a. I've attached a plot of the surface. Each face is parameterized by
• [tex]\mathbf s_1(x,y)=x\,\mathbf i+y\,\mathbf j[/tex] with [tex]0\le x\le2[/tex] and [tex]0\le y\le6-x[/tex];
• [tex]\mathbf s_2(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf k[/tex] with [tex]0\le u\le2[/tex] and [tex]0\le v\le\frac\pi2[/tex];
• [tex]\mathbf s_3(y,z)=y\,\mathbf j+z\,\mathbf k[/tex] with [tex]0\le y\le 6[/tex] and [tex]0\le z\le2[/tex];
• [tex]\mathbf s_4(u,v)=u\cos v\,\mathbf i+(6-u\cos v)\,\mathbf j+u\sin v\,\mathbf k[/tex] with [tex]0\le u\le2[/tex] and [tex]0\le v\le\frac\pi2[/tex]; and
• [tex]\mathbf s_5(u,y)=2\cos u\,\mathbf i+y\,\mathbf j+2\sin u\,\mathbf k[/tex] with [tex]0\le u\le\frac\pi2[/tex] and [tex]0\le y\le6-2\cos u[/tex].
b. Assuming you want outward flux, first compute the outward-facing normal vectors for each face.
[tex]\mathbf n_1=\dfrac{\partial\mathbf s_1}{\partial y}\times\dfrac{\partial\mathbf s_1}{\partial x}=-\mathbf k[/tex]
[tex]\mathbf n_2=\dfrac{\partial\mathbf s_2}{\partial u}\times\dfrac{\partial\mathbf s_2}{\partial v}=-u\,\mathbf j[/tex]
[tex]\mathbf n_3=\dfrac{\partial\mathbf s_3}{\partial z}\times\dfrac{\partial\mathbf s_3}{\partial y}=-\mathbf i[/tex]
[tex]\mathbf n_4=\dfrac{\partial\mathbf s_4}{\partial v}\times\dfrac{\partial\mathbf s_4}{\partial u}=u\,\mathbf i+u\,\mathbf j[/tex]
[tex]\mathbf n_5=\dfrac{\partial\mathbf s_5}{\partial y}\times\dfrac{\partial\mathbf s_5}{\partial u}=2\cos u\,\mathbf i+2\sin u\,\mathbf k[/tex]
Then integrate the dot product of f with each normal vector over the corresponding face.
[tex]\displaystyle\iint_{S_1}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{6-x}f(x,y,0)\cdot\mathbf n_1\,\mathrm dy\,\mathrm dx[/tex]
[tex]=\displaystyle\int_0^2\int_0^{6-x}0\,\mathrm dy\,\mathrm dx=0[/tex]
[tex]\displaystyle\iint_{S_2}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,0,u\sin v)\cdot\mathbf n_2\,\mathrm dv\,\mathrm du[/tex]
[tex]\displaystyle=\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=-8[/tex]
[tex]\displaystyle\iint_{S_3}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^6\mathbf f(0,y,z)\cdot\mathbf n_3\,\mathrm dy\,\mathrm dz[/tex]
[tex]=\displaystyle\int_0^2\int_0^60\,\mathrm dy\,\mathrm dz=0[/tex]
[tex]\displaystyle\iint_{S_4}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,6-u\cos v,u\sin v)\cdot\mathbf n_4\,\mathrm dv\,\mathrm du[/tex]
[tex]=\displaystyle\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=\frac{40}3+6\pi[/tex]
[tex]\displaystyle\iint_{S_5}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^{\frac\pi2}\int_0^{6-2\cos u}\mathbf f(2\cos u,y,2\sin u)\cdot\mathbf n_5\,\mathrm dy\,\mathrm du[/tex]
[tex]=\displaystyle\int_0^{\frac\pi2}\int_0^{6-2\cos u}12\,\mathrm dy\,\mathrm du=36\pi-24[/tex]
c. You can get the total flux by summing all the fluxes found in part b; you end up with 42π - 56/3.
Alternatively, since S is closed, we can find the total flux by applying the divergence theorem.
[tex]\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\iiint_R\mathrm{div}\mathbf f(x,y,z)\,\mathrm dV[/tex]
where R is the interior of S. We have
[tex]\mathrm{div}\mathbf f(x,y,z)=\dfrac{\partial(3x)}{\partial x}+\dfrac{\partial(x+y+2z)}{\partial y}+\dfrac{\partial(3z)}{\partial z}=7[/tex]
The integral is easily computed in cylindrical coordinates:
[tex]\begin{cases}x(r,t)=r\cos t\\y(r,t)=6-r\cos t\\z(r,t)=r\sin t\end{cases},0\le r\le 2,0\le t\le\dfrac\pi2[/tex]
[tex]\displaystyle\int_0^2\int_0^{\frac\pi2}\int_0^{6-r\cos t}7r\,\mathrm dy\,\mathrm dt\,\mathrm dr=42\pi-\frac{56}3[/tex]
as expected.
Q9 A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistance. Find (a) the height of the tabletop above the floor; (b) the horizontal distance from the edge of the table to the point where the book strikes the floor; (c) the horizontal and vertical components of the book's velocity, and the magnitude and direction of its velocity, just before the book reaches the floor.
Answer:
(a) 0.613 m
(b) 0.385 m
(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s
v = 3.68 m/s², θ = 72.6° below the horizontal
Explanation:
(a) Take down to be positive.
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 0.350 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²
Δy = 0.613 m
(b) Given in the x direction:
v₀ = 1.10 m/s
a = 0 m/s²
t = 0.350 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²
Δx = 0.385 m
(c) Find: vₓ and vᵧ
vₓ = aₓt + v₀ₓ
vₓ = (0 m/s²) (0.350 s) + 1.10 m/s
vₓ = 1.10 m/s
vᵧ = aᵧt + v₀ᵧ
vᵧ = (10 m/s²) (0.350 s) + 0 m/s
vᵧ = 3.50 m/s
The magnitude is:
v² = vₓ² + vᵧ²
v = 3.68 m/s²
The direction is:
θ = atan(vᵧ / vₓ)
θ = 72.6° below the horizontal
Design a voltage divider to provide the following approximate voltages with respect to ground using a 30 V source: 8.18 V, 14.7 V, and 24.6 V. The current drain on the source must be limited to no more than 1 mA. The number of resistors, their values, and their wattage ratings must be specified. A schematic showing the circuit arrangement and resistor placement must be provided
Answer:
R₁ = 14.7 10³ Ω , R₂ = 8.18 10³ Ω , R₃ = 1.72 10³ Ω , R₄ = 5.4 10³ Ω 1/8 W resistor
Explanation:
For this exercise we must use a series circuit since the sum of the voltage on each resin is equal to the source voltage (V = 30 V)
Therefore we build a circuit with 4 resistors in series, in such a way that
V = i R
let the voltage
1st resistance
V = i R
R₁ = V / i
R₁ = 14.7 / 1 10⁻³
R₁ = 14.7 10³ Ω
power is
P = V i
P = 14.7 1 10⁻³
P = 14.7 10⁻³ W = 0.0147 W
a resistance of ⅛ W is indicated
2nd resistance
R₂ = 8.18 / 1 10⁻³
R₂ = 8.18 10³ Ω
Power
P = 8.18 1 10⁻³
P = 0.00818W
a 1/8 W resistor
3rd resistance
this resistance is calculated in such a way that
V₁ + V₂ + V₃ = 24.6
V₃ = 24.6 - V₁ -V₂
V₃ = 24.6 - 14.7 - 8.18
V₃ = 1.72 V
R₃ = 1.72 / 1 10⁻³
R₃ = 1.72 10³ Ω
power
P = Vi
P = 1.72 10⁻³
P = 0.00172 W
a resistance of ⅛ W
To obtain the voltage of 24.6 we use this three resistors together
4th resistance
The value of this resistance is calculated so that the sum of all the voltages reaches the source voltage
30 = V₁ + V₂ + V₃ + V₄
V₄ = 30 - V₁ -V₂ -V₃
V₄ = 30 -14.7 - 8.18 - 1.72
V₄ = 5.4 V
R₄ = 5.4 / 1 10⁻³
R₄ = 5.4 10³ Ω
Power
P = V i
P = 5.4 10⁻³
P = 0.0054 W
⅛ W resistance
The values of these resistance are commercially
Let's check the consumption of the circuit
R_total = R₁ + R₂ + R₃ + R₄
R_total = (14.7 + 8.18 + 1.72 + 5.4) 10³
R_total = 30 10³
the current circulating in the circuit is
i = V / R_total
i = 30/30 10³
i = 1 10⁻³ A
therefore it is within the order requirement.
for connections see attached diagram
within which type of system is the total mass conserved but not the total energy
In a closed system the mass is conserved, but the energy is not conserved.
To find the answer, we have to study about different systems in thermodynamics.
What is thermodynamic system?A system, which can be expressed in terms of thermodynamic coordinates is called Thermodynamic system.Open system: System can exchange both energy and matter, thus, both energy and matter is not conserved here.Closed system can exchange energy with its surroundings (as heat or work), but not matter.Isolated system: A system that is open to the environment can interchange energy and matter, but a system that is insulated from it cannot.Thus, we can conclude that, in closed system the mass is conserved, but the energy is not conserved.
Learn more about Thermodynamic system here:
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A plastic balloon that has been rubbed with wool will stick to a wall.
a. Can you conclude that the wall is charged? If not, why not? If so, where does the charge come from?
b. Draw a series of charge diagrams showing how the balloon is held to the wall.
Answer:
Explanation:
When plastic balloon is rubbed with wool , charges are created on both balloon and silk in equal amount . Rubber balloon will acquire negative charge and silk will acquire positive charge .
Now when balloon is brought near a wall , there is induction of charge on the wall due to charge on the balloon . On the near surface of wall positive charge is produced and on the surface deep inside the wall negative charge is produced . The charge deep inside goes inside the earth but the positive charge near the surface of wall can not escape . It remains trapped by negative charge on the balloon .
hence there is mutual attraction between balloon and surface of wall is just like attraction between opposite charges . But once the ballon due to mutual attraction comes in contact with the wall , the charge on balloon and on wall neutralises each other and hence after some time the balloon falls off from the wall on the ground . It does not remain attracted to wall for ever . It happens due to neutralisation of charges on balloon and wall .
Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consider points along the line connecting the two sources.Required:a. At what distance from source A is there constructive interference between points A and B?b. At what distances from source A is there destructive interference between points A and B?
Answer:
a
[tex]z= 2.5 \ m[/tex]
b
[tex]z = (1 \ m , 4 \ m )[/tex]
Explanation:
From the question we are told that
Their distance apart is [tex]d = 5.00 \ m[/tex]
The wavelength of each source wave [tex]\lambda = 6.0 \ m[/tex]
Let the distance from source A where the construct interference occurred be z
Generally the path difference for constructive interference is
[tex]z - (d-z) = m \lambda[/tex]
Now given that we are considering just the straight line (i.e points along the line connecting the two sources ) then the order of the maxima m = 0
so
[tex]z - (5-z) = 0[/tex]
=> [tex]2 z - 5 = 0[/tex]
=> [tex]z= 2.5 \ m[/tex]
Generally the path difference for destructive interference is
[tex]|z-(d-z)| = (2m + 1)\frac{\lambda}{2}[/tex]
=> [tex]|2z - d |= (0 + 1)\frac{\lambda}{2}[/tex]
=> [tex]|2z - d| =\frac{\lambda}{2}[/tex]
substituting values
[tex]|2z - 5| =\frac{6}{2}[/tex]
=> [tex]z = \frac{5 \pm 3}{2}[/tex]
So
[tex]z = \frac{5 + 3}{2}[/tex]
[tex]z = 4\ m[/tex]
and
[tex]z = \frac{ 5 -3 }{2}[/tex]
=> [tex]z = 1 \ m[/tex]
=> [tex]z = (1 \ m , 4 \ m )[/tex]
For a beam of light in air (n = 1) reflecting off glass (n = 1.5), what is Brewster's angle to the nearest degree?
Answer: 56°
Explanation:
Brewster's angle refers to the angle at the point where light of a certain polarization passes through a transparent dielectric surface and is transmitted perfectly such that no reflection is made.
The formula is;
[tex]= Tan^{-1} (\frac{n_{2} }{n_{1}} )[/tex]
[tex]= Tan^{-1} (\frac{1.5 }{1} )[/tex]
= 56.30993247
= 56°
A 5.0-µC point charge is placed at the 0.00 cm mark of a meter stick and a -4.0-µC point charge is placed at the 50 cm mark. At what point on a line joining the two charges is the electric field due to these charges equal to zero?
Answer:
Electric field is zero at point 4.73 m
Explanation:
Given:
Charge place = 50 cm = 0.50 m
change q1 = 5 µC
change q2 = 4 µC
Computation:
electric field zero calculated by:
[tex]E1 =k\frac{q1}{r^2} \\\\E2 =k\frac{q2}{R^2} \\\\[/tex]
Where electric field is zero,
First distance = x
Second distance = (x-0.50)
So,
E1 = E2
[tex]k\frac{q1}{r^2}=k\frac{q2}{R^2} \\\\[/tex]
[tex]\frac{5}{x^2}=\frac{4}{(x-50)^2} \\\\[/tex]
x = 0.263 or x = 4.73
So,
Electric field is zero at point 4.73 m
Heat and thermodynamics Numerical
Answer:
K = 227.04 W/m.°C
Explanation:
First we need to find the heat required to melt the ice:
q = m H
where,
q = heat required = ?
m = mass of the ice = 8.5 g = 8.5 x 10⁻³ kg
H = Latent heat of fusion of ice = 3.34 x 10⁵ J/kg
Therefore,
q = (8.5 x 10⁻³ kg)(3.34 x 10⁵ J/kg)
q = 2839 J
Now, we find the heat transfer rate through rod:
Q = q/t
where,
t = time = (10 min)(60 s/1 min) = 600 s
Q = Heat Transfer Rate = ?
Therefore,
Q = 2839 J/600 s
Q = 4.73 W
From Fourier's Law of Heat Conduction:
Q = KA ΔT/L
where,
K = Thermal Conductivity = ?
A = cross sectional area = 1.25 cm² = 1.25 x 10⁻⁴ m²
L = Length of rod = 60 cm = 0.6 m
ΔT = Difference in temperature = 100°C - 0°C = 100°C
Therefore,
4.73 W = K(1.25 X 10⁻⁴ m²)(100°C)/0.6 m
K = (4.73 W)/(0.0208 m.°C)
K = 227.04 W/m.°C
Light of wavelength 519 nm passes through two slits. In the interference pattern on a screen 4.6 m away, adjacent bright fringes are separated by 5.2 mm in the general vicinity of the center of the pattern. What is the separation of the two slits?
Answer:
The separation of the two slits is 0.456 mm.
Explanation:
Given the wavelength of light = 519 nm
The indifference pattern = 4.6 m
Adjacent bright fringes = 5.2 mm
In the interference, the equation required is Y = mLR/d
Here, d sin theta = mL
L = wavelgnth
For bright bands, m is the order = 1,2,3,4
For dark bands, m = 1.5, 2.5, 3.5, 4.5
R = Distance from slit to screen (The indifference pattern)
Y = Distance from central spot to the nth order fringe or fringe width
Thus, here d = mLR/Y
d = 1× 519nm × 4.6 / 5.2mm
d = 0.459 mm
15.Restore the battery setting to 10 V. Now change the number of loops from 4 to 3. Explain what happens to the magnitude and direction of the magnetic field. Now change to 2 loops, then to 1 loop. What do you observe the relationship to be between the magnitude of the magnetic field and the number of loops for the same current
Answer:
we see it is a linear relationship.
Explanation:
The magnetic flux is u solenoid is
B = μ₀ N/L I
where N is the number of loops, L the length and I the current
By applying this expression to our case we have that the current is the same in all cases and we can assume the constant length. Consequently we see that the magnitude of the magnetic field decreases with the number of loops
B = (μ₀ I / L) N
the amount between paracentesis constant, in the case of 4 loop the field is worth
B = cte 4
N B
4 4 cte
3 3 cte
2 2 cte
1 1 cte
as we see it is a linear relationship.
In addition, this effect for such a small number of turns the direction of the field that is parallel to the normal of the lines will oscillate,
A rigid container holds 4.00 mol of a monatomic ideal gas that has temperature 300 K. The initial pressure of the gas is 6.00 * 104 Pa. What is the pressure after 6000 J of heat energy is added to the gas?
Answer:
The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.
Explanation:
When a container is rigid, the process is supposed to be isochoric, that is, at constant volume. Then, the equation of state for ideal gases can be simplified into the following expression:
[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}[/tex]
Where:
[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Initial and final pressures, measured in pascals.
[tex]T_{1}[/tex], [tex]T_{2}[/tex] - Initial and final temperatures, measured in Kelvins.
In addtion, the specific heat at constant volume for monoatomic ideal gases, measured in joules per mole-Kelvin is given by:
[tex]\bar c_{v} = \frac{3}{2}\cdot R_{u}[/tex]
Where:
[tex]R_{u}[/tex] - Ideal gas constant, measured by pascal-cubic meters per mole-Kelvin.
If [tex]R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}[/tex], then:
[tex]\bar c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{2}}{mol\cdot K} \right)[/tex]
[tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex]
And change in heat energy ([tex]Q[/tex]), measured by joules, by:
[tex]Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})[/tex]
Where:
[tex]n[/tex] - Molar quantity, measured in moles.
The final temperature of the monoatomic ideal gas is now cleared:
[tex]T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}[/tex]
Given that [tex]T_{1} = 300\,K[/tex], [tex]Q = 6000\,J[/tex], [tex]n = 4\,mol[/tex] and [tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex], the final temperature is:
[tex]T_{2} = 300\,K + \frac{6000\,J}{(4\,mol)\cdot \left(12.471\,\frac{J}{mol\cdot K} \right)}[/tex]
[tex]T_{2} = 420.279\,K[/tex]
The final pressure of the system is calculated by the following relationship:
[tex]P_{2} = \left(\frac{T_{2}}{T_{1}}\right) \cdot P_{1}[/tex]
If [tex]T_{1} = 300\,K[/tex], [tex]T_{2} = 420.279\,K[/tex] and [tex]P_{1} = 6.00\times 10^{4}\,Pa[/tex], the final pressure is:
[tex]P_{2} = \left(\frac{420.279\,K}{300\,K} \right)\cdot (6.00\times 10^{4}\,Pa)[/tex]
[tex]P_{2} = 8.406\times 10^{4}\,Pa[/tex]
The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.
A beam of light from a laser illuminates a glass how long will a short pulse of light beam take to travel the length of the glass.
Answer:
The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]
Explanation:
Given that,
A beam of light from a laser illuminates a glass.
Suppose, the length of piece is [tex]L=25.21\times10^{-2}\ m[/tex]
Index of refraction is 2.83.
We need to calculate the speed of light pulse in glass
Using formula of speed
[tex]v=\dfrac{c}{\mu}[/tex]
Put the value into the formula
[tex]v=\dfrac{3\times10^{8}}{2.83}[/tex]
[tex]v=1.06\times10^{8}\ m/s[/tex]
We need to calculate the time of short pulse of light beam
Using formula of velocity
[tex]v=\dfrac{d}{t}[/tex]
[tex]t=\dfrac{d}{v}[/tex]
Put the value into the formula
[tex]t=\dfrac{25.21\times10^{-2}}{1.06\times10^{8}}[/tex]
[tex]t=2.37\times10^{-9}\ sec[/tex]
Hence, The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]
Just wondering if I did this right
Yeah
All they are all correct
As you finish listening to your favorite compact disc (CD), the CD in the player slows down to a stop. Assume that the CD spins down with a constant angular acceleration. If the CD rotates clockwise (let's take clockwise rotation as positive) at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angular speed in 2.60 s with constant angular acceleration, the angular acceleration of the CD, as it spins to a stop at -20.1 rad/s 2. How many revolutions does the CD make as it spins to a stop?
Answer:
10.8rev
Explanation:
Using
Wf²-wf = 2 alpha x theta
0²- 56.36x56.36/ 2(-20.13) x theta
Theta = 68.09 rad
But 68.09/2π
>= 10.8 revolutions
Explanation:
Air bags greatly reduces the chance og injury in a car accident.explain how they do si in terms of energy transfer
Answer:
in an accident, when the body collides with the air bags, the collision time of impact between the two bodies will increase due to the presence of air bags in the car. Larger is the impact time smaller is the transformation of energy between the body and air bag. That is why air bags greatly reduce the chance of injury in a car accident.
A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3 degrees angle. What is the y-component of the velocity of the second ball?
Answer:
v_{1fy} = - 0.4549 m / s
Explanation:
This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved
initial. Before the crash
p₀ = m v₁₀
final. After the crash
[tex]p_{f}[/tex] = m [tex]v_{1f}[/tex] + m v_{2f}
Recall that velocities are a vector so it has x and y components
p₀ = p_{f}
we write this equation for each axis
X axis
m v₁₀ = m v_{1fx} + m v_{2fx}
Y Axis
0 = -m v_{1fy} + m v_{2fy}
the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components
sin 23.3 = v_{2fy} / v_{2f}
cos 23.3 = v_{2fx} / v_{2f}
v_{2fy} = v_{2f} sin 23.3
v_{2fx} = v_{2f} cos 23.3
we substitute in the momentum conservation equation
m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3
0 = - m v_{1f} sin θ + m v_{2f} sin 23.3
1.83 = v_{1f} cos θ + 1.15 cos 23.3
0 = - v_{1f} sin θ + 1.15 sin 23.3
1.83 = v_{1f} cos θ + 1.0562
0 = - v_{1f} sin θ + 0.4549
v_{1f} sin θ = 0.4549
v_{1f} cos θ = -0.7738
we divide these two equations
tan θ = - 0.5878
θ = tan-1 (-0.5878)
θ = -30.45º
we substitute in one of the two and find the final velocity of the incident ball
v_{1f} cos (-30.45) = - 0.7738
v_{1f} = -0.7738 / cos 30.45
v_{1f} = -0.8976 m / s
the component and this speed is
v_{1fy} = v1f sin θ
v_{1fy} = 0.8976 sin (30.45)
v_{1fy} = - 0.4549 m / s
A velocity selector in a mass spectrometer uses a 0.100-T magnetic field. (a) What electric field strength is needed to select a speed of 4.00 . 106 m/s
Answer:
The electric field strength needed is 4 x 10⁵ N/C
Explanation:
Given;
magnitude of magnetic field, B = 0.1 T
velocity of the charge, v = 4 x 10⁶ m/s
The velocity of the charge when there is a balance in the magnetic and electric force is given by;
[tex]v = \frac{E}{B}[/tex]
where;
v is the velocity of the charge
E is the electric field strength
B is the magnetic field strength
The electric field strength needed is calculated as;
E = vB
E = 4 x 10⁶ x 0.1
E = 4 x 10⁵ N/C
Therefore, the electric field strength needed is 4 x 10⁵ N/C
A city of Punjab has a 15 percent chance of wet weather on any given day. What is the probability that it will take a week for it three wet weather on 3 separate days?
Answer: 0.0617
Explanation:
Given: The probability of wet weather on any given day in a city of Punjab : p=15%=0.15
Let X be a binomial variable that represents the number of days having wet weather.
Binomial probability formula : [tex]P(X=x)=^nC_xp^x(1-p)^x[/tex], where n= total outcomes, p = probability of success in each outcomes.
Here, n= 7 ( 1 week = 7 days)
The probability that it will take a week for it three wet weather on 3 separate days:
[tex]P(X=3)^=\ ^7C_3(0.15)^3(1-0.15)^{7-3}\\\\=\dfrac{7!}{3!(7-3)!}(0.15)^3(0.85)^4\\\\=\dfrac{7\times6\times5}{3\times2}\times 0.003375\times0.52200625\approx0.0617[/tex]
Hence, the required probability =0.0617
Which unbalanced force accounts for the direction of the net force of the rocket?
a. Air resistance
b. Friction
c. Gravity
d. Thrust of rocket engine
It depends on what stage of the mission you're talking about.
==> While it's sitting on the pad before launch, the forces on the rocket are balanced, so there's no net force on it.
==> When the engines ignite, their thrust (d) is greater than the force of gravity. So the net force on the rocket is upward, and the spacecraft accelerates upward.
==> After the engines shut down, the net force acting on the rocket is due to Gravity (c).
. . . If the rocket has enough vertical speed, it escapes the Earth completely, and just keeps going.
. . . If it has enough horizontal speed, it enters Earth orbit.
. . . If it doesn't have enough vertical or horizontal speed, it falls back to Earth.
A rocket will preserve to speed up so long as there's a resultant pressure upwards resulting from the thrust of the rocket engine.
What unbalanced force bills for the course of the internet pressure of the rocket?A rocket launches whilst the pressure of thrust pushing it upwards is greater than the burden force because of gravity downwards. This unbalanced pressure reasons a rocket to accelerate upwards. A rocket will maintain to hurry up so long as there's a resultant force upwards resulting from the thrust of the rocket engine.
What's the net pressure of unbalanced?
If the forces on an item are balanced, the net pressure is zero. If the forces are unbalanced forces, the results do not cancel each difference. Any time the forces acting on an object are unbalanced, the net pressure is not 0, and the movement of the item modifications.
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When the magnet falls toward the copper block, the changing flux in the copper creates eddy currents that oppose the change in flux. The resulting braking force between the magnet and the copper block always opposes the motion of the magnet, slowing it as it falls. The braking force on the magnet is nearly equal to its weight, so it falls very slowly. The rate of the fall produces a rate of flux change sufficient to produce a current that provides the braking force. If the magnet is pushed, forcefully, toward the block, the rate of change of flux is much higher than this. When the magnet is moving much more quickly than it will fall unaided, what is the direction of the net force on the magnet?
Answer:
The net force is directed downwards.
Explanation:
Since the magnet is falling much more faster than it would unaided, then there is a net force that is accelerating the magnet downwards. We know that acceleration is due to a force acting on a mass, and in this case, the magnet is the mass. Also, the acceleration is always in the direction of the force producing it, which means that the net force on the magnet is vertically downwards.
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be observed when the detector is 3 meters from the sample?
Using the sample in above question how many counts per second would be observed when the detector is 10 meters away from the sample?
Answer:
At 3 meter distance, the per-second count is 222.22 and at a 10 meter distance, the per-second count is 20.
Explanation:
The number of particles (N) counts are inversely proportional to the distance between the source and the detector.
By using the below formula we can find the number of counts.
[tex]N2 = \frac{(D1)^2}{(D2)^2} \times N1 \\N1 = 2000 \\D 1 = 1 \ meter \\D2 = 3 \\[/tex]
The number of count per second, when the distance is 3 meters.
[tex]= \frac{1}{3^2} \times 2000 \\= 222.22[/tex]
Number of count per second when the distance is 10 meters.
[tex]= \frac{1}{10^2} \times 2000 \\= 20[/tex]
If this is the only water being used in your house, how fast is the water moving through your house's water supply line, which has a diameter of 0.021 m (about 3/4 of an inch)?
Answer:
0.273m/s
Explanation:
first find out the meaning of 0.90×10−4m3/s
literally, that is 0.9x6 = 5.4m3/s = 3•5.4m/s or 16.2 m/s
1.5 gal/min = 0.00009464 m³/s, perhaps that is what you mean?
cross-sectional area of pipe is πr² = 0.0105²π = 0.0003464 m²
so you have a a flow of 0.00009464 m³/s flowing through an area of 0.0003464 m²
they divide to 0.00009464 m³/s / 0.0003464 m² = 0.273 m/s
An electromagnetic flowmeter is useful when it is desirable not to interrupt the system in which the fluid is flowing (e.g. for the blood in an artery during heart surgery). Such a device is illustrated. The conducting fluid moves with velocity v in a tube of diameter d perpendicular to which is a magnetic field B. A voltage V is induced between opposite sides of the tube. Given B = 0.120 T, d = 1.2 cm., and a measured voltage of 2.88 mV, determine the speed of the blood.
Answer:
2 m/s
Explanation:
The electromagnetic flow-metre work on the principle of electromagnetic induction. The induced voltage is given as
[tex]E = Blv[/tex]
where [tex]E[/tex] is the induced voltage = 2.88 mV = 2.88 x 10^-3 V
[tex]l[/tex] is the distance between the electrodes in this field which is equivalent to the diameter of the tube = 1.2 cm = 1.2 x 10^-2 m
[tex]v[/tex] is the velocity of the fluid through the field = ?
[tex]B[/tex] is the magnetic field = 0.120 T
substituting, we have
2.88 x 10^-3 = 0.120 x 1.2 x 10^-2 x [tex]v[/tex]
2.88 x 10^-3 = 1.44 x 10^-3 x [tex]v[/tex]
[tex]v[/tex] = 2.88/1.44 = 2 m/s
How many turns of wire are needed in a circular coil 13 cmcm in diameter to produce an induced emf of 5.6 VV
Answer:
Number of turns of wire(N) = 3,036 turns (Approx)
Explanation:
Given:
Diameter = 13 Cm
emf = 5.6 v
Note:
The given question is incomplete, unknown information is as follow.
Magnetic field increases = 0.25 T in 1.8 (Second)
Find:
Number of turns of wire(N)
Computation:
radius (r) = 13 / 2 = 6.5 cm = 0.065 m
Area = πr²
Area = (22/7)(0.065)(0.065)
Area = 0.013278 m²
So,
emf = (N)(A)(dB / dt)
5.6 = (N)(0.013278)(0.25 / 1.8)
5.6 = (N)(0.013278)(0.1389)
N = 3,036.35899
Number of turns of wire(N) = 3,036 turns (Approx)
In a LRC circuit, a second capacitor is connected in parallel with the capacitor previously in the circuit. What is the effect of this change on the impedance of the circuit
Answer:
Impedance increases for frequencies below resonance and decreases for the frequencies above resonance
Explanation:
See attached file
Explanation:
You are walking around your neighborhood and you see a child on top of a roof of a building kick a soccer ball. The soccer ball is kicked at 31° from the edge of the building with an initial velocity of 15 m/s and lands 63 meters away from the wall. How tall, in meters, is the building that the child is standing on?
Answer:
69.58 m tall
Explanation:
Pls see attached file
Simple harmonic oscillations can be modeled by the projection of circular motion at constant angular velocity onto the diameter of a circle. When this is done, the analog along the diameter of the acceleration of the particle executing simple harmonic motion is
Answer:
the analog along the diameter of the acceleration of the particle executing simple harmonic motion is the projection along the diameter of the centripetal acceleration of the particle in the circle
An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.165×105 km and the planet has a radius of 4175 km and a mass of 6.70×1022 kg , how long (in days) does it take the moon to make one revolution around the planet? The gravitational constant is 6.67×10−11N·m2/kg2 .
Answer:
364days
Explanation:
Pls see attached file
Explanation:
The moon will take 112.7 days to make one revolution around the planet.
What is Kepler's third law?The period of the satellite around any planet only depends upon the distance between the planet's center and satellite and also depends upon the planet's mass.
Given, the distance from the moon's center to the planet's surface,
h = 2.165 × 10⁵ km,
The radius of the planet, r = 4175 km
The mass of the planet = 6.70 × 10²² kg
The total distance between the moon's center to the planet's center:
a = r +h = 2.165 × 10⁵ + 4175
a = 216500 + 4175
a = 220675
a = 2.26750 × 10⁸ m
The period of the planet can be calculated as:
[tex]T =2\pi \sqrt{\frac{a^3}{Gm} }[/tex]
[tex]T =2\3\times 3.14 \sqrt{\frac{(2.20675 \times 10^8)^3}{(6.67\times 10^{-11}).(6.70\times 10^{22})} }[/tex]
T = 9738253.26 s
T = 112.7 days
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g A projectile is fired from the ground at an angle of θ = π 4 toward a tower located 600 m away. If the projectile has an initial speed of 120 m/s, find the height at which it strikes the tower
Answer:
The projectile strikes the tower at a height of 354.824 meters.
Explanation:
The projectile experiments a parabolic motion, which consist of a horizontal motion at constant speed and a vertical uniformly accelerated motion due to gravity. The equations of motion are, respectively:
Horizontal motion
[tex]x = x_{o}+v_{o}\cdot t \cdot \cos \theta[/tex]
Vertical motion
[tex]y = y_{o} + v_{o}\cdot t \cdot \sin \theta +\frac{1}{2} \cdot g \cdot t^{2}[/tex]
Where:
[tex]x_{o}[/tex], [tex]x[/tex] - Initial and current horizontal position, measured in meters.
[tex]y_{o}[/tex], [tex]y[/tex] - Initial and current vertical position, measured in meters.
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]t[/tex] - Time, measured in seconds.
The time spent for the projectile to strike the tower is obtained from first equation:
[tex]t = \frac{x-x_{o}}{v_{o}\cdot \cos \theta}[/tex]
If [tex]x = 600\,m[/tex], [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]\theta = \frac{\pi}{4}[/tex], then:
[tex]t = \frac{600\,m-0\,m}{\left(120\,\frac{m}{s} \right)\cdot \cos \frac{\pi}{4} }[/tex]
[tex]t \approx 7.071\,s[/tex]
Now, the height at which the projectile strikes the tower is: ([tex]y_{o} = 0\,m[/tex], [tex]t \approx 7.071\,s[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex])
[tex]y = 0\,m + \left(120\,\frac{m}{s} \right)\cdot (7.071\,s)\cdot \sin \frac{\pi}{4}+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (7.071\,s)^{2}[/tex]
[tex]y \approx 354.824\,m[/tex]
The projectile strikes the tower at a height of 354.824 meters.
A student uses a spring scale attached to a textbook to compare the static and kinetic friction between the textbook and the top of a lab
table. If the scale measures 1,580 g while the student is pulling the sliding book along the table, which reading on the scale could have been
possible at the moment the student overcame the static friction? (1 point)
1,860 g
820 g
1,580 g
1,140 g
Answer:
1,860 g
Explanation:
In a system, the coefficient of static friction is usually higher than the coefficient of kinetic friction. This means that the kinetic friction is usually less than the static friction. From the question, since the book is already sliding, it means that kinetic friction is the friction in play. This means that before the reading on the scale that could have been possible at the moment the student overcame the static friction must be greater than the reading on the scale during sliding. The only option above 1580 g is 1860 g
