Answer:
A. [tex]H=\frac{v_{0}^{2}Sin^{2}\theta }{2g}[/tex]
B. 5556.1 m
Explanation:
A.
Launch speed, vo
Angle of projection = θ
The value of vertical component of velocity at maximum height is zero. Let the maximum height is H.
Use third equation of motion in vertical direction
[tex]v_{y}^{2}=u_{y}^{2}+2a_{y}H[/tex]
[tex]0^{2}=\left (v_{0}Sin\theta \right )^{2}-2gH[/tex]
[tex]H=\frac{v_{0}^{2}Sin^{2}\theta }{2g}[/tex]
B.
u = 330 m/s
Let it goes upto height H.
V = 0 at maximum height
Use third equation of motion in vertical direction
[tex]v^{2}=u^{2}+2as[/tex]
[tex]0^{2}=330^{2}-2\times 9.8\times H[/tex]
H = 5556.1 m
An airplane is flying at altitude 2 000[m] through a thundercloud. If +40[C] is concentrated at altitude 3 000[m] , directly above the airplane, and −40[C] is concentrated at altitude 1 000[m] directly below it, what Electric field intensity (magnitude) is at the aircraft?
Answer:
The electric field intensity at the aircraft is [tex]7.18\times10^{5}\ N/C[/tex]
Explanation:
Given that,
Altitude = 2000 m
Charge = +40 C
At the position of the airplane, the electric field will be due to both the positive and negative charges and will be downwards due to both the charges.
The position is equidistant from both the charges.
We need to calculate the resultant electric field
The contribution due to the positive charge at 3000 m altitude
Using formula of electric field
[tex]E_{+}=k\dfrac{|q|}{r^2}[/tex]
Put the value into the formula
[tex]E_{+}=8.99\times10^{9}\times\dfrac{40}{(3000-2000)^2}[/tex]
[tex]E_{+}=3.59\times10^{5}\ N/C[/tex]
The contribution due to the negative charge at 1000 m altitude
Using formula of electric field
[tex]E_{-}=8.99\times10^{9}\times\dfrac{40}{(2000-1000)^2}[/tex]
[tex]E_{-}=3.59\times10^{5}\ N/C[/tex]
We need to calculate the electric field intensity at the aircraft
The resultant electric field is
[tex]E=E_{+}+E_{-}[/tex]
Put the value into the formula
[tex]E=3.59\times10^{5}+3.59\times10^{5}[/tex]
[tex]E=7.18\times10^{5}\ N/C[/tex]
Hence, The electric field intensity at the aircraft is [tex]7.18\times10^{5}\ N/C[/tex]
How do resistors in parallel affect the total resistance?
Answer:
They're going to increase the total resistance as [tex]R_{T} = \sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1}[/tex]
Explanation:
If the resistors are in parallel, the potential difference is the same for each resistor. But the total current is the sum of the currents that pass through each of the resistors. Then
[tex]I = I_1 + I_2 + ... + I_N[/tex]
where
[tex]I_i = \frac{V_i}{R_i}[/tex]
but
[tex]V_i = V_j = V[/tex] for [tex]i,j= 1, 2,..., N[/tex]
so
[tex]I = \frac{V}{R_1}+ \frac{V}{R_2} + ... + \frac{V}{R_N} = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)V = \frac{V}{R_T}[/tex]
where
[tex]R_T = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)^{-1} =\sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1} [/tex]
The current in a 100 watt lightbulb is 0.890 A. The filament inside the bulb is 0.280 mm in diameter. What is the current density in the filament? Express your answer to three significant figures and include the appropriate units. What is the electron current in the filament? Express your answer using three significant figures.
Answer:
Current density [tex]j=1.44\times 10^7A/m^2[/tex]
Electron density [tex]=5.55\times 10^{18}electron/sec[/tex]
Explanation:
We have given power = 100 watt
Current = 0.89 A
Diameter d = 0.280 mm
So radius [tex]r=\frac{d}{2}=\frac{0.28}{2}=0.14mm=0.14\times 10^{-3}m[/tex]
Area [tex]A=\pi r^2=3.14\times (0.14\times 10^{-3})^2=0.016\times 10^{-6}m^2[/tex]
We know that current density [tex]J=\frac{I}{A}=\frac{0.89}{0.016\times 10^{-6}}=1.44\times 10^7A/m^2[/tex]
Now we have to calculate the electron density
We have current i = 0.89 A = 0.89 J/sec
Charge on 1 electron [tex]1.6\times 10^{-19}C/electron[/tex]
So electron density [tex]=\frac{0.89j/sec}{1.6\times 10^{-19}C/electron}=5.55\times 10^{18}electron/sec[/tex]
Final answer:
The current density in the filament is 14.4 x 10^6 A/m^2. The electron current in the filament is 1.42 x 10^-19 C/s.
Explanation:
The current density in a filament can be calculated by dividing the current in the lightbulb by the cross-sectional area of the filament. To find the current density, we first need to find the cross-sectional area of the filament. The diameter of the filament is given as 0.280 mm, so the radius is 0.140 mm (0.140 mm = 0.280 mm / 2).
Using the formula for the area of a circle (A = πr^2), we can find the cross-sectional area of the filament:
A = π(0.140 mm)^2 = 0.0616 mm^2
Now, we can calculate the current density by dividing the current (0.890 A) by the cross-sectional area (0.0616 mm^2), and converting mm to m:
Current density = 0.890 A / (0.0616 mm^2) * (1 m / 1000 mm)^2 = 14.4 x 10^6 A/m^2
The electron current in the filament can be calculated using the formula:
Electron current = charge * number of charges passing per second
The charge of an electron is approximately 1.6 x 10^-19 C, and the current is given as 0.890 A, so:
Electron current = (1.6 x 10^-19 C) * (0.890 A) = 1.42 x 10^-19 C/s
n Section 12.3 it was mentioned that temperatures are often measured with electrical resistance thermometers made of platinum wire. Suppose that the resistance of a platinum resistance thermometer is 125 Ω when its temperature is 20.0°C. The wire is then immersed in boiling chlorine, and the resistance drops to 99.6 Ω. The temperature coefficient of resistivity of platinum is α = 3.72 × 10−3(C°)−1. What is the temperature of the boiling chlorine?
The temperature of the boiling chlorine is calculated using the resistance change of a platinum wire thermometer and the temperature coefficient of resistivity. After calculations, the boiling point of chlorine is found to be -34.6°C.
Explanation:The temperature of the boiling chlorine can be deduced using the relation between temperature change and resistance change for a platinum wire thermometer. The equation we use is:
R = R0 (1 + αΔT)
Where R represents the resistance at the new temperature, R0 the resistance at a reference temperature (20°C in this case), α the temperature coefficient of resistivity, and ΔT the change in temperature. We can rearrange this equation to solve for the new temperature:
ΔT = ∂(T₂ - T₁)
Giving us:
(99.6 ohms / 125 ohms - 1) = 3.72 × 10-3°C-1 ∂T
Simplifying:
ΔT = (0.7968 - 1) / 3.72 × 10-3°C-1
ΔT = -0.2032 / 3.72 × 10-3°C-1
ΔT = -54.6°C
Thus, the boiling point of chlorine is:
20°C – 54.6°C = -34.6°C
A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. a) At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of v0, h, and g .
b) What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?
Express your answer in terms of the variable v0 and appropriate constants.
c) For what value of h does the collision occur at the instant when the first ball is at its highest point?
Express your answer in terms of the variable v0 and appropriate constants.
In this exercise we have to have knowledge about the horizontal launch, so we have to use the known formulas to find that:
a) [tex]h*(1 - 1/2 g * h/v_0^2)[/tex]
b)[tex]h = v_0^2/ g[/tex]
c)[tex]h = v_0^2/ g[/tex]
So we have to remember some famous equations like the position and velocity of an object moving in a constant, like this:
[tex]y = y_0 + v_0*t + 1/2 * a * t^2\\v = v_0 + a * t[/tex]
where:
y = height at time ty0 = initial heightv0 = initial velocitya = accelerationt = timev = velocitya) When the balls collide, h1 = h2. Then,
[tex]h_1 = h_2\\v_0 * t - 1/2 g * t^2 = h - 1/2 * g * t^2\\v_0 * t = h\\t = h / v_0[/tex]
Replacing in the equation of the height of the first ball:
[tex]h_1 = v_0 * h/v_0 - 1/2g * h^2/v_0^2\\h_1 = h - 1/2 g * h^2/ v_0^2\\h_1 = h*(1 - 1/2 g * h/v_0^2)[/tex]
b) that the balls collide at t = h/v0. Then:
[tex]h/ v_0 = -v_0/-g\\h = v_0^2/ g[/tex]
c) The value of h for which the collision occurs at the instant when the first ball is at its highest point is the maximum value,Then:
[tex]h = v_0^2/ g[/tex]
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The surplus energy theory of play suggests that
A. children are the link between animals and human beings
B. children have too much energy and play will rid them of that energy.
C. play is necessary to reenergize human cognition.
D. play provides children with an opportunity to practice adult activities
Answer:
B. children have too much energy and play will rid them of that energy.
Explanation:
The surplus energy theory of play suggests that human being have excess of energy that must be released through active play.
This theory is given by Friedreich Schiller. Therefore, the correct answer is children have too much energy and play will rid them of that energy. rest of the option are examples of other theories, whereas Option B is the example of surplus energy theory.
2.85 A police car is traveling at a velocity of 18.0 m/s due north, when a car zooms by at a constant velocity of 42.0 m/s due north. The police officer begins to pursue the speeder - first there is a 0.800 s reaction time when the officer has no change in speed, then the officer accelerates at 5.00 m/s2. Including the reaction time, how long does it take for the police car to reach the same position as the speeding car
Answer:
11.1 s
Explanation:
Speed of the police car as given = v = 18 m/s
Speed of the car = V = 42 m/s
Reaction time = t = 0.8 s
Distance traveled by the police car during the reaction time = d₁= 0.8 x 18 = 14.4 m
Distance traveled by speeding car = d₂ =0.8 x 42 = 33.6 m
Acceleration of the police car = a = 5 m/s/s
The police car can catch the speeding car only if it travels a distance equal to the speeding car in a time t.
Distance traveled by the police car = D = d₁ + v t +0.5 at², according to the kinematic equation.
⇒ D = 14.4 + 18 t + 0.5 (5) t²
⇒ D = 14.4 + 18 t+2.5 t² → (1)
For the speeding car, distance traveled is D = 33.6 + 42 t, since it is constant velocity. Substitute for D from the above equation (1).
⇒ 14.4 + 18 t+2.5 t²= 33.6 + 42 t
⇒ 2.5 t² -24 t - 19.2 = 0
⇒ t = 10.3 s
Total time = t +0.8 s
⇒ Time taken for the police car to reach the speeding car = 10.3+0.8= 11.1 s
A 7.94-nC charge is located 1.77 m from a 4.14-nC point charge. (a) Find the magnitude of the electrostatic force that one charge exerts on the other. (b) is the force attractive or repulsive?
Answer:
F=94.32*10⁻⁹N , The force F is repusilve because both charges have the same sign (+)
Explanation:
Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
F=K*q₁*q₂/d² Formula (1)
F: Electric force in Newtons (N)
K : Coulomb constant in N*m²/C²
q₁,q₂:Charges in Coulombs (C)
d: distance between the charges in meters(m)
Equivalence
1nC= 10⁻⁹C
Data
K=8.99x10⁹N*m²/C²
q₁ = 7.94-nC= 7.94*10⁻⁹C
q₂= 4.14-nC= 4.14 *10⁻⁹C
d= 1.77 m
Magnitude of the electrostatic force that one charge exerts on the other
We apply formula (1):
[tex]F=8.99x10^{9} *\frac{7.94*10^{-9} *4.14 *10^{-9} }{1.77^{2} }[/tex]
F=94.32*10⁻⁹N , The force F is repusilve because both charges have the same sign (+)
An electron with a kinetic energy of 22.5 eV moves into a region of uniform magnetic field B of magnitude 4.55 x 104 T. The angle between the directions of the magnetic field and the electron's velocity is 65.5 degrees. What is the pitch of the helical path taken by the electron?
The pitch of the helical path followed by an electron in a magnetic field is given by the formula p = v * cos(theta) * T, where v is the velocity of the electron, theta is the angle between the direction of the magnetic field and the electron's velocity, and T is the period of the electron's motion in the magnetic field. By substituting the given values into this formula, we can calculate the pitch.
Explanation:The pitch of a helical path can be determined by considering how the charged particle reacts in a magnetic field. The helical path of an electron in a magnetic field is caused by the perpendicular and parallel components of the electron's velocity. In particular, the parallel component of the velocity, v(cos(theta)), is responsible for the linear movement of the electron along the field lines. This leads to the corkscrew-like helical path, and the pitch of this helix is equivalent to the linear distance moved by the electron in one revolution.
The pitch (p) can be calculated by the formula p = v * cos(theta) * T, where v is the velocity of the electron, theta is the angle between the direction of the magnetic field and the velocity of the electron, and T is the period of circular motion of the electron in the magnetic field. From the provided data, we can first calculate the velocity of the electron using its kinetic energy, 22.5 eV: v = sqrt((2 * KE) / m), where KE is the kinetic energy and m is the mass of the electron (9.11 * 10^-31 kg). Next, we calculate the period using the formula T = 2 * pi * m / (q * B), where q is the charge of the electron (-1.6 * 10^-19 C) and B is the magnetic field intensity (4.55 * 10^4 T).
Finally, we substitute these values into the pitch formula and calculate the pitch of the helical path taken by the electron in the magnetic field.
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The pitch of the helical path taken by the electron is approximately [tex]\( 9.20 \times 10^{-11} \)[/tex] meters.
To find the pitch of the helical path taken by the electron, we need to follow these steps:
1. Calculate the speed of the electron:
The kinetic energy (K.E.) of the electron is given by:
[tex]\[ \text{K.E.} = \frac{1}{2}mv^2 \][/tex]
where [tex]\( m \)[/tex] is the mass of the electron [tex](\( 9.11 \times 10^{-31} \) kg)[/tex] and [tex]\( v \)[/tex] is the speed of the electron.
First, we need to convert the kinetic energy from electron volts (eV) to joules (J):
[tex]\[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \][/tex]
[tex]\[ 22.5 \, \text{eV} = 22.5 \times 1.602 \times 10^{-19} \, \text{J} = 3.6045 \times 10^{-18} \, \text{J} \][/tex]
Now, using the kinetic energy formula:
[tex]\[ \text{K.E.} = \frac{1}{2}mv^[/tex]
To find the pitch of the helical path taken by the electron, we need to determine the components of the electron's velocity and the motion along the magnetic field.
1. Calculate the speed of the electron:
The kinetic energy (KE) of the electron is given by:
[tex]\[ \text{KE} = \frac{1}{2} mv^2 \][/tex]
where m is the mass of the electron [tex](\( 9.11 \times 10^{-31} \) kg)[/tex] and v is the speed of the electron.
First, convert the kinetic energy from electron volts (eV) to joules (J):
[tex]\[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \][/tex]
[tex]\[ 22.5 \, \text{eV} = 22.5 \times 1.602 \times 10^{-19} \, \text{J} = 3.6045 \times 10^{-18} \, \text{J} \][/tex]
Now, solve for the speed \( v \) using the kinetic energy formula:
[tex]\[ \frac{1}{2} mv^2 = 3.6045 \times 10^{-18} \, \text{J} \][/tex]
[tex]\[ v^2 = \frac{2 \times 3.6045 \times 10^{-18}}{9.11 \times 10^{-31}} \][/tex]
[tex]\[ v^2 = 7.9137 \times 10^{12} \][/tex]
[tex]\[ v = \sqrt{7.9137 \times 10^{12}} \approx 2.81 \times 10^6 \, \text{m/s} \][/tex]
2. Determine the components of the velocity:
The angle between the velocity and the magnetic field is [tex]\( 65.5^\circ \)[/tex]. The component of the velocity parallel to the magnetic field [tex](\( v_{\parallel} \))[/tex] and the component perpendicular to the magnetic field[tex](\( v_{\perp} \))[/tex] can be found using trigonometry:
[tex]\[ v_{\parallel} = v \cos(65.5^\circ) \][/tex]
[tex]\[ v_{\perp} = v \sin(65.5^\circ) \][/tex]
Calculate these components:
[tex]\[ v_{\parallel} = 2.81 \times 10^6 \, \text{m/s} \times \cos(65.5^\circ) \approx 2.81 \times 10^6 \times 0.417 \approx 1.17 \times 10^6 \, \text{m/s} \][/tex]
[tex]\[ v_{\perp} = 2.81 \times 10^6 \, \text{m/s} \times \sin(65.5^\circ) \approx 2.81 \times 10^6 \times 0.909 \approx 2.55 \times 10^6 \, \text{m/s} \][/tex]
3. Find the cyclotron frequency [tex](\( f \))[/tex] and the period [tex](\( T \))[/tex] of the electron's circular motion:
The cyclotron frequency is given by:
[tex]\[ f = \frac{qB}{2\pi m} \][/tex]
where q is the charge of the electron [tex](\( 1.602 \times 10^{-19} \) C)[/tex], B is the magnetic field [tex](\( 4.55 \times 10^4 \) T)[/tex], and m is the mass of the electron.
[tex]\[ f = \frac{1.602 \times 10^{-19} \times 4.55 \times 10^4}{2\pi \times 9.11 \times 10^{-31}} \][/tex]
[tex]\[ f \approx \frac{7.29 \times 10^{-15}}{5.72 \times 10^{-31}} \approx 1.27 \times 10^{16} \, \text{Hz} \][/tex]
The period [tex](\( T \))[/tex] is the inverse of the frequency:
[tex]\[ T = \frac{1}{f} \approx \frac{1}{1.27 \times 10^{16}} \approx 7.87 \times 10^{-17} \, \text{s} \][/tex]
4. Calculate the pitch of the helical path:
The pitch is the distance the electron travels parallel to the magnetic field in one period:
[tex]\[ \text{Pitch} = v_{\parallel} \times T \][/tex]
[tex]\[ \text{Pitch} \approx 1.17 \times 10^6 \, \text{m/s} \times 7.87 \times 10^{-17} \, \text{s} \approx 9.20 \times 10^{-11} \, \text{m} \][/tex]
A 10-cm-long thin glass rod uniformly charged to 11.0 nC and a 10-cm-long thin plastic rod uniformly charged to - 11.0 nC are placed side by side, 4.10 cm apart. What are the electric field strengths E1 to E3 at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods?
The electric field strengths at various points around a charged object can be calculated using Coulomb's Law. They depend on the charge of the object and the distance from it. Fields produced by multiple charges need to be added vectorially.
Explanation:You're asking about the electric field strengths at varying distances from a charged glass rod, located next to a plastic rod with a charge of equal magnitude, but of opposite sign. The electric field strength E at a point in the vicinity of a charged object can be derived using Coulomb's Law. This law states that the force F between two charges q1 and q2 is proportional to the product of the charges and inversely proportional to the square of the distance r between them.
Given this, the electric field E created by a charge q at a distance r from the charge is given by E = k|q| / r², where k is Coulomb's constant, approximately 9 × 10⁹ N•m²/C². The direction of the electric field is determined by the sign of the charge. When it's positive, the field lines are going outward, and when it's negative, they are directed inward.
To calculate the net electric field at a point on the line connecting the midpoints of the rods, we consider the electric fields produced by both rods. Since they're opposite in sign, the fields will have opposite directions, and they should be added vectorially. The magnitudes and directions of the fields will also depend on the specific distances (1.0 cm, 2.0 cm, and 3.0 cm) from the glass rod.
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An airplane flies northwest for 250 mi and then west for 150 mi. What is the resultant displacement of the plane after this time?
Answer:371.564 mi
Explanation:
Given
Airplane flies northwest for 250 mi and then travels west 150 mi
That is first it travels 250cos45 in - ve x direction and simultaneously 250sin45 in y direction
after that it travels 150 mi in -ve x direction
So its position vector is given by
[tex]r=-250cos45\hat{i}-150\hat{i}+250sin45\hat{j}[/tex]
[tex]r=-\left ( 250cos45+150\right )\hat{i}+250sin45\hat{j}[/tex]
so magnitude of displacement is
[tex]|r|=\sqrt{\left ( \frac{250}{\sqrt{2}}+150\right )^2+\left ( \frac{250}{\sqrt{2}}\right )^2}[/tex]
|r|=371.564 mi
A particle moves along the x axis according to the equation x = 2.08 + 3.06t − 1.00t^2, where x is in meters and t is in seconds. (a) Find the position of the particle at t = 3.30 s. (b) Find its velocity at t = 3.30 s. (c) Find its acceleration at t = 3.30 s.
Explanation:
The position of a particle along x - axis is given by :
[tex]x=2.08+3.06t-1t^2[/tex]
(a) Position at t = 3.3 s
[tex]x=2.08+3.06(3.3)-1(3.3)^2[/tex]
x = 1.288 m
(b) Velocity at t = 3.3 s
[tex]v=\dfrac{dx}{dt}[/tex]
[tex]v=\dfrac{d(2.08+3.06t-1t^2)}{dt}[/tex]
[tex]v=3.06-2t[/tex]
at t = 3.3 s
[tex]v=3.06-2(3.3)[/tex]
v = -3.54 m/s
(c) Acceleration,
[tex]a=\dfrac{dv}{dt}[/tex]
[tex]a=\dfrac{d(3.06-2t)}{dt}[/tex]
[tex]a=-2\ m/s^2[/tex]
Hence, this is the required solution.
Final answer:
At t = 3.30 s, the particle is at the position of 1.288 meters, moves with a velocity of -3.54 m/s, and is subjected to a constant acceleration of -2 m/s^2.
Explanation:
The student's question about a particle's motion along the x-axis can be answered using kinematic equations from physics, which relate to the position, velocity, and acceleration of the particle.
Position at t = 3.30 s
To find the position of the particle at t = 3.30 s, we substitute the time into the position function x = 2.08 + 3.06t - 1.00t2.
2)>
Calculating this gives x(3.30 s) = 2.08 + 10.098 - 10.89 = 1.288 meters. The particle is at 1.288 meters from the origin at 3.30 seconds.
Velocity at t = 3.30 s
To find the velocity, we differentiate the position function with respect to time: v(t) = 3.06 - 2.00t.
The particle's velocity at 3.30 seconds is -3.54 meters per second.
Acceleration at t = 3.30 s
The acceleration is the second derivative of the position, which in this case is a constant because the t2 term has a constant coefficient: a(t) = -2.00 m/s2.
Therefore, the acceleration of the particle at any time, including at t = 3.30 s, is -2 meters per second squared.
If a negative charge is placed in an electric field, what direction will it be accelerated? a) In the direction of the field.
b) In the opposite direction of the field.
c) Perpendicular to the direction of the field.
d) It will not be accelerated.
e) None of the above.
Answer:
option B
Explanation:
the correct answer is option B
when negative charge is placed in electric field then direction will be accelerated in opposite direction of the field.
charged electric particle produce electric field and they exert force on other charged particle.
when a charged particle is positive it will accelerate in the direction of the electric field if the charge is negative then particle will accelerate in opposite direction of electric field.
The electric output of a power plant is 669 MW. Cooling water is the main way heat from the powerplant is rejected, and it flows at a rate of 1.17x 108 L/Hr. The water enters the plant at 23.6°C and exits at 29.8°C. (a) What is the power plant's total thermal power? (b) What is the efficiency of the power plant? (a) MWT (Megawatt thermal) O Answer part (a) (b) Answer part (b)
Answer:
a) 1511 MW
b) 44%
Explanation:
The thermal power will be the electric power plus the heat taken away by the cooling water.
Qt = P + Qc
The heat taken away by the water will be:
Qc = G * Cp * (t1 - t0)
The Cp of water is 4180 J/(kg K)
The density of water is 1 kg/L
Then
G = 1.17 * 10^8 L/h * 1 kg/L * 1/3600 h/s = 32500 kg/s
Now we calculate Qc
Qc = 32500 * 4180 * (29.8 - 23.6) = 842*10^6 W = 842 MW
The total thermal power then is
Qt = 669 + 842 = 1511 MW
The efficiency is
η = P / Qt
η = 669 / 1511 = 44%
What wavelength photon is required to excite a hydrogen atom from the n = 1 state to the n = 3 state?
Answer:
The wavelength required is 102.9 nm.
Explanation:
The energy levels for the hydrogen atom are
[tex]E(n) = \frac{-13.6 \ eV}{n^2}[/tex]
So, for a transition from the first level to the third level we got
[tex]\Delta E = E(3) - E (1)[/tex]
[tex]\Delta E = \frac{-13.6 \ eV}{ 3 ^2} - \frac{-13.6 \ eV}{1^2}[/tex]
[tex]\Delta E = \frac{-13.6 \ eV}{ 9} - \frac{-13.6 \ eV}{1}[/tex]
[tex]\Delta E = \frac{8}{9} 13.6 \ eV[/tex]
[tex]\Delta E = 12.09 \ eV[/tex]
[tex]\Delta E = 12.09 \ eV * \frac{1.6 \ 10^-19 \ Joules}{1 \ eV}[/tex]
[tex]\Delta E = 1.93 \ 10^-18 \ Joules[/tex]
So we need a photon with this energy.
The energy of a photon its given by
[tex]E = h \nu = h \frac{c}{\lambda}[/tex]
So, the wavelength will be
[tex]\lambda = \frac{h c}{E}[/tex]
[tex]\lambda = \frac{6.62 \ 10^{-34} \ \frac{m^2 kg}{s} \ * 3.00 \ 10^8 \ \frac{m}{s}}{1.93 \ 10^-18 \ Joules}[/tex]
[tex]\lambda = 10.29 \ 10^{-8} m[/tex]
[tex]\lambda = 1.029 \ 10^{-7} m[/tex]
[tex]\lambda = 102.9 \ nm[/tex]
When 19.3 J was added as heat to a particular ideal gas, the volume of the gas changed from 56.7 cm^3 to 104 cm^3 while the pressure remained constant at 0.947 atm. (a) By how much did the internal energy of the gas change? If the quantity of gas present is 1.97 x 10^-3 mol, find the molar specific heat of the gas at (b) constant pressure and (c) constant volume.
Answer:
a)[tex] C_p=35.42\ \rm J/mol.K[/tex]
b)[tex] C_v=27.1\ \rm J/mol.K[/tex]
Explanation:
Given:
Heat given to the gas, [tex]Q=19.3\ \rm J[/tex]Initial volume of the gas,[tex]V_i=56.7\ \rm cm^3[/tex]Final volume of the gas, [tex]V_f=104\ \rm cm^3[/tex]Constant pressure of the gas,[tex]P=0.947\ \rm atm[/tex]Number of moles of the gas,[tex]n=1.97\times10^{-3}[/tex]Let R be the gas constant which has value [tex]R=8.31432\times10^3\ \rm N\ m\ kmol^{-1}K^{-1}[/tex]
Work done in the process
[tex]W=PdV\\W=0.947\times(104-56.7)\times10^{-3}\ \rm L\ atm\\W=44.8\times10^{-3}\times101.33\ \rm J\\W=4.43\ \rm J[/tex]
Now Using First Law of thermodynamics
[tex]Q=W+\Delta U\\19.3=4.43+\Delta U\\\Delta U=14.77\ \rm J[/tex]
Let[tex]C_v[/tex] be the molar specific heat of the gas at constant volume given by
[tex]\Delta U=nC_v\Delta T\\14.77=\dfrac{C_v}{R}{PdV}\\14.77=\dfrac{C_v}{R}(0.987\times10^5(104-56.7)\times10^-6})\\C_v=3.26R\\C_v=27.1\ \rm J/mol.K[/tex]
Also we know that
Let[tex]C_p[/tex] be the molar specific heat of the gas at constant pressure given by
[tex]C_p-C_v=R\\C_p=R+C_v\\C_p=4.26R\\C_p=35.42\ \rm J/mol.K[/tex]
A particular spiral galaxy can be approximated by a thin disk-like volume 62 Thousand Light Years in radius and 7 Hundred Light Years thick. If this Galaxy contains 1,078 Billion stars, estimate the average distance between the stars in this galaxy. Hint: calculate the average volume per star in cubic Light Years, and then estimate the approximate linear dimension across such a volume. (Indicate your answer to one decimal place.)
Answer:
Approximate linear dimension is 2 light years.
Explanation:
Radius of the spiral galaxy r = 62000 LY
Thickness of the galaxy h = 700 LY
Volume of the galaxy = πr²h
= (3.14)(62000)²(700)
= (3.14)(62)²(7)(10)⁸
= 84568×10⁸
= [tex]8.45\times 10^{12}[/tex] (LY)³
Since galaxy contains number of stars = 1078 billion stars ≈ [tex]1.078\times 10^{12}[/tex]
Now volume covered by each star of the galaxy = [tex]\frac{\text{Total volume of the galaxy}}{\text{Number of stars}}[/tex]
= [tex]\frac{8.45\times 10^{12} }{1.078\times 10^{12}}[/tex]
= 7.839 Light Years
Now the linear dimension across the volume
= [tex](\text{Average volume per star})^{\frac{1}{3}}[/tex]
= [tex](7.839)^{\frac{1}{3}}[/tex]
= 1.99 LY
≈ 2 Light Years
Therefore, approximate linear dimension is 2 light years.
A guy wire 1005 feet long is attached to the top of a tower. When pulled taut, it touches level ground 552 feet from the base of the tower. What angle does the wire make with the ground? Express your answer using degree measure rounded to one decimal place.
Answer:
56.7°
Explanation:
Imagine a rectangle triangle.
The triangle has 3 sides.
One side is the height of the tower, let's name it A.
Another side is the distance from the base of the tower to the point where the waire touches the ground. Let's name that B.
Sides A and B are perpendicular.
The other side is the length of the wire. Let's name it C.
From trigonometry we know that:
cos(a) = B / C
Where a is the angle between B anc C, between the wire and the ground.
Therefore
a = arccos(B/C)
a = arccos(552/1005) = 56.7°
[tex]56.7^\circ[/tex] angle the wire makes with the ground.
Given :
A guy wire 1005 feet long is attached to the top of a tower.
When pulled taut, it touches level ground 552 feet from the base of the tower.
Solution :
We know that,
[tex]\rm cos \theta = \dfrac{Base}{Hypotanuse}[/tex]
Given that,
base = 552 ft
hypotanuse = 1005 ft
Therefore,
[tex]\rm cos\theta = \dfrac{552}{1005}[/tex]
[tex]\rm \theta = cos^-^1 \dfrac{552}{1005}[/tex]
[tex]\theta = 56.7^\circ[/tex]
[tex]56.7^\circ[/tex] angle the wire makes with the ground.
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A free positive charge released in an electric field will:
accelerate in the direction opposite the electric field.
accelerate along a circular path.
accelerate in the direction in which the electric field is pointing.
remain at rest.
accelerate in a direction perpendicular to the electric field.
Answer:
accelerate in the direction in which the electric field is pointing.
Explanation:
The positive charge feels a force in the same direction as the electric field
F=Eq
F and E are vectors, q is a scalar
(if it were a negative charge the force would be in the opposite direction)
that force will produce an acceleration in the same direction, that acceleration will cause the particle to move in the same direction, ie the direction of the electric field.
A free positive charge in an electric field will accelerate in the direction in which the electric field is pointing due to electrostatic forces.
Explanation:In the context of physics and electric fields, a free positive charge released in an electric field will accelerate in the direction in which the electric field is pointing. This is because electric field lines are drawn from positive to negative, indicating the direction that a positive test charge would move. Therefore, when a positive charge is placed in an electric field, it is repelled from the positive source charge and attracted to the negative source charge, causing it to accelerate along the electric field lines in the direction they are pointing.
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You and a friend are driving to the beach during spring break. You travel 14.5 km east and then 66.5 km south in a total time of 50 minutes. (a) What is the average speed of the trip? (b) What is the magnitude of the average velocity? -In this question for part B, why isnt it necessay to find the unit vectors for displacement?
Answer:
(a). The average speed is 97.23 km/hr.
(b). The average velocity is 81.70 km/min.
Explanation:
Given that,
Distance in east = 14.5 km
Distance in south = 66.5 km
Time = 50 min = 0.833 hr
(a). We need to calculate the average speed
Using formula of average speed
[tex]v_{avg}=\dfrac{D}{T}[/tex]
Where, D = total distance
T = total time
Put the value into the formula
[tex]v_{avg}=\dfrac{14.5+66.5}{0.833}[/tex]
[tex]v_{avg}=97.23\ Km/hr[/tex]
(b). We need to calculate the displacement
Using Pythagorean theorem
[tex]d=\sqrt{(d_{e})^2+(d_{s})^2}[/tex]
Put the value into the formula
[tex]d=\sqrt{(14.5)^2+(66.5)^2}[/tex]
[tex]d=68.06\ km[/tex]
(b). We need to calculate the average velocity
Using formula of average velocity
[tex]v_{avg}=\dfrac{d}{t}[/tex]
Where, d = displacement
t = time
Put the value into the formula
[tex]v_{avg}=\dfrac{68.06}{0.833}[/tex]
[tex]v_{avg}=81.70\ km/min[/tex]
Here, We can not necessary to find the unit vectors for displacement because we need to the displacement for find the average velocity.
Hence, (a). The average speed is 97.23 km/hr.
(b). The average velocity is 81.70 km/hr.
To make a secure fit, rivets that are larger than the rivet hole are often used and the rivet is cooled (usually in dry ice) before it is placed in the hole. A steel rivet 1.871 cm in diameter is to be placed in a hole 1.869 cm in diameter in a metal at 24 ∘C.To what temperature must the rivet be cooled if it is to fit in the hole?
Answer:
-65 degC
Explanation:
ΔD=[tex]D_{0}[/tex]αΔT
1.869-1.871=1.871(1.2E-5)ΔT
ΔT≈-89 degC
24-89=-65 degC
A ball is thrown vertically upward with a speed of 15.0 m/s. (a) How high does it rise? m (b) How long does it take to reach its highest point? s (c) How long does the ball take to hit the ground after it reaches its highest point? s (d) What is its velocity when it returns to the level from which it started? (Assume the positive direction is upward. Indicate the direction with the sign of your answer.) m/s
Answer:
a) 0.76 m
b) 1.53 s
c) 0.39 s
d) 3.8 m/s
Explanation:
This is a problem in which we need to use the equations pertaining Uniformly Accelerated Motion, as the acceleration during all this process is constant: The gravitational pull on the ball, 9.8 m/s².
To make things easier, we can divide this process in two parts: The first one (A) is from the moment the ball is thrown, until the moment it reaches it highest point and momentarily stops. The second one (B) from the moment it starts descending until it hits the ground.
a) During part A, we use the formula Vf²=Vi² +2*a*x . Where Vf is the final velocity (0 m/s, as the ball stopped in midair), Vi is the initial velocity (15 m/s), a is the acceleration (-9.8 m/s², it has a minus sign, as it goes against the direction of the movement) and x is the distance; thus, we're left with:
0 m/s=(15.0 m/s)²+2*(-9.8 m/s²)*x
We solve for x
x = 0.76 m
b) The formula is Vf=Vi +a*t, where t is the time. We're left with:
0 m/s=15.0 m/s + (-9.8 m/s²)*t
We solve for t
t= 1.53 s
c) Now we focus on part B, and use the formula x=Vi * t + [tex]\frac{at^{2}}{2}[/tex] , with the difference that the Vi is 0 m/s. We already know the value of x in exercise a). Note that a does not have a negative sign, as the direction of movement is opposite to the direction of part A
0.76 m=0 m/s * t +[tex]\frac{9.8\frac{m}{s^{2} } *t^{2} }{2}[/tex]
Solve for t
0.76=4.9t²
t=0.39 s
d) Once again we use the formula Vf=Vi +a*t, using the value of t previously calculated in exercise c).
Vf=0 m/s + 9.8 m/s² * 0,39 s
Vf=3.8 m/s
The ball rises to a height of 11.48 m, takes 1.53 s to reach its highest point, the same amount of time to fall back down, and has a velocity of -15.0 m/s when it returns to the starting level.
The question involves concepts from physics, specifically the kinematics of one-dimensional motion under constant acceleration due to gravity. Here is how we can solve each part of the given problem:
(a) How high does it rise?
To find the maximum height reached by the ball, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:
v² = u² + 2as
At the highest point, the final velocity (v) is 0 m/s, the initial velocity (u) is 15.0 m/s (upward), the acceleration due to gravity (a) is -9.8 m/s² (downward), and s represents the height. Solving for s:
0 = (15.0)² + 2(-9.8)s
s = (15.0)² / (2 * 9.8)
s = 11.48 m
(b) How long does it take to reach its highest point?
To find the time (t) it takes for the ball to reach its highest point, we can use the equation:
v = u + at
Since the final velocity at the highest point is 0 m/s:
0 = 15.0 + (-9.8)t
t = 15.0 / 9.8
t = 1.53 s
(c) How long does the ball take to hit the ground after it reaches its highest point?
The time for the ball to fall back down is the same as the time taken to reach the highest point, so this is also 1.53 s.
(d) What is its velocity when it returns to the level from which it started?
The velocity on returning to the starting level will be the same magnitude as the initial velocity but in the opposite direction, so it will be -15.0 m/s, with the negative sign indicating the downward direction.
A cube whose sides are of length = 1.8 m is placed in a uniform electric field of magnitude E = 5.8 ✕ 10^3 N/C so that the field is perpendicular to two opposite faces of the cube. What is the net flux through the cube (in N · m^2/C)?
Answer:
Zero
Explanation:
Electric flux is defined as the number of electric field lines which passes through any area in the direction of area vector.
The formula for the electric flux is given by
[tex]\phi =\overrightarrow{E}.\overrightarrow{dS}[/tex]
Here, E is the strength of electric field and dS be the area vector.
It is a scalar quantity.
According to the Gauss's theorem, the electric flux passing through any surface is equal to the [tex]\frac{1}{\epsilon _{0}}[/tex] times the total charge enclosed through the surface.
Here, the charge enclose is zero, so the total flux is also zero.
Take the following measured gauge (or gage) pressures and convert them to absolute pressures in both kPa and psi units for an ambient pressure equal to 98.10 kPa a) pgage 152 kPa, b) Pgage=67.5 Torr, c) pvac 0.10 bar, d) pvac 0.84 atm
Answer:
Explanation:
Given
ambient Pressure =98.10 kPa
(a)gauge pressure 152 kPa
we know
Absolute pressure=gauge pressure+Vacuum Pressure
[tex]P_{abs}[/tex]=152+98.10=250.1 kPa or 36.27 psi
(b)[tex]P_{gauge}[/tex]=67.5 Torr or 8.99 kpa
as 1 Torr is 0.133 kPa
[tex]P_{abs}[/tex]=8.99+98.10=107.09 kPa or 15.53 psi
(c)[tex]P_{vaccum}[/tex]=0.1 bar or 10 kPa
Thus absolute pressure=98.10-10=88.10 kPa or 12.77 psi
as 1 kPa is equal to 0.145 psi
(d)[tex]P_{vaccum}[/tex]=0.84 atm or 85.113 kPa
as 1 atm is equal to 101.325 kPa
[tex]P_{abs}[/tex]=98.10-85.11=12.99 kPa or 1.88 psi
A 50-V de voltage source was connected in series with a resistor and capacitor. Calculate the current in A to two significant figures) after 5.0 s if the resistance was 25 MΩ and the capacitance 0.10μF.
Answer:
Current through circuit will be [tex]0.2706\times 10^6A[/tex]
Explanation:
We have given source voltage v = 50 volt
Resistance [tex]R=25Mohm=25\times 10^6ohm[/tex]
Capacitance [tex]C=0.1\mu F=0.1\times 10^{-6}F[/tex]
Time t = 5 sec
Time constant of RC circuit is given by [tex]\tau =RC=25\times 10^6\times 0.1\times 10^{-6}=2.5sec[/tex]
We know that voltage across capacitor is given by [tex]v_c=v_s(1-e^{\frac{-t}{\tau }})[/tex]
[tex]v_c=50(1-e^{\frac{-5}{2.5 }})=43.2332v[/tex]
So current will be [tex]=\frac{v_s-v_c}{R}=\frac{50-43.2332}{25\times 10^{-6}}=0.2706\times 10^6A[/tex]
So current through circuit will be [tex]0.2706\times 10^6A[/tex]
A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is 1. After how many seconds does the ball strike the ground?
Answer:
[tex]t=6.96s[/tex]
Explanation:
From this exercise, our knowable variables are hight and initial velocity
[tex]v_{oy}=96ft/s[/tex]
[tex]y_{o}=112ft[/tex]
To find how much time does the ball strike the ground, we need to know that the final position of the ball is y=0ft
[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]
[tex]0=112ft+(96ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}[/tex]
Solving for t using quadratic formula
[tex]t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}[/tex]
[tex]a=-\frac{1}{2} (32.2)\\b=96\\c=112[/tex]
[tex]t=-0.999s[/tex] or [tex]t=6.96s[/tex]
Since time can't be negative the answer is t=6.96s
How do the significant figures in a measurement affect the significant figures in a calculation?
Answer:
Explanation:
The digits which are reliable and first uncertain digit is called significant figure.
In any measurement, if the significant figure is more than the accuracy of the measurement is also more.
If there are some numbers in a calculations and they having different numbers of significant figures, then in the final result the number of significant figures is the least number of significant figures which are in the individual number.
When performing calculations with measurements that have different numbers of significant figures, the final answer will be rounded to the same number of significant figures as the measurement with the fewest significant figures. This applies to both addition and subtraction, as well as multiplication and division.
Explanation:For addition and subtraction, the result should be rounded to the same decimal place as the least precise measured value. For example, if one measurement has two decimal places and another measurement has three decimal places, the final answer should be rounded to two decimal places.
For multiplication and division, the final answer should have the same number of significant figures as the measurement with the fewest significant figures. For example, if one measurement has three significant figures and another measurement has four significant figures, the final answer should have three significant figures.
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Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of mA = 18.5 kg and an initial velocity of v0A = 8.15 m/s, due east. Object B, however, has a mass of mB = 30.5 kg and an initial velocity of v0B = 5.00 m/s, due north. Find the magnitude of the final velocity of the two-object system after the collision.
Answer:
v =4.36 m/s
Explanation:
given,
mass of object A = 18.5 Kg
initial velocity of object A = 8.15 m/s in east
mass of object B = 30.5 kg
initial velocity of object B = 5 m/s
[tex]P = P_A+P_B[/tex]
[tex]P = m_Av_A\widehat{i} + m_B v_B\widehat{j}[/tex]
[tex]P = 18.5\times 8.15 \widehat{i} + 30.5\times 5\widehat{j}[/tex]
[tex]P = 150.775 \widehat{i} + 152.5 \widehat{j}[/tex]
[tex]P = \sqrt{150.775^2+152.5^2}[/tex]
P = 214. 45 N s
velocity after collision is equal to
[tex]v =\dfrac{214.45}{18.5+30.5}[/tex]
v =4.36 m/s
hence, velocity after collision is equal to 4.36 m/s
Answer:
The magnitude of the final velocity of the two-object system is [tex]v=4.37\frac{m}{s}[/tex]
Explanation:
As the Momentum is conserved, we can compare the instant before the collision, and the instant after. Also, we have to take in account the two components of the problem (x-direction and y-direction).
To do that, we put our 0 of coordinates where the collision takes place.
So, for the initial momentum we have that
[tex]p_{ix}=m_{a}v_{0a}+0[/tex]
[tex]p_{iy}=0+m_{b}v_{0b}[/tex]
Now, this is equal to the final momentum (in each coordinate)
[tex]p_{fx}=(m_{a}+m_{b}) v_{fx}[/tex]
[tex]p_{fy}=(m_{a}+m_{b}) v_{fy}[/tex]
So, we equalize each coordinate and get each final velocity
[tex]m_{a}v_{0a}=(m_{a}+m_{b}) v_{fx} \Leftrightarrow v_{fx}=\frac{m_{a}v_{0a}}{(m_{a}+m_{b})}[/tex]
[tex]m_{b}v_{0b}=(m_{a}+m_{b}) v_{fy} \Leftrightarrow v_{fy}=\frac{m_{b}v_{0b}}{(m_{a}+m_{b})}[/tex]
Finally, to calculate the magnitude of the final velocity, we need to calculate
[tex]v_{f}=\sqrt{(v_{fx})^{2}+(v_{fy})^{2}}[/tex]
which, replacing with the previous results, is
[tex]v_{f}=\sqrt{(v_{fx})^{2}+(v_{fy})^{2}}=(\sqrt{(\frac{18.5*8.15}{49})^{2}+(\frac{30.5*5.00}{49})^{2}})\frac{m}{s}[/tex]
Therefore, the outcome is
[tex]v_{f}=4.37\frac{m}{s}[/tex]
A 6.0-m board is leaning against a wall. A 28° angle is formed with the WALL What is the height above the ground where the board makes contact with the wall? (In Meters)
Answer:
the height above the ground where the board makes contact with the wall equals 5.297 meters above ground.
Explanation:
The wall and ladder are shown in the attached figure
In the figure we can see that
[tex]cos(28^{o})=\frac{H}{L}\\\\cos(28^{o})=\frac{H}{6}\\\\\therefore H=6.0\times cos(28^{o})==5.297meters[/tex]
Assuming that 70 percent of the Earth’s surface is covered with water at average depth of 0.95 mi, estimate the mass of the water on Earth. One mile is approximately 1.609 km and the radius of the earth is 6.37 × 10^6 m. Answer in units of kg. Your answer must be within ± 20.0%
Answer:
The mass of the water on earth is [tex]5.4537\times10^{20}\ kg[/tex]
Explanation:
Given that,
Average depth h= 0.95 mi
[tex]h=0.95\times1.609[/tex]
[tex]h =1.528\ km[/tex]
[tex]h=1.528\times10^{3}\ m[/tex]
Radius of earth [tex]r= 6.37\times10^{6}\ m[/tex]
Density = 1000 kg/m³
We need to calculate the area of surface
Using formula of area
[tex]A =4\pi r^2[/tex]
Put the value into the formula
[tex]A=4\pi\times(6.37\times10^{6})^2[/tex]
[tex]A=5.099\times10^{14}\ m^2[/tex]
We need to calculate the volume of earth
[tex]V = Area\times height[/tex]
Put the value into the formula
[tex]V=5.099\times10^{14}\times1.528\times10^{3}[/tex]
[tex]V=7.791\times10^{17}\ m^3[/tex]
Now, 70 % volume of the total volume
[tex]V= 7.791\times10^{17}\times\dfrac{70}{100}[/tex]
[tex]V=5.4537\times10^{17}\ m^3[/tex]
We need to calculate the mass of the water on earth
Using formula of density
[tex]\rho = \dfrac{m}{V}[/tex]
[tex]m = \rho\times V[/tex]
Put the value into the formula
[tex]m=1000\times5.4537\times10^{17}[/tex]
[tex]m =5.4537\times10^{20}\ kg[/tex]
Hence, The mass of the water on earth is [tex]5.4537\times10^{20}\ kg[/tex]