy sack of sand is dropped onto it while it coasts at constant speed. This outcome is due to the conservation of momentum in the system, where the increase in mass leads to a decrease in speed to conserve the initial momentum.
Explanation:
When a heavy sack of sand is dropped onto a small boat that coasts at constant speed under a bridge, the initial impact will be an external force acting on the system of the sand and the boat. According to Newton's third law of motion, the sack of sand will exert a force on the boat when it hits, but we should consider the conservation of momentum to understand the speed of the boat after the sand lands in it. If we assume the system is closed and there are no external forces after the impact, the total momentum of the system must be conserved.
Before the sack is dropped, the momentum of the system is just the momentum of the boat since the sack is at rest with respect to the bridge. When the sack lands on the boat, their combined mass increases. Since momentum must be conserved (the product of mass and velocity before and after the event must be equal), the speed of the boat decreases to compensate for the greater mass of the combined system. Therefore, the correct answer is (d) decreases.
A transmission line that has a resistance per unit length of 2.40 10-4 Ω/m is to be used to transmit 5.00 MW across 400 mi (6.44 105 m). The output voltage of the source is 4.50 kV. (a) What is the line loss if a transformer is used to step up the voltage to 506 kV? (b) What fraction of the input power is lost to the line under these circumstances? (c) What difficulties would be encountered in attempting to transmit the 5.00 MW at the source voltage of 4.50 kV?
Answer:
(a) 30.174 kW
(b) [tex]6.035\times 10^{- 3}\ N[/tex]
(c) I = 1.112 kA
Solution:
As per the question:
Resistance per unit length, [tex]\sigma = 2.40\times 10^{- 4}\Omega/ m[/tex]
Power transmitted, P = 5.00 MW = [tex]5\times 10^{6}\ W[/tex]
Length, L = [tex]6.44\times 10^5\ m[/tex]
Output voltage, V = 4.50 kV
Stepped-up Voltage, V' = 506 kV
Now,
(a) The line loss can be calculated as:
Current, I = [tex]\frac{P}{V'} = \frac{5\times 10^{6}}{506\times 10^{3}} = 9.88\ A[/tex]
To calculate the power loss:
[tex]P_{L} = I^{2}R[/tex]
where
R = Resistance in the transmission line
R = [tex]2\sigma L = 2\times 2.40\times 10^{- 4}\times 6.44\times 10^5 = 309.12\Omega [/tex]
Thus
[tex]P_{L} = 9.88^{2}\times 309.12 = 30.174\ kW[/tex]
(b) Fraction of the input power lost is given by:
[tex]f = \frac{P_{L}}{P} = \frac{30.174\times 10^{3}}{5\times 10^{6}} = 6.035\times 10^{- 3}\ W[/tex]
(c) To calculate the current at the sending end at ther source voltage of 4.50 kV:
[tex]I = \frac{P}{V} = \frac{5\times 10^{6}}{4.50\times 10^{3}} = 1.112\ kA[/tex]
This value of current is very high and sending such a high watt power is impossible, even if the circuit at the other end is open then also this current can be destructive.
To calculate the line loss, we first need to calculate the value of current flowing through the transmission line. The line loss is approximately 2.08 * 10^5 A and the fraction of input power lost to the line is approximately 4.33 * 10^4 MW. Transmitting the power at the source voltage of 4.50 kV would result in high power losses.
Explanation:To calculate the line loss, we first need to calculate the value of current flowing through the transmission line. Using Ohm's Law, we have:V = IR
where V is the voltage, I is the current, and R is the resistance. Rearranging the equation, we get:I = V/R
Substituting the given values, we have:I = (506,000 V)/((2.40 * 10^-4) Ω/m * 6.44 * 10^5 m)
Simplifying, we find that the current is approximately 1.23 * 10^3 A. To find the line loss, we use the formula:P_loss = I^2 * R * L
where P_loss is the line loss, I is the current, R is the resistance per unit length, and L is the length of the transmission line. Substituting the given values, we find that the line loss is approximately 2.08 * 10^5 A.
The fraction of input power lost to the line can be found using the formula:
Fraction = P_loss / P_input
where P_input is the input power. Substituting the given values, we find that the fraction of input power lost to the line is approximately 4.33 * 10^4 MW.
Lastly, if we were to transmit the 5.00 MW at the source voltage of 4.50 kV, we would encounter difficulties due to the high power losses in the transmission lines. The power losses increase with higher currents, so the voltage is stepped up to reduce the current and subsequently decrease the power losses.
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In a certain region the average horizontal component of Earth's magnetic field is 22 μT, and the average inclination or "dip" is 75°. What is the corresponding magnitude of Earth's magnetic field in microteslas?
Answer:
B= 82.105 μT
Explanation:
Given that
Dip angle ,θ = 75°
Horizontal component of [tex]B_h[/tex] = 22 μT
The magnitude of earth magnetic filed given as
[tex]B= B_h tan\theta[/tex]
Noe by putting values in the above equation we get
[tex]B= 22\ \mu\ tan75^{\circ}[/tex]
B= 82.105 μT
Therefore magnetic filed in the microteslas will be 82.105 μT .
The magnitude of Earth's total magnetic field in the given region, calculated from the given horizontal component and inclination, is approximately 84.85 microteslas.
Explanation:The question relates to Earth's magnetic field and its components. The horizontal component of Earth's magnetic field and the inclination (or 'dip') are given, and you are asked to find the total magnetic field.
We use the formula for the total magnetic field (B), which can be calculated from the horizontal component (B_h) and the inclination (I) as follows:
B = B_h / cos(I)
In this case, B_h = 22 μT and I = 75°. Substituting the given values into the formula, we calculate the total magnetic field as:
B = 22 μT / cos(75°) = 84.85 μT
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The magnitude of a component of a vector must be:
a. equal to the magnitude of the vector.
b. greater than or equal to the magnitude of the vector.
c. less than, equal to, or greater than the magnitude of the vector.
d. less than or equal to the magnitude of the vector.
Answer:
(d) Less than or equal to the magnitude of vector
Explanation:
The magnitude of any vector is vector sum of magnitude of its component
As magnitude of vector is sum of its component so magnitude of its component never be greater than the magnitude of vector
It can be equal to the magnitude of vector in one case when the magnitude of other component of the vector is zero
So it can be less or equal to the magnitude of vector
The magnitude of a component of a vector must be less than or equal to the magnitude of the vector.
Explanation:The magnitude of a component of a vector must be less than or equal to the magnitude of the vector.
When a vector is resolved into its components, the magnitude of each component represents the projection of the vector onto that specific axis. The maximum magnitude of any component can be equal to the magnitude of the vector itself, but it cannot exceed it.
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A typical wind turbine extracts 40% of the kinetic energy of the wind that blows through the area swept by the blades. For a large turbine, 110,000 kg of air moves past the blades at 13 m/s every second. If the wind turbine extracts 40% of this kinetic energy, and if 80% of this energy is converted to electric energy, what is the power output of the generator?
Answer:
2.9744 MW
Explanation:
The air blowing through the wind turbine at 110000kg at speed of 13m/s for every second would have the kinetic energy of
[tex]E_k = 0.5mv^2 = 0.5*110000*13^2 = 9295000J/s[/tex]
If the wind turbine is able to extract only 40% of this kinetic energy, and 80% of this is converted to electricity, then the power output of the generator must be
[tex]E = 80\%*40\%*E_k = 0.8*0.4*9295000= 2974400J/s[/tex]
or 2.9744 MegaWatt
The total power output of the wind turbine is calculated by first finding the kinetic energy of the wind, and then determining the amount of that energy that is harvested and converted into electricity. The wind turbine extracts 40% of the kinetic energy, and 80% of this energy is converted into electricity, resulting in a power output of 2.974 gigawatts.
Explanation:To calculate the power output of the generator, we'll first need to compute the kinetic energy of the air moving through the turbine. The equation for kinetic energy is given by: KE = 0.5*mass*velocity^2.
So, the kinetic energy of the air per second is KE = 0.5*110,000 kg*(13 m/s)^2 = 9.295 gigajoules per second. Since the wind turbine extracts 40% of this energy, the energy extracted would be 3.718 gigajoules per second.
Of this extracted energy, 80% is converted to electric energy, which corresponds to 2.974 gigajoules per second. Because power is defined as the energy transferred or converted per unit of time, and here the energy is given per second, the power output is therefore 2.974 gigawatts.
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: A 10 g bullet travelling at 300 m/s hits a 500 g wooden block that is initially stationary. The bullet becomes embedded in the block, and both travel together along a frictioneless surface. Please answer each of the following questions a) What is the kinetic energy of the bullet before it hits the block? b) What is the velocity of the bullet+block after the collision?
Answer:
450 J
5.88235 m/s
Explanation:
The kinetic energy is given by
[tex]K=\frac{1}{2}m_1u_1^2\\\Rightarrow K=\frac{1}{2}\times 0.01\times 300^2\\\Rightarrow K=450\ J[/tex]
The kinetic energy of the bullet before it hits the block is 450 J
[tex]m_1[/tex] = Mass of bullet = 0.01 kg
[tex]m_2[/tex] = Mass of block = 0.5 kg
[tex]u_1[/tex] = Initial Velocity of bullet = 300 m/s
[tex]u_2[/tex] = Initial Velocity of second block = 0 m/s
v = Velocity of combined mass
In this system the linear momentum is conserved
[tex]m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{0.01\times 300 + 0.5\times 0}{0.01 + 0.5}\\\Rightarrow v=5.88235\ m/s[/tex]
The velocity of the bullet+block after the collision is 5.88235 m/s