Answer:
[tex]v=177.95m/s[/tex]
Explanation:
First, determine circle's radius between Earth's pole and the location. This can be calculated as:
[tex]r=R_e_a_r_t_hCos(90\frac{3}{4})\\R_e_a_r_t_h=6.37\times10^6m\\r=6.37\times10^6\times Cos67.5\textdegree\\r=2,437,693.46m\\[/tex]
The angular speed of earth is constant and is :
[tex]w=\frac{2\pi}{T}=\frac{2\pi}{24\times 3600}\\=7.3\times10^{-5}rad/s[/tex]
Velocity is:
[tex]v=wr\\=7.3\times10^{-5}\times 2,437,693.46\\v=177.95m/s[/tex]
An isolated charged insulating solid sphere with radius of 20 cm is carrying charge of -4.8 x 10-16 A solid sphere is surrounded by a hollow charged to +4.8 x 10-16 C conducting sphere with inner radius of r; = 40 cm and outer radius ro = 50 cm Find an Electric field at
a) r= 30 cm
b) = 45 cm
c) = 60 cm Use following constants: K = 8.99 x 109 Nm2/C2, Ep = 8.854 x 10-12 C2/Nm2
Answer:
a) E = -4.8 10⁻⁵ N / C , b) E = 0 , c) E = 0
Explanation:
For this exercise let's use Gauss's law
Ф = E. dA = [tex]q_{int}[/tex] / ε₀
As a Gaussian surface we use a sphere, whereby the electric field lines are parallel to the normal area, and the scalar product is reduced to the algebraic product
a) the field for r = 30 cm
At this point we place our Gaussian surface and see what charge there is inside, which is the charge of the solid sphere (r> 20cm), the charge on the outside does not contribute to the flow
E = q_{int} / A ε₀
q_{int} = -4.8 10-16 C
The area of a sphere is
A = 4π R²
We replace
E = -4.8 10⁻¹⁶ / 4π 0.30² 8.85 10⁻¹²
E = -4.8 10⁻⁵ N / C
b) r = 45 cm
This point is inside the spherical conductor shell, as in an electric conductor in electrostatic equilibrium the charges are outside inside the shell there is no charge for which the field is zero
E = 0
c) R = 60 cm
This part is outside the two surfaces
The chare inside is
q_{int} = -4.8 10⁻¹⁶ + 4.8 10⁻¹⁶
q_{int} = 0
Therefore the electric field is
E = 0
When a garden hose with an output diameter of 20 mm is directed straight upward, the stream of water rises to a height of 0.13m . You then use your thumb to partially cover the output opening so that its diameter is reduced to 10 mm.
Part A
How high does the water rise now? Ignore drag and assume that the smaller opening you create with your thumb is circular.
Express your answer with the appropriate units
h=
Answer: h = 0.52m
Explanation:
Using the equation of out flow;
A1 × V1 = A2 ×V2
Where A1 = area of the first nozzle
A2 = area of the second nozzle
V1= velocity of flow out from the first nozzle
V2 = velocity of flow out from 2nd nozzle
But AV= area of nozzle × velocity of water = volume of water per second(m³/s).
Now we can set A×V = Area of nozzle × height of rise.
Henceb A1× h1 = A2 × h2 ( note the time cancel on both sides)
D1 = 20mm= 0.02m; h1 = 0.13m
D2 = 10mm = 0.01m; h2= ?
h2 = π(D1/2)²× h1 /π(D2/2)²
h2 = (0.02/2)² × 0.13/(0.01/2)²
= (0.01)² ×0.13 /(0.005)²
= 1.3 × 10^-5/(5 × 10^-3)²
= 1.3 × 10^-5/25 × 10^-6
= (1.3/25) 10^-5 × 10^6
= 0.052× 10
= 0.52m
An ideal gas at 27°C is contained in a piston that ensures that its pressure will always be constant. Raising the tem- perature of the gas causes it to expand. At what tempera- ture will the gas take up twice its original volume?
Answer:
T = 600K
Explanation:
See attachment below.
During a long jump, an Olympic champion's center of mass rose about 1.2 m from the launch point to the top of the arc. 1) What minimum speed did he need at launch if he was traveling at 6.7 m/s at the top of the arc
Answer:
1) [tex]v_{A} \approx 8.272\,\frac{m}{s}[/tex]
Explanation:
1) Let assume that the campion begins running at a height of zero. The movement of the Olympic champion is modelled after the Principle of Energy Conservation:
[tex]K_{A} = K_{B} + U_{g,B}[/tex]
[tex]\frac{1}{2}\cdot m \cdot v_{A}^{2} = \frac{1}{2}\cdot m \cdot v_{B}^{2} + m \cdot g \cdot h_{B}[/tex]
[tex]\frac{1}{2} \cdot v_{A}^{2} = \frac{1}{2} \cdot v_{B}^{2} + g \cdot h_{B}[/tex]
The minimum speed is obtained herein:
[tex]v_{A}=\sqrt{v_{B}^{2} + 2 \cdot g \cdot h}[/tex]
[tex]v_{A} = \sqrt{(6.7\,\frac{m}{s} )^{2}+2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (1.2\,m)}[/tex]
[tex]v_{A} \approx 8.272\,\frac{m}{s}[/tex]
The minimum speed he need at launch point is [tex]8.27m/s[/tex]
Energy conservation :Energy is neither be created nor be destroyed just change into one form to another form.
[tex]\frac{1}{2}mv^{2} =\frac{1}{2}mv_{a}^{2} +mgh\\ \\v=\sqrt{v_{a}^{2}+2gh }[/tex]
Where,
[tex]v[/tex] is velocity at launch.[tex]v_{a}[/tex] is velocity at the top of the arc[tex]g[/tex] is gravitational acceleration, [tex]g=9.8m/s^{2}[/tex][tex]h[/tex] is height of center of mass.Given that, [tex]v_{a}=6.7m/s,h=1.2m,g=9.8m/s^{2}[/tex]
Substitute all values in above relation.
[tex]v=\sqrt{(6.7)^{2}+2*9.8*1.2 } \\\\v=\sqrt{68.41}=8.27m/s[/tex]
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Problem 8: Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and then injected into a region of constant magnetic field with a field strength of 0.65 T.
What is the potential difference in volts required in the first part of the experiment to accelerate electrons to a speed of 6 1 × 107 m/s?
Answer:
10581.59 V
Explanation:
We are given that
Magnetic field=B=0.65 T
Speed of electron=[tex]v=6.1\times 10^7m/s[/tex]
Charge on electron, [tex]q=e=1.6\times 10^{-19} C[/tex]
Mass of electron,[tex]m_e=9.1\times 10^{-31} kg[/tex]
We have to find the potential difference in volts required in the first part of the experiment to accelerate electrons.
[tex]V=\frac{v^2m_e}{2e}[/tex]
Where V=Potential difference
[tex]m_e=[/tex]Mass of electron
v=Velocity of electron
Using the formula
[tex]V=\frac{(6.1\times 10^7)^2\times 9.1\times 10^{-31}}{2\times 1.6\times 10^{-19}}[/tex]
[tex]V=10581.59 V[/tex]
Hence, the potential difference=10581.59 V
Final answer:
To accelerate electrons to a speed of 6.1 × 10^7 m/s in a constant magnetic field with a strength of 0.65 T, the potential difference required in the first part of the experiment is approximately 88.6 volts.
Explanation:
To accelerate electrons to a speed of 6.1 × 10^7 m/s in a constant magnetic field with a strength of 0.65 T, we need to calculate the potential difference required in the first part of the experiment. The formula for the potential difference is given by:
V = (1/2)m*(v^2)/(q * B)
Where V is the potential difference, m is the mass of the electron (9.11 × 10^-31 kg), v is the velocity of the electron (6.1 × 10^7 m/s), q is the charge of the electron (-1.6 × 10^-19 C), and B is the magnetic field strength (0.65 T).
Plugging in the values into the formula, we get:
V = (1/2)(9.11 × 10^-31 kg)(6.1 × 10^7 m/s)^2/(-1.6 × 10^-19 C)(0.65 T)
Simplifying the expression, we find that the potential difference required is approximately 88.6 volts.
If you pay him $200 up front, Freddie promises to paint your garage. But you'd be crazy to pay him anything up front because Freddie lies all the time, and every dime he gets his hands on he uses to feed his methamphetamine habit.a. False dichotomyb. Accident.c. Argument against the person, abusived. False cause.e. Argument against the person, circumstantial.
Answer: e
Explanation:
Argument against the person, circumstantial.
Clearly, Atwood’s machine has a lot of systematic error that would not be present if we were to simplify the experiment. What is one reason we might expect to get better results using Atwood’s machine rather than following Galileo’s example and just dropping objects off of tall buildings?
Answer:
Better Equilibrium Maintenance for better accuracy...
Explanation:
In the Galileo's experiment, there is no utilization of two equal masses at a time. However, as we can see in a Atwood Machine, there are two equal masses involved that make the whole system to be in a state of equilibrium and ultimately the better measurements of acceleration due to gravity.
The electric potential, when measured at a point equidistant from two particles that have charges equal in magnitude but of opposite sign, isA) equal to the net electric field B) smaller than zero C) equal to zero D) equal to the averages of the two distances times the charge E) larger than zero
Answer:
C) equal to zero
Explanation:
Electric potential is calculated by multiplying constant and charge, then dividing it by distance. The location that we want to measure is equidistant from two particles, mean that the distance from both particles is the same(r2=r1). The charges of the particle have equal strength of magnitude but the opposite sign(q2=-q1). The resultant will be:V = kq/r
ΔV= V1 + V2= kq1/r1 + kq2/r2
ΔV= V1 + V2= kq1/r1 + k(-q1)/(r)1
ΔV= kq1/r1 - kq1/r1
ΔV=0
The electric potential equal to zero
What are the radius and height above the ground of a circular geosynchronous orbit around the Earth (in m and Earth radii)? Is this a high or low orbit? How does its height compare with the height of the orbit of the International Space Station (about 360 km)? [HINT: You'll need to derive an algebraic relation between orbit radius and period, instead of just radius and speed.] (b)
Answer: radius r = 42360.7km
Height above ground = 35950.7km
The height of the satellite above the ground is about 100 times the height of the ISS above the ground.
This is a high orbit.
Explanation: a synchronous satellite of mass m, revolving around earth with angular speed w, having a radius of travel r will experience centripetal force F = m*r*w^2*
But w = 2¶/T
F = m*r*(2¶/T)^2
F = (4*m*r*¶^2)/T^2
For the same body on the surface of the earth of radius R, the force F will be F =mg
According to newton's law,
(4*m*r*¶^2)/T^2 is proportional to 1/r^2
also mg is proportional to 1/R^2
Therefore,
(4*m*r*¶^2)/T^2 = K/r^2,
mg = K/R^2
Equating the two we get
K = gR^2 = (4*r^3*¶^2)/T^2 (where K is a constant equal to the product of mass of earth M and gravitational constant.)
r^3 = (g*R^2*T^2)/(4x3.142^2)
Substituting values of g=9.81m/s2
R = 6400000m (radius of earth)
T = 60x60x24 = 86400s (synchronous orbit has period equal one day)
r = 42350775.04m = 42350.7km
Height above ground H = r - R
H = 42350.7 - 6400 = 35950.7km
Please verify with calculator. Thanks
Two plates with area A are held a distance d apart and have a net charge +Q, and -Q, respectively. Assume that all the charge is uniformly distributed on the inner surfaces of the plates.
The left plate has charge -Q, the right plate has charge +Q, separated by distance d.
1) Find the charge density on the plates.
2) Find the electric potential difference between the plates.
3) Show that the capacitance of the enlarged plates in this case is the same as the capacitance in a case where
Answer:
Explanation:
1 )
Charge density of left plate
= - Q / A
Charge density of right plate
= + Q / A
2 )
capacitance c = ε₀ A / d
potential difference = charge / capacitance
= Q / [ ε₀ A / d ]
= Q d / ε₀ A
Final answer:
The charge density on each plate is ±Q/A, the electric potential difference between the plates is calculated using the electric field and the separation distance, and the capacitance C=ε₀A/d demonstrates that the size of the plates does not affect their capacitance as long as their proportion remains constant.
Explanation:
When considering two parallel plates each with area A and charges of +Q and -Q respectively, separated by a distance d, we can address the posed questions systematically.
Finding the Charge Density on the Plates
The surface charge density σ is defined as charge per unit area. Given the total charge +Q or -Q and the area A of each plate, the charge density on each plate is σ = ±Q/A. This is a direct result of the uniform distribution assumption of the charges across the plates.
Finding the Electric Potential Difference Between the Plates
The electric field E created between the plates by the charge distribution is uniform and can be represented as E = σ/ε₀, where σ is the surface charge density and ε₀ is the vacuum permittivity. Consequently, the electric potential difference V between the plates can be derived from the relation V = Ed, linking the electric field and the separation of the plates.
Demonstrating the Capacitance of Enlarged Plates Remains Constant
The capacitance C of a parallel-plate capacitor is given by C = ε₀A/d, which is independent of the charge on the plates. This formula illustrates that the capacitance is solely dependent on the physical characteristics of the capacitor (i.e., the area of the plates A, the distance between them d, and the permittivity of free space ε₀), and does not change with the amount of charge nor with the size of the plates as long as their proportional relationship remains constant.
The slope of a line on a distance-time graph represents _____. distance time displacement speed
Answer:
speed
Explanation:
The slope of a line on any distance-time graph represents the speed of the object.
Velocity only comes in when there is speed of the object in a particular direction.
A generator is constructed by rotating a coil of N turns in a magnetic field B at a frequency f. The internal resistance of the coil is R and the cross sectional area of the coil is A.
The average induced EMF doubles if the area A isdoubled.
a. True
b. False
The average induced EMF doubles if the frequency f isdoubled.
a. True
b. False
The maximum induced EMF occurs when the coil is rotatedabout an axis perpendicular to area A.
a. True
b. False
The average induced EMF doubles if the resistance R isdoubled.
a. True
b. False
The average induced EMF doubles if the magnetic field Bis doubled.
a. True
b. False
The induced electromotive force (EMF) in a generator coil is directly proportional to the area of the coil, frequency of rotation, and the magnetic field strength, but not to the coil's internal resistance. Changing any of these factors as described, except resistance, results in a proportional change in the induced EMF.
Explanation:Understanding the Factors Affecting Induced EMF in a Generator CoilThe behavior of induced electromotive force (EMF) in a generator is explained by Faraday's law of electromagnetic induction, which states that the induced EMF in a coil is proportional to the rate of change of magnetic flux through the coil.
The average induced EMF doubles if the area A is doubled: True. By doubling the area A, the magnetic flux through the coil also doubles, leading to a doubling of the induced EMF.The average induced EMF doubles if the frequency f is doubled: True. Increasing the frequency of rotation leads to a faster rate of change of magnetic flux, thus doubling the induced EMF.The maximum induced EMF occurs when the coil is rotated about an axis perpendicular to area A: True. This orientation results in the maximum change in magnetic flux, thus inducing the maximum EMF.The average induced EMF doubles if the resistance R is doubled: False. Internal resistance does not affect the induced EMF, which is solely dependent on the change in magnetic flux.The average induced EMF doubles if the magnetic field B is doubled: True. Doubling the magnetic field strength doubles the magnetic flux, which leads to a doubling of the induced EMF.These principles help us to understand the operation of an electric generator, where mechanical work is converted into electrical energy.
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A charged particle isinjected into a uniform magnetic field such that its velocityvector is perpendicular to themagnetic field vector. Ignoring the particle's weight, the particlewill
A) follow a spiralpath.
B) move in a straight line.
C) move along a parabolic path.
D)follow a circular path.
A charged particle is injected into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vector. Ignoring the particle's weight, the particle will follow a circular path.
Option D
Explanation:
Magnetic force causes charged particles to move in spiral paths. The Particle accelerators keep the protons to follow circular paths when it is in the magnetic field. Velocity has a change in direction but magnitude remains the same when this condition exists.
The magnetic force exerted on the charged particle is given by the formula:
[tex]F=q v B \sin \theta[/tex]
where
q is the charge
v is the velocity of the particle
B is the magnetic field
[tex]\theta[/tex] is the angle
In this problem, the velocity is perpendicular to the magnetic field vector, hence
[tex]\theta[/tex] = [tex]90^{\circ}[/tex] and sin[tex]\boldsymbol{\theta}[/tex] =sin 90 degree = 1.
So applying the formula,
the force is simply [tex]F=q v B[/tex]
Also, the force is perpendicular to both B and v and so according to the right-hand rule, we have:
a force that is always perpendicular to the velocity, va force which is constant in magnitude (because the magnitude of v or B does not change)This means that the force acts as a centripetal force, so it will keep the charged particle in a uniform circular motion.
For your senior project, you would like to build a cyclotron that will accelerate protons to 10% of the speed of light. The largest vacuum chamber you can find is 60 cm in diameter.
The given question is incomplete. The complete question is as follows.
For your senior project, you would like to build a cyclotron that will accelerate protons to 10% of the speed of light. The largest vacuum chamber you can find is 60 cm in diameter.
What magnetic field strength will you need?
Explanation:
Formula for the strength of magnetic field is as follows.
B = [tex]\frac{mv}{qr}[/tex]
Here, m = mass of proton = [tex]1.67 \times 10^{-27}[/tex] kg
v = velocity = 10% of [tex]3 \times 10^{8}[/tex] = [tex]3 \times 10^{7}[/tex] m/s
q = charge of proton = [tex]1.6 \times 10^{-19} C[/tex]
r = radius = [tex]\frac{60}{2}[/tex] = 30 cm = 0.30 m (as 1 m = 100 cm)
Therefore, magnetic field will be calculated as follows.
B = [tex]\frac{mv}{qr}[/tex]
= [tex]\frac{1.67 \times 10^{-27} \times 3 \times 10^{7}}{1.6 \times 10^{-19} C \times 0.30 m}[/tex]
= [tex]\frac{5.01 \times 10^{-20}}{0.48 \times 10^{-19}}[/tex]
= 1.0437 T
Thus, we can conclude that magnetic field strength is 1.0437 T.
The student queries about the construction of a cyclotron to accelerate protons to a specific velocity. The radius and rotational period of protons within the cyclotron are calculated using the magnetic field strength and the desired velocity.
Explanation:The student is interested in building a cyclotron that can accelerate protons to 10% of the speed of light, with specific constraints on the vacuum chamber dimensions. In physics, particularly in the field of particle accelerators, a cyclotron is a type of particle accelerator that uses a combination of an electric field and a constant magnetic field to increase the kinetic energy of charged particles. The radius of the cyclotron, which determines the maximum orbit size for the particles being accelerated, is a critical design parameter and can be calculated based on the desired kinetic energy of the particles and the strength of the magnetic field.
In the context of a cyclotron, the student might need to calculate the rotational period and maximum radius of proton orbits within given specifications such as the strength of the magnetic field and desired velocity. Understanding the principles behind cyclotrons and particle acceleration is essential for this project, which falls under the umbrella of advanced physics topics.
Two radio antennas A and B radiate in phase. Antenna B is a distance of 120 m to the right of antenna A. Consider point Q along the extension of the line connecting the antennas, a horizontal distance of 40.0 m to the right of antenna B. The frequency, and hence the wavelength, of the emitted waves can be varied. What is the longest wavelength for which there will be destructive interference at point Q?
Answer:
240 m
120 m
Explanation:
d = Path difference = 120 m
For destructive interference
Path difference
[tex]d=\dfrac{\lambda}{2}\\\Rightarrow \lambda=2d\\\Rightarrow \lambda=2\times 120\\\Rightarrow \lambda=240\ m[/tex]
The longest wavelength is 240 m
For constructive interference
[tex]d=\lambda\\\Rightarrow 120\ m=\lambda[/tex]
The longest wavelength is 120 m
Final answer:
The longest wavelength for destructive interference at point Q, where the path difference is 120 m, is 240 m.
Explanation:
The longest wavelength for which there will be destructive interference at point Q can be found by considering the path difference between the waves from the two antennas at point Q. The waves must differ by an odd multiple of ½ wavelengths for destructive interference to occur. The distance from antenna A to point Q is 120 m + 40 m = 160 m, and the distance from antenna B to point Q is 40 m. Therefore, the path difference is 120 m.
For the first occurrence of destructive interference, the path difference should be ½ wavelength, so we calculate the longest wavelength (λ) as:
½ λ = 120 m
λ = 240 m
This is the longest wavelength that causes destructive interference at point Q.
The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23.5 m/s is h = 2 + 23.5t − 4.9t2 after t seconds. (Round your answers to two decimal places.) (a) Find the velocity after 2 s and after 4 s. v(2) = m/s v(4) = m/s
Answer:
a) [tex]v(2) = 3.9\,\frac{m}{s}[/tex], b) [tex]v(4) = -15.7\,\frac{m}{s}[/tex]
Explanation:
a) The equation for vertical velocity is obtained by deriving the function with respect to time:
[tex]v(t) = 23.5 -9.8\cdot t[/tex]
The velocities at given instants are, respectivelly:
[tex]v(2) = 3.9\,\frac{m}{s}[/tex]
[tex]v(4) = -15.7\,\frac{m}{s}[/tex]
During a baseball game, a baseball is struck at ground level by a batter. The ball leaves the baseball bat with an initial velocity v0 = 26 m/s at an angle θ = 17° above horizontal. Let the origin of the Cartesian coordinate system be the ballʼs position at impact. Air resistance may be ignored throughout this problem.
a) express the magnitude of the ball's initial horizontal velocity, v0x, in terms of v0 and theta.
b) express the magnitude of the ball's inital vertical velocity, v0y, in terms of v0 and theta.
c) find the ball's maximum vertical height, hmax, in meters above ground.
d) create an expression in terms of v0, theta, and g for the time (tmax) it takes the ball to travel to its maximum vertical height.
e) calculate the horizontal distance, xmax, in meters the ball has traveled when it returns to ground level.
Answer: a) vox = vo × cos θ, b) voy =vo× sin θ,
c) H=2.94 m, d) t = vo sinθ / g, e) R = 38.57 m
Explanation:
A)
The velocity v0 is at angle θ to the horizontal.
The horizontal component of vo (vox), vo and the vertical component of vo (voy) all form a right angle triangle.
With vo as the hypotenus, vox as the adjacent and voy as the opposite.
To get vox, we relate vo and vox ( hypotenus and adjacent)
From trigonometry
Cos θ relates hypotenus and adjacent, hence we have that
Cos θ = vox/vo
vox = vo × cos θ
B)
To get the vertical component of vo, we relate vo and voy ( hypotenus and opposite).
According to trigonometry, sin θ relates hypotenus and opposite, hence we have that
Sin θ = voy/vo
voy =vo× sin θ
C)
The formulae for the maximum height of a projectile motion is given as
H = vo² (sin θ)²/2g
Where g = acceleration due to gravity = 9.8 m/s²
By substituting the parameters, we have that
H = 26² × (sin 17)²/2(9.8)
H = 676 × 0.0854/19.6
H = 57.7304/ 19.6
H = 2.94 m
D)
This is the motion of a projectile and the conditions at maximum height are vy = 0 and ay = - g
From the equation of motion
vy = voy - gt
0 = voy - gt
But voy = vo sinθ
0 = vo sinθ - gt
gt = vo sinθ
t = vo sinθ / g
E)
The horizontal distance covered formulae is given by
R = u² sin2θ/g
R = 26² × sin 2(17)/9.8
R = 676 × sin 34/ 9.8
R = 378.014/ 9.8
R = 38.57 m
The correct Answer is:
a) vox = vo × cos θ, b) voy =vo× sin θ,c) H=2.94 m, d) t = vo sinθ / g, e) R = 38.57 mA) When The velocity v0 is at angle θ to the horizontal.When The horizontal component of vo (vox), vo, and also that the vertical component of vo (voy) all form a right angle triangle.Although when With vo as the hypotenuse, vox as the adjacent and voy as the opposite.Then To get vox, we relate vo and also that vox ( hypotenuse and adjacent)When From trigonometryAlso, Cos θ relates hypotenuse and adjacent, hence we have thatThen Cos θ = vox/voThen vox = vo × cos θB) When To get the vertical component of vo, we relate to and also voy ( hypotenuse and also opposite).According to trigonometry, sin θ relates to the hypotenuse and also that opposite, hence we have thatThen Sin θ = voy/voThen voy =vo× sin θ
C) When The formulae for the maximum height of a projectile motion is given asH = vo² (sin θ)²/2gWhere that g = acceleration due to gravity = 9.8 m/s²By substituting the parameters, we have thatThen H = 26² × (sin 17)²/2(9.8)Then H = 676 × 0.0854/19.6Then H = 57.7304/ 19.6Then H = 2.94 m
D) When This is the motion of a projectile and the conditions at maximum height are vy = 0 and ay = - g From the equation of motionvy = voy - gt0 = voy - gtBut voy = vo sinθ0 = vo sinθ - gtgt = vo sinθt = vo sinθ / g
E) When The horizontal distance covered formulae is given byR = u² sin2θ/gR = 26² × sin 2(17)/9.8R = 676 × sin 34/ 9.8R = 378.014/ 9.8R = 38.57 m
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Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N, the second a force of 95.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 21.0 kg.
(c) Calculate the acceleration.
m/s2
(d) What would the acceleration be if friction is 20.0 N?
Final answer:
The acceleration of the child in the wagon can be calculated using Newton's Second Law of Motion. The sum of the forces acting on the system is calculated by adding the forces exerted by the children and subtracting the force of friction. Using the given values, the acceleration can be determined by dividing the sum of the forces by the mass of the system. If the force of friction is increased to 20.0 N, the acceleration can be recalculated using the new value.
Explanation:
To calculate the acceleration of the child in the wagon, we can use Newton's Second Law of Motion:
ΣF = ma
Where ΣF is the sum of all the forces acting on the system, m is the mass of the system, and a is the acceleration.
In this case, the forces acting on the system are the forces exerted by the two children and the force of friction.
Using the given values:
Force1 = 75.0 N
Force2 = 95.0 N
Friction = 12.0 N
Mass = 21.0 kg
The sum of the forces acting on the system is:
ΣF = Force1 + Force2 - Friction
Substituting the values:
ΣF = 75.0 N + 95.0 N - 12.0 N
ΣF = 158.0 N
Now we can calculate the acceleration:
a = ΣF / m
Substituting the values:
a = 158.0 N / 21.0 kg
a = 7.52 m/s²
Therefore, the acceleration of the child in the wagon is 7.52 m/s².
(d) If friction is 20.0 N, we can recalculate the acceleration using the new value:
ΣF = 75.0 N + 95.0 N - 20.0 N
ΣF = 150.0 N
a = ΣF / m
a = 150.0 N / 21.0 kg
a = 7.14 m/s²
Therefore, if friction is 20.0 N, the acceleration of the child in the wagon would be 7.14 m/s².
The force that generates the heat and light produced by the sun and other stars is 1. the electromagnetic force. 2. the weak force. 3. the strong force. 4. the gravitational force.
The electromagnetic force generates the heat and light produced by the sun and other stars.
Explanation:The force that generates the heat and light produced by the sun and other stars is the electromagnetic force. This force is responsible for holding atoms together and producing electromagnetic radiation, which includes heat and light. It is much stronger compared to the weak force and gravity.
A rotating space station is said to create "artificial gravity"—a loosely-defined term used for an acceleration that would be crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments.If the space station is 200 m in diameter, what angular velocity would produce an "artificial gravity" of 9.80 m/s2 at the rim?
Answer:
The required angular velocity (ω) will be [tex]0.313~rads^{-1}[/tex].
Explanation:
Due to the rotation of the space station the astronauts experience a centripetal acceleration towards the centre of the space station. If '[tex]\large{a_{c}}[/tex]', 'ω' and 'R' represent the centripetal acceleration, angular velocity of the space station and the radius of the space station respectively, then
[tex]a_{c} = \omega^{2}.R[/tex]
As according to the problem the space station has to rotate in such an angular velocity that it produces the same "artificial gravity" as Earth's surface, we can write
[tex]a_{c} = g = 9.8 ms^{-2}[/tex]
Also given [tex]R = \dfrac{diameter~of~the~space~station}{2} = \dfrac{200 m}{2} = 100 m[/tex]
Therefore we can write,
[tex]&& a_{c} = g = \omega^{2}.R\\&or,& \omega = \sqrt{\dfrac{g}{R}} = \sqrt{\dfrac{9.8 ms^{-1}}{100 m}} = 0.313~rads^{-1}[/tex]
The 500-kg cylindrical drum is supported by a metal cable (A-B) and a rigid plate (B-C). If the contact between the drum and all surfaces is frictionless and its radius is 500 mm, compute the elongation of the cable (A-B). The elastic modulus of A-B is 200 GPa. A, B, and C are pinned joints. Note that the cross section of Cable A-B is 5 mm2 .
Explanation:
Beloware attachments containing the complete question and solution.
g Two hollow conducting spheres (radius ????) with a uniformly distributed charge are placed a distance ???? apart center to center. A thin wire with a switch ???? is connected to the surface of each sphere. The switch is initially open. a. What is the potential between points ???? and ????? b. If the switch is then closed, what is the charge on each sphere at time ???? → [infinity]? c. What is the potential between points ???? and ???? after the sphere reaches its steady state?
The given question is incomplete. The complete question is as follows.
Two hollow conducting spheres (radius r) with a uniformly distributed charge are placed a distance d apart center to center. A thin wire with a switch S is connected to the surface of each sphere. The switch is initially open.
a. What is the potential between points a and b?
b. If the switch is then closed, what is the charge on each sphere at time [tex]t \rightarrow \infty[/tex].
c. What is the potential between points a and b after the sphere reaches its steady state?
Explanation:
(a) In order to bring a positive test charge from infinity to a point 'a', the work done is equal to the potential energy of the charge at point 'a'.
Hence, [tex]V_{a} = \frac{1}{4 \pi \epsilon_{o}} \frac{q}{a}[/tex]
Now, work done in bringing a positive test charge from infinity to point 'b' is equal to the potential energy of the charge at point 'b'.
[tex]V_{b} = \frac{1}{4 \pi \epsilon_{o}} \frac{-q}{b}[/tex]
[tex]V_{a} - V_{b} = \frac{q}{4 \pi \epsilon_{o}}(\frac{1}{a} + \frac{1}{b})[/tex]
Therefore, the potential between points a and b is as follows.
[tex]V_{a} - V_{b} = \frac{q}{4 \pi \epsilon_{o}}(\frac{1}{a} + \frac{1}{b})[/tex]
(b) As the spheres are connected through a conducting wire then charges will flow from one sphere to another unless and until the charge on both the sphere will become equal. In this case, it is equal to zero.
(c) Since, the charge of both the spheres is equal to zero so, no work is necessary to bring another charge to a and b. Therefore, potential difference between the points will also become equal to zero.
A planet in elliptical orbit around a star moves from the point in its orbit furthest from the star (A) to the closest point (P). Choose the planet to be your system. The work done by the force of gravity during this movement is:
Answer:
Zero work done,since the body isn't acting against or by gravity.
Explanation:
Gravitational force is usually considered as work done against gravity (-ve) and work by gravity ( +ve ) and also When work isn't done by or against gravity work done in this case is zero.
Gravitational force can be define as that force that attracts a body to any other phyical body or system that have mass.
The planet been considered as our system in this case is assumed to have mass, and ought to demonstrate such properties associated with gravitational force in such system. Such properties include the return of every object been thrown up as a result of gravity acting downwards. The orbiting nature of object along an elliptical part when gravitational force isn't acting on the body and it is assumed to be zero.
A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turntable (initially at rest) begins to rotate with its rate of rotation constantly increasing. 1)What is the first event that will occur
Answer:
The answer to the question is
The ladybug begins to slide
Explanation:
To solve the question we assume that the frictional force of the ladybug and the gentleman bug are the same
Where the frictional force equals [tex]F_{Friction}[/tex] = μ×N = m×g×μ
and the centripetal force is given by m·ω²·r
If we denote the properties of the ladybug as 1 and that of the gentleman bug as 2, we have
m₁×g×μ = m₁·ω²·r₁ ⇒ g×μ = ω²·r₁
and for the gentleman bug we have
m₂×g×μ = m₂·ω²·r₂ ⇒ g×μ = ω²·r₂
But r₁ = 2×r₂
Therefore substituting the values of r₁ =2×r₂ we have
g×μ = ω²·r₁ = g×μ = ω²·2·r₂
Therefore ω²·r₂ = 0.5×g×μ for the ladybug. That is the ladybug has to overcome half the frictional force experienced by the gentleman bug before it start to slide
The ladybug begins to slide
A spherical tank with radius 3 m is half full of a liquid that has a density of 900 kg/m3. The tank has a 1 m spout at the top. Find the work W required to pump the liquid out of the spout. (Use 9.8 m/s2 for g.)
Approximately 1.6 million joules of work is required to pump the liquid out of the spherical tank through a 1 m long spout at the top.
Explanation:The work done to pump the liquid out of a tank can be calculated using the formula W = ρgVh, where ρ is the density of the liquid, g is the acceleration due to gravity, V is the volume of the liquid, and h is the height up to which the liquid is pumped. Since the tank is spherical and half full, the volume of the liquid is ½(4/3πr³), or 2πr³. Substituting the given values: ρ = 900 kg/m³, g = 9.8 m/s², r = 3 m, and h is the radius of the sphere plus the length of the spout (3 m + 1 m = 4 m), we get W = 900 kg/m³ * 9.8 m/s² * 2π(3 m)³ * 4 m ≈ 1.6 * 10⁶ J. Therefore, approximately 1.6 million joules of work is required to pump the liquid out of the tank.
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The three forces (in units of N) given below are acting on a 20 kg mass. Calculate the magnitude of the acceleration of the mass. stack F subscript 1 with rightwards harpoon with barb upwards on top equals 3 i with hat on top space space space space space stack F subscript 2 with rightwards harpoon with barb upwards on top equals 5 j with hat on top space space space space stack F subscript 3 with rightwards harpoon with barb upwards on top equals open parentheses i with hat on top minus 3 j with hat on top close parentheses space space A. 0.2 m/s^2 B. 0.224 m/s^2 C. 0.1 m/s^2 D. 1.0 m/s^2
Answer:
B. 0.224 m/s²
Explanation:
Given:
Mass of the object (m) = 20 kg
The forces acting on the object are:
[tex]\vec{F_1}=3\vec{i}\ N\\\\\vec{F_2}=5\vec{j}\ N\\\\\vec{F_3}=(\vec{i}-3\vec{j})\ N[/tex]
Now, the net force acting on the object is equal to the vector sum of the forces acting on it. Therefore,
[tex]\vec{F_{net}}=\vec{F_1}+\vec{F_2}+\vec{F_3}\\\\\vec{F_{net}}=3\vec{i}+5\vec{j}+\vec{i}-3\vec{j}\\\\\vec{F_{net}}=(3+1)\vec{i}+(5-3)\vec{j}\\\\\vec{F_{net}}=(4\vec{i}+2\vec{j})\ N[/tex]
Now, the magnitude of the net force is equal to the square root of the sum of the squares of its components and is given as:
[tex]|\vec{F_{net}}|=\sqrt{4^2+2^2}\\\\|\vec{F_{net}}|=\sqrt{20}\ N[/tex]
Now, from Newton's second law, the magnitude of acceleration is equal to the ratio of the magnitude of net force and mass. So,
Magnitude of acceleration is given as:
[tex]|\vec{a}|=\dfrac{|\vec{F_{net}}|}{m}\\\\|\vec{a}|=\frac{\sqrt{20}\ N}{20\ kg}\\\\|\vec{a}|=0.224\ m/s^2[/tex]
Therefore, option (B) is correct.
A rectangular tank that is 4 meters long, 3 meters wide and 6 meters deep is filled with a rubbing alcohol that has density 786 kilograms per cubic meter. In each part below, assume that the tank is initially full, and that gravity is 9.8 meters per second squared. Your answers must include the correct units.
(a) How much work is done pumping all of the liquid out over the top of the tank?
(b) How much work is done pumping all of the liquid out of a spout 2 meters above the top of the tank?
(c) How much work is done pumping two-thirds of the liquid out over the top of the tank?
(d) How much work is done pumping two-thirds of the liquid out of a spout 2 meters above the top of the tank?
The work done pumping the liquid out of the tank relies on the weight of the mass being lifted, the gravity, and the height it is being lifted to. Calculations are given for lifting all the fluid and two-thirds of it to two different heights.
Explanation:To calculate the work done pumping the liquid out of the tank, we first need to find the volume of the tank, which is 4 meters * 3 meters * 6 meters, giving a total of 72 cubic meters of rubbing alcohol. Multiplying this by the density of the alcohol (786 kg/m^3) gives us the total mass of the alcohol, 56,592 kg. The work done by gravity when an object is lifted is equal to the weight of the object (mass*gravity) multiplied by the distance it is lifted (height). Therefore, we can calculate the work done pumping the liquid out of the tank:
(a) The height is 6 m, so the work done is 56,592 kg * 9.8 m/s^2 * 6m = 3,331,723.2 J (b) The height is 8m (6m + 2m), so the work done is 56,592 kg * 9.8 m/s^2 * 8m = 4,442,297.6 J (c) Two thirds of the liquid is 37,728 kg, height is 6m, so the work done is 37,728 kg * 9.8 m/s^2 * 6m = 2,221,148.8 J (d) Two thirds of the liquid is 37,728 kg, height is 8m, so the work done is 37,728 kg * 9.8 m/s^2 * 8m = 2,961,531.2 J. Learn more about Work Done here:
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The work done to pump all of the liquid out over the top of the tank is calculated as 1665568.8 J, and 4441516.8 J through a spout 2 meters above the tank. To pump out two-thirds of the liquid over the top, the work is 1109203.2 J, and through the spout, it is 2957875.2 J. Each part uses the weight of the liquid and the distance it needs to be moved to calculate the work done.
Part a :
To find the work done, we need to calculate the force required to move the liquid to the top of the tank and the distance it needs to be moved.
Calculate the volume of the tank:
Volume = length × width × depth = 4 m × 3 m × 6 m = 72 m³.
Calculate the mass of the liquid:
Mass = density × volume = 786 kg/m³ × 72 m³ = 56652 kg.
Calculate the weight of the liquid:
Weight = mass × gravity = 56652 kg × 9.8 m/s² = 555189.6 N.
Calculate the center of mass:
The center of mass for a full tank is at half the depth:
3 meters.
Calculate the work done:
Work = weight × height = 555189.6 N × 3 m = 1665568.8 J (Joules).
Part b :
Here, the height the liquid needs to be moved is 6 m (tank height) + 2 m (spout height):
6 + 2 = 8 meters.
Work = weight × height
Work = 555189.6 N × 8 m = 4441516.8 J.
Part c :
Calculate the volume of two-thirds of the tank:
Volume = (2/3) × 72 m³ = 48 m³.
Calculate the mass of two-thirds of the liquid:
Mass = 786 kg/m³ × 48 m³ = 37728 kg.
Calculate the weight of two-thirds of the liquid:
Weight = 37728 kg × 9.8 m/s² = 369734.4 N.
Since the tank is still of uniform depth, the center of mass for the remaining liquid will be halfway up the tank's current depth:
1.5 meters (half of the 3 meters left).
Work = weight × height
Work = 369734.4 N × 3 m = 1109203.2 J.
Part d :
Height the liquid needs to be moved = 6 m (tank height) + 2 m (spout height) = 8 meters.
Work = weight × height
Work = 369734.4 N × 8 m = 2957875.2 J.
Calculate the noise voltage spectrum Su(f) on a paraller RC circuit with R=1 kOhm and C=1 nF. (Don't forget to provide the unit!) (3 points) - How much it is at 1 Hz ?
Answer: 10^-3 V^2/Hz
Explanation:
1 Hz:
Su(f) = No * |H(f)|^2
= 10^-3 * 1/(1+(2*pi*f*R*C)^2)
= 10^-3 V^2/Hz
A block of mass 3.5 kg, sliding on a horizontal plane, is released with a velocity of 1.6 m/s. The block slides and stops at a distance of 1.6 m beyond the point where it was released. How far would the block have slid if its initial velocity were increased by a factor of 3.5
Answer:
19.6 m
Explanation:
The work-energy theorem applies here,
The theorem states that the change in momentum of a particle between two points is equal to the work done in moving the force between the two distance.
ΔK.E = W
ΔK.E = (final kinetic energy) - (initial kinetic energy)
Kinetic energy = (1/2)(m)(v²)
m = 3.5 kg, v = 1.6 m/s
Final kinetic energy = 0 J, since the block of mass comes to rest.
Initial kinetic energy = (1/2)(3.5)(1.6²) = 4.48 J
ΔK.E = - 4.48 J
The workdone on the block of mass is done by the frictional force, F, which acts opposite to the direction of the displacement.
W = - Fd = - 1.6 F
ΔK.E = W
- 4.48 = - 1.6 F
F = 2.8 N
when the initial velocity is increased by a factor 3.5,
v = 1.6×3.5 = 5.6 m/s
Final kinetic energy = 0 J, since the block of mass comes to rest.
Initial kinetic energy = (1/2)(3.5)(5.6²) = 4.48 J
ΔK.E = - 54.88 J
The workdone on the block of mass is done by the frictional force, F, which acts opposite to the direction of the displacement. The frictional force is the same as above.
W = - Fd = - (2.8) d
ΔK.E = W
- 54.88 = - 2.8 d
d = 19.6 m
Compared to 1.6 m, 19.6 m is an increase by a factor 12.25.
Two loudspeakers, 4 meters apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point towards one of the speakers, you encounter a minimum of sound intensity when you have moved 0.25 meters. Velocity of the sound is 343 m/s
A.What is the frequency of the sound?
B.If the frequency is then increased while you remain 0.35m from the center, what is the first frequency for which that location will be a maximum of sound intensity?
m/s
Answer:
Explanation:
Minimum intensity occurs due to destructive interference of sound. For it to take place ,
path difference = odd multiple of half wave length
When moved by .025 m ,
path difference created = ( x + .25 ) - ( x - .25 )
= 2 x .25
= .5 m
So .5 = half wave length
wave length = 2 x .5
= 1 m
frequency = velocity / wave length
= 343 / 1
= 343 Hz
B )
Now when frequency is increased , wave length will be decreased . For maximum intensity , constructive interference will have to take place .
For that
path difference = integral multiple of wave length
0.5 = 1 x wave length
wave length = .5
frequency = 343 / .5
= 686 Hz