Answer:
The force on the foundation wall is [tex]F_f = 191394 \ N[/tex]
Explanation:
From the question we are told that
The depth of the hole's vertical side is [tex]d = 2.10 \ m[/tex]
The width of the hole is [tex]b = 8.90 \ m[/tex]
The distance of the concrete wall from the front of the cellar is [tex]c = 0.189 \ m[/tex]
Generally the area which the water from the drainage covers is mathematically represented as
[tex]A = d * b[/tex]
substituting values
[tex]A = 2.10 * 8.90[/tex]
[tex]A = 18.69 \ m^2[/tex]
Now the gauge pressure exerted on the foundation wall is mathematically evaluated as
[tex]P_g = \rho * d_{avg} * g[/tex]
Here is the average height foundation wall where the pressure of the water is felt and it is evaluated as
[tex]d_{avg} = \frac{h_1 + h_2 }{2}[/tex]
where [tex]h_1[/tex] at the height at bottom of the hole which is equal to [tex]h_1 = 0[/tex]
and [tex]h_2[/tex] is the height at the top of the hole [tex]h_2 = d = 2.10[/tex]
[tex]d_{avg} = \frac{0 + 2.10 }{2}[/tex]
[tex]d_{avg} = 1.05[/tex]
Where [tex]\rho[/tex] is the density of water with constant value [tex]\rho = 1000 \ kg/m^3[/tex]
substituting values
[tex]P_g = 1000 * 1.05 * 9.8[/tex]
[tex]P_g = 10290 \ Pa[/tex]
Then the force exerted by the water on the foundation wall mathematically represented as
[tex]F_f = P_g * A[/tex]
substituting values
[tex]F_f = 10290 * 18.69[/tex]
[tex]F_f = 191394 \ N[/tex]
1. The frequency of a wave defines
O A. the minimum height of a wave.
O B. the maximum height of a wave.
O C. how fast the wave is moving in cycles per second.
D. the height of the wave at a given point.
Answer:
The answer is CExplanation:
Frequency, in physics, the number of waves that pass a fixed point in unit time; also, the number of cycles or vibrations undergone during one unit of time by a body in periodic motion. ... See also angular velocity; simple harmonic motion.
A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If the current in the solenoid increases uniformly from 0 to 5.0 A in 0.60 s, what will be the induced emf in the short coil during this time
Answer:
The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V
Explanation:
The magnetic field at the center of the solenoid is given by;
B = μ(N/L)I
Where;
μ is permeability of free space
N is the number of turn
L is the length of the solenoid
I is the current in the solenoid
The rate of change of the field is given by;
[tex]\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s[/tex]
The induced emf in the shorter coil is calculated as;
[tex]E = NA\frac{\delta B}{\delta t}[/tex]
where;
N is the number of turns in the shorter coil
A is the area of the shorter coil
Area of the shorter coil = πr²
The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m
Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²
[tex]E = NA\frac{\delta B}{\delta t}[/tex]
E = 14 x 0.000491 x 0.02514
E = 1.728 x 10⁻⁴ V
Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V
The induced emf in the coil at the center of the longer solenoid is [tex]1.725\times10^{-4}V[/tex]
Induced EMF:The induced emf is produced in a coil when the magnetic flux through the coil is changing. It opposes the change of magnetic flux. Mathematically it is represented as the negative rate of change of magnetic flux at follows:
[tex]E=-\frac{\delta\phi}{\delta t}[/tex]
where E is the induced emf,
[tex]\phi[/tex] is the magnetic flux through the coil.
The changing current varies the magnetic flux through the coil at the center of the long solenoid, which is given by:
[tex]\phi = \frac{\mu_oNIA}{L}[/tex]
so;
[tex]\frac{\delta\phi}{\delta t}=\frac{\mu_oNA}{L} \frac{\delta I}{\delta t}[/tex]
where N is the number of turns of longer solenoid, A is the cross sectional area, I is the current and L is the length of the coil.
[tex]\frac{\delta\phi}{\delta t}=\frac{4\pi \times10^{-7} \times600 \times \pi \times(1.25\times10^{-2})^2}{25\times10^{-2}} \frac{5}{60}\\\\\frac{\delta\phi}{\delta t}=1.23\times10^{-7}Wb/s[/tex]
The emf produced in the coil at the center of the solenoid which has 14 turns will be:
[tex]E=N\frac{\delta \phi}{\delta t}\\\\E=14\times1.23\times10^{-7}V\\\\E=1.725\times10^{-4}V[/tex]
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Estimate the volume of a human heart (in mL) using the following measurements/assumptions:_______.
1. Blood flow through the aorta is approximately 11.2 cm/s
2. The diameter of the aorta is approximately 3.0 cm
3. Assume the heart pumps its own volume with each beat
4. Assume a pulse rate of 67 beats per minute.
Answer:
Explanation:
radius of aorta = 1.5 cm
cross sectional area = π r²
= 3.14 x 1.5²
= 7.065 cm²
volume of blood flowing out per second out of heart
= a x v , a is cross sectional area , v is velocity of flow
= 7.065 x 11.2
= 79.128 cm³
heart beat per second = 67 / 60
= 1.116666
If V be the volume of heart
1.116666 V = 79.128
V = 70.86 cm³.
Calculate the maximum kinetic energy of electrons ejected from this surface by electromagnetic radiation of wavelength 236 nm.
Answer:
Explanation:
Using E= hc/wavelength
6.63 x10^-34 x3x10^8/ 236nm
19.86*10^-26/236*10^-9
=0.08*10^-35Joules
A solenoid 26.0 cm long and with a cross-sectional area of 0.580 cm^2 contains 490 turns of wire and carries a current of 90.0 A.
Calculate:
(a) the magnetic field in the solenoid;
(b) the energy density in the magnetic field if the solenoid is filled with air;
(c) the total energy contained in the coil’s magnetic field (assume the field is uniform);
(d) the inductance of the solenoid.
Answer:
A.21.3T
B.1.8x 10^6J/m^3
C.0.27x10^2J
D.6.6x10^-3H
Explanation:
Pls see attached file
which example describes a nonrenewable resource?
A. everyone in our neighborhood uses solar panels to generate electricity to run their pool pumps.
B. once up and running, the power plant will convert the energy from tides and waves into electricity.
C. there is a long stretch of land in the desert with many windmills that are able to generate enough electricity to run the town.
D. there are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity.
Answer:
D. There are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity
Explanation:
Solar panels are a renewable resource because the sun will not run out. The power plant uses water, so it is also a renewable resource. Windmills use wind, and wind will not run out so it is a renewable resource. However, natural gas and oil are not renewable resources because they will run out one day.
Two parallel plates have charges of equal magnitude but opposite sign. What change could be made to increase the strength of the electric field between the plates
Answer:
The electric field strength between the plates can be increased by decreasing the length of each side of the plates.
Explanation:
The electric field strength is given by;
[tex]E = \frac{V}{d}[/tex]
where;
V is the electric potential of the two opposite charges
d is the distance between the two parallel plates
[tex]E =\frac{V}{d} = \frac{\sigma}{\epsilon _o} \\\\(\sigma = \frac{Q}{A} )\\\\E = \frac{Q}{A\epsilon_o} \\\\E = \frac{Q}{L^2\epsilon_o}[/tex]
Where;
ε₀ is permittivity of free space
L is the length of each side of the plates
From the equation above, the electric field strength can be increased by decreasing the length of each side of the plates.
Therefore, decreasing the length of each side of the plates, could be made to increase the strength of the electric field between the plates
A variable force of 6x−2 pounds moves an object along a straight line when it is x feet from the origin. Calculate the work done in moving the object from x = 1 ft to x = 18 ft. (Round your answer to two decimal places.) ft-lb
Answer:
931.00ft-lb
Explanation:
Pls see attached file
The work done in moving the object from x = 1 ft to x = 18 ft is 935 ft-lb.
What is work?
Work is the product of the displacement's magnitude and the component of force acting in that direction. It is a scalar quantity having only magnitude and Si unit of work is Joule.
Given that force = 6x - 2 pounds.
So, work done in moving the object from x = 1 ft to x = 18 ft is = [tex]\int\limits^{18}_1 {(6x-2)} \, dx[/tex]
= [ 3x² - 2x]¹⁸₁
= 3(18² - 1² ) - 2(18-1) ft-lb
= 935 ft-lb.
Hence, the work done is 935 ft-lb.
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A 1000-turn toroid has a central radius of 4.2 cm and is carrying a current of 1.7 A. The magnitude of the magnetic field along the central radius is
Answer:
0.0081T
Explanation:
The magnetic field B in the toroid is proportional to the applied current I and the number of turns N per unit length L of the toroid. i.e
B ∝ I [tex]\frac{N}{L}[/tex]
B = μ₀ I [tex]\frac{N}{L}[/tex] ----------------(i)
Where;
μ₀ = constant of proportionality called the magnetic constant = 4π x 10⁻⁷N/A²
Since the radius (r = 4.2cm = 0.042m) of the toroid is given, the length L is the circumference of the toroid given by
L = 2π r
L = 2π (0.042)
L = 0.084π
The number of turns N = 1000
The current in the toroid = 1.7A
Substitute these values into equation (i) to get the magnetic field as follows;
B = 4π x 10⁻⁷ x 1.7 x [tex]\frac{1000}{0.084\pi }[/tex] [cancel out the πs and solve]
B = 0.0081T
The magnetic field along the central radius is 0.0081T
A block of mass m is suspended by a vertically oriented spring. If the mass of a block is increased to 4m, how does the frequency of oscillation change, if at all
Answer:
The frequency will be reduced by a factor of √2/2
Explanation:
Pls see attached file
The new frequency of oscillation will be half the original frequency of oscillation of spring-block system.
Let the initial mass of block be m.
And new mass is, 4m.
The frequency of oscillating motion is defined as the number of complete oscillation made during the time interval of 1 second. The mathematical expression for the frequency of oscillation of block-spring system is given as,
[tex]f = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]
Here,
k is the spring constant.
If the mass of block increased to 4m, then the new frequency of oscillation of spring will be,
[tex]f' = \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{4m}}\\\\\\f' =\dfrac{1}{2} \times \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{m}}\\\\\\f' =\dfrac{1}{2} \times f[/tex]
Thus, we can conclude that the new frequency of oscillation will be half the original frequency of oscillation of spring-block system.
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An air conditioner connected to a 103 V rms AC line is equivalent to a 20 resistance and a 1.68 inductive reactance in series. a) What is the impedance of the air conditioner
Answer:
20.07ohms
Explanation:
Impedance is defined as the opposition to the flow of current through the elements of the circuit.
Impedance for R-L AC circuit is expressed as Z = √R²+XL²
R is the resistance
XL is the inductive reactance.
Given resistance of the air condition = 20 ohms
Inductive reactance XL = 1.68 ohms
Z = √20²+1.68²
Z = √400+2.8224
Z = √402.8224
Z = 20.07 ohms
Hence the impedance of the air conditioner is 20.07ohms
A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite from the planet is 6600 N. What is the kinetic energy of the satellite
Answer:
The kinetic energy is [tex]KE = 7.59 *10^{10} \ J[/tex]
Explanation:
From the question we are told that
The radius of the orbit is [tex]r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m[/tex]
The gravitational force is [tex]F_g = 6600 \ N[/tex]
The kinetic energy of the satellite is mathematically represented as
[tex]KE = \frac{1}{2} * mv^2[/tex]
where v is the speed of the satellite which is mathematically represented as
[tex]v = \sqrt{\frac{G M}{r^2} }[/tex]
=> [tex]v^2 = \frac{GM }{r}[/tex]
substituting this into the equation
[tex]KE = \frac{ 1}{2} *\frac{GMm}{r}[/tex]
Now the gravitational force of the planet is mathematically represented as
[tex]F_g = \frac{GMm}{r^2}[/tex]
Where M is the mass of the planet and m is the mass of the satellite
Now looking at the formula for KE we see that we can represent it as
[tex]KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r[/tex]
=> [tex]KE = \frac{ 1}{2} *F_g * r[/tex]
substituting values
[tex]KE = \frac{ 1}{2} *6600 * 2.3*10^{7}[/tex]
[tex]KE = 7.59 *10^{10} \ J[/tex]
A flashlight is held at the edge of a swimming pool at a height h = 1.6 m such that its beam makes an angle of θ = 38 degrees with respect to the water's surface. The pool is d = 1.75 m deep and the index of refraction for air and water are n1 = 1 and n2 = 1.33, respectively.
Required:
What is the horizontal distance from the edge of the pool to the bottom of the pool where the light strikes? Write your answer in meters.
one of the answers that i found was 5.83 m i did some more research and it showed the same answer again. good luck with it. hope i was able to help you.
Suppose a particle moves back and forth along a straight line with velocity v(t), measured in feet per second, and acceleration a(t). What is the meaning of ^120∫60 |v(t)| dt?
Answer:
The meaning of the integral (120, 60)∫ |v(t)| dt is simply the distance covered by the particle from time t = 60 seconds to time t = 120 seconds
Explanation:
We are told that the particle moves back and forth along a straight line with velocity v(t).
Now, velocity is the rate of change of distance with time. Thus, the integral of velocity of a particle with respect to time will simply be the distance covered by the particle.
Thus, the meaning of the integral (120, 60)∫ |v(t)| dt is simply the distance covered by the particle from time t = 60 seconds to time t = 120 seconds
Two slits are separated by 0.370 mm. A beam of 545-nm light strikes the slits, producing an interference pattern. Determine the number of maxima observed in the angular range−26.0° ≤ θ ≤ 26.0°.
Answer:
There are 586maxima
Explanation:
Pls see attached file
A 18.0 kg electric motor is mounted on four vertical springs, each having a spring constant of 24.0 N/cm. Find the period with which the motor vibrates vertically.
Answer:
Explanation:
Total mass m = 18 kg .
Spring are parallel to each other so total spring constant
= 4 x 24 = 96 N/cm = 9600 N/m
Time period of vibration
[tex]T=2\pi\sqrt{\frac{m}{k} }[/tex]
Putting the given values
[tex]T=2\pi\sqrt{\frac{18}{9600} }[/tex]
= .27 s .
Rick spends four hours researching on the internet and does 1090 J of work. In the process, his internal energy decreases by 2190 J. Determine the value of Q, including the algebraic sign.
Answer:
Q = -3280J
Explanation:
From the First Law of Thermodynamics, energy cannot be created nor destroyed but it can be converted from one form to another with the interaction of heat. Mathematically, this can be expressed as:
ΔU = Q + W ----------(i)
Where;
ΔU = total change in internal energy of a system.
Q = heat exchanged between the system and the surrounding
W = work done by or on the system.
If heat is lost into the surrounding, then Q = -ve, else Q = +ve
If work is done on the system, then W = -ve, else W = -ve
=> From the question, Rick is the system and does a work of
W = +1090J [since Rick does the work, W = +ve]
=>Also, the internal energy decreases by 2190J, therefore,
ΔU = -2190J [since there is a decrease in internal energy]
Substitute the values of W and ΔU into equation (i) as follows;
-2190 = Q + 1090
=> Q = -2190 - 1090
=> Q = -3280J
Therefore, the value of Q = -3280J
In the 1980s, the term picowave was used to describe food irradiation in order to overcome public resistance by playing on the well-known safety of microwave radiation. Find the energy in MeV of a photon having a wavelength of a picometer.
Answer:
E = 1.24MeV
Explanation:
The photon travels at the speed of light, 3.0 × [tex]10^{8}[/tex] m/s, and given that its frequency = 1 picometer = 1.0 × [tex]10^{-12}[/tex] m.
Its energy can be determined by;
E = hf
= (hc) ÷ λ
where E is the energy, h is the Planck's constant, 6.626 × [tex]10^{-34}[/tex] Js, c is the speed of the light and f is its frequency.
Therefore,
E = (6.626 × [tex]10^{-34}[/tex]× 3.0 × [tex]10^{8}[/tex]) ÷ 1.0 × [tex]10^{-12}[/tex]
= 1.9878 × [tex]10^{-25}[/tex] ÷ 1.0 × [tex]10^{-12}[/tex]
E = 1.9878 × [tex]10^{-13}[/tex] J
But, 1 eV = 1.6 × [tex]10^{-19}[/tex] J. So that;
E = [tex]\frac{1.9878*10^{-13} }{1.6*10^{-19} }[/tex]
= 1242375 eV
∴ E = 1.24MeV
The energy of the photon is 1.24MeV.
Velocity of a Hot-Air Balloon A hot-air balloon rises vertically from the ground so that its height after t sec is given by the following function.
h=1/2t2+1/2t
(a) What is the height of the balloon at the end of 40 sec?
(b) What is the average velocity of the balloon between t = 0 and t = 30?
ft/sec
(c) What is the velocity of the balloon at the end of 30 sec?
ft/sec
Answer:
Explanation:
Given the height reached by a balloon after t sec modeled by the equation
h=1/2t²+1/2t
a) To calculate the height of the balloon after 40 secs we will substitute t = 40 into the modeled equation and calculate the value of t
If h(t)=1/2t²+1/2t
h(40) = 1/2(40)²+1/2 (40)
h(40) = 1600/2 + 40/2
h(40) = 800 + 20
h(40) = 820 feet
The height of the balloon after 40 secs is 820 feet
b) Velocity is the change of displacement of a body with respect to time.
v = dh/dt
v(t) = 2(1/2)t²⁻¹ + 1/2
v(t) = t + 1/2
when v = 0sec
v(0) = 0 + 1/2
v(0) = 1/2 ft/sec
at v = 30secs
v(30) = 30 + 1/2
v(30) = 30 1/2 ft/sec
average velocity = v(30) - v(0)
average velocity = 30 1/2 - 1/2
average velocity of the balloon between t = 0 and t = 30 = 30 ft/sec
c) Velocity is the change of displacement of a body with respect to time.
v = dh/dt
v(t) = 2(1/2)t²⁻¹ + 1/2
v(t) = t + 1/2
The velocity of the balloon after 30secs will be;
v(30) = 30+1/2
v(30) = 30.5ft/sec
The velocity of the balloon after 30 secs is 30.5 feet/sec
A) The height of the balloon at the end of 40 sec is 820 feet.
B) The average velocity of the balloon is 30 ft/sec.
C) The velocity of the balloon at the end of 30 sec is
VelocityGiven :
h=1/2t²+1/2tPart A)
The height of the balloon after 40 secs is :
h(t)=1/2t²+1/2t
h(40) = 1/2(40)²+1/2 (40)
h(40) = 1600/2 + 40/2
h(40) = 800 + 20
h(40) = 820 feet
The height of the balloon after 40 secs is 820 feet
Part B)
The average velocity of the balloon is :
v = dh/dt
v(t) = 2(1/2)t²⁻¹ + 1/2
v(t) = t + 1/2
when v = 0 sec
v(0) = 0 + 1/2
v(0) = 1/2 ft/sec
When at v = 30secs
v(30) = 30 + 1/2
v(30) = 30 1/2 ft/sec
average velocity = v(30) - v(0)
average velocity = 30 1/2 - 1/2
average velocity of the balloon = 30 ft/sec
The average velocity of the balloon is 30 ft/sec.
Part C)
The velocity of the balloon at the end of 30 sec is :
v = dh/dt
v(t) = 2(1/2)t²⁻¹ + 1/2
v(t) = t + 1/2
The velocity of the balloon after 30secs will be;
v(30) = 30+1/2
v(30) = 30.5ft/sec
The velocity of the balloon after 30 secs is 30.5 feet/sec.
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A sound wave of frequency 162 Hz has an intensity of 3.41 μW/m2. What is the amplitude of the air oscillations caused by this wave? (Take the speed of sound to be 343 m/s, and the density of air to be 1.21 kg/m3.)
Answer:
I believe it is 91
Explanation:
At t=0 a 2150kg rocketship in outer space fires the engine which exerts a force=At2, and F(1.25s)=781.25N in the x direction. Find the impulse J during the interval t=2.00s and t=3.5s
Answer:
5.81 X 10^3 Ns
Explanation:
Given that
F = At² and F at t = 1.25 s is 781.25 N ?
A = F/t² at t = 1.25 s => F = 781.25/(1.25)² = 500 N/s²
d(Impulse) = Fdt
Impulse = ∫Fdt =∫At²dt evaluated in the interval 2.00 s ≤ t ≤ 3.50 s
Impulse = At³/3 = (500/3)(t³) = 166.7t³ between t = 2.00 s and t = 3.50 s
Impulse = 166.7[3.5³ - 2³] = 166.7[42.875 - 8] = 166.7[34.875] = 5813.7 Ns
5.81 X 10^3 N.s
A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and the branch is 12.0 m long.
(a) If the mass of the branch is negligible, what force must be exerted on the free end to just barely lift the rock?
(b) What is the mechanical advantage of this lever system?
Answer:
a
[tex]F =326.7 \ N[/tex]
b
[tex]M = 6[/tex]
Explanation:
From the question we are told that
The mass of the rock is [tex]m_r = 200 \ kg[/tex]
The length of the small object from the rock is [tex]d = 2 \ m[/tex]
The length of the small object from the branch [tex]l = 12 \ m[/tex]
An image representing this lever set-up is shown on the first uploaded image
Here the small object acts as a fulcrum
The force exerted by the weight of the rock is mathematically evaluated as
[tex]W = m_r * g[/tex]
substituting values
[tex]W = 200 * 9.8[/tex]
[tex]W = 1960 \ N[/tex]
So at equilibrium the sum of the moment about the fulcrum is mathematically represented as
[tex]\sum M_f = F * cos \theta * l - W cos\theta * d = 0[/tex]
Here [tex]\theta[/tex] is very small so [tex]cos\theta * l = l[/tex]
and [tex]cos\theta * d = d[/tex]
Hence
[tex]F * l - W * d = 0[/tex]
=> [tex]F = \frac{W * d}{l}[/tex]
substituting values
[tex]F = \frac{1960 * 2}{12}[/tex]
[tex]F =326.7 \ N[/tex]
The mechanical advantage is mathematically evaluated as
[tex]M = \frac{W}{F}[/tex]
substituting values
[tex]M = \frac{1960}{326.7}[/tex]
[tex]M = 6[/tex]
The tune-up specifications of a car call for the spark plugs to be tightened to a torque of 38N⋅m38N⋅m. You plan to tighten the plugs by pulling on the end of a 25-cm-long wrench. Because of the cramped space under the hood, you'll need to put at an angle of 120∘with respect to the wrench shaft. With what force must you pull?
Answer:
F= 175.5N
Explanation:
Given:
Torque which can also be called moment is defined as rotational equivalent of linear force. It is the product of the external force and perpendicular distance
torque of 38N⋅m
angle of 120∘
Torque(τ): 38Nm
position r relative to its axis of rotation: 25cm , if we convert to metre for consistency we have 0.25m
Angle: 120°
To find the Force, the torque equation will be required which is expressed below
τ = Frsinθ
We need to solve for F, if we rearrange the equation, we have the expression below
F= τ/rsinθ
Note: the torque is maximum when the angle is 90 degrees
But θ= 180-120=60
F= 38/0.25( sin(60) )
F= 175.5N
select the example that best describes a renewable resource.
A.after a shuttle launch, you can smell the jet fuel for hours.
B.solar panels generate electricity that keeps the satellites running.
C.tractor trailers are large trucks that run on diesel fuel.
D. we use our barbeque every night; it cooks with propane.
Answer:
B.solar panels generate electricity that keeps the satellites running.
Explanation:
Solar panels are a renewable resource because they take energy from the sun.
A resistor, capacitor, and switch are all connected in series to an ideal battery of constant terminal voltage. Initially, the switch is open and the capacitor is uncharged. What is the voltage across the resistor and the capacitor at the moment the switch is closed
Answer:
The voltage across the resistor is zero, and the voltage across the capacitor is equal to the terminal voltage of the battery.
Explanation:
This is because when a capacitor is charged no current or voltage flows through it so it will have a voltage equal to the terminal voltage of the battery
Resistance and Resistivity: The length of a certain wire is doubled while its radius is kept constant. What is the new resistance of this wire?
Answer:
Explanation:
The formula for calculating the resistance of a material in terms of its resistivity is expressed as [tex]R = \rho L/A[/tex] where;
R is the resistance of the material
[tex]\rho[/tex] is the resistivity of the material
L is the length of the wire
A is the area = πr² with r being the radius
[tex]R = \rho L/\pi r^{2}[/tex]
If the length of a certain wire is doubled while its radius is kept constant, then the new length of the wire L₁ = 2L
The new resistance of the wire R₁ will be expressed as [tex]R_1 = \frac{\rho L_1}{A_1}[/tex]
since the radius is constant, the area will also be the same i.e A = A₁ and the resistivity also will be constant. The new resistance will become
[tex]R_1 = \frac{\rho(2L)}{A}[/tex]
[tex]R_1 = \frac{2\rho L}{\pi r^2}[/tex]
Taking the ratio of both resistances, we will have;
[tex]\frac{R_1}{R} = \frac{2\rho L/\pi r^2}{\rho L/ \pi r^2} \\\\\frac{R_1}{R} = \frac{2\rho L}{\pi r^2} * \frac{\pi r^2}{ \rho L} \\\\\frac{R_1}{R} = \frac{2}{1}\\\\R_1 = 2R[/tex]
This shoes that the new resistance of the wire will be twice that of the original wire
A 900 kg roller coaster car starts from rest at point A. rolls down the track, goes
around a loop (points B and C) and then flies off the inclined part of the track (point D),
Figure 2.
The dimensions are: H =80 m.
r= 15m, h=10m and theta =9.30°
Calculate the
(a) gravitational potential energy at point A.
(b) velocity at point C, if the work done to move the roller coaster from point B to C is 264870 J.
c) distance of the car land (in the horizontal direction) from point D if given the
velocity at point D is 37.06 m/s
I
Answer:
gravitational potential energy at point A.
A) The gravitational potential energy at point A is; 705600 J
B) The velocity at point C, if the work done to move the roller coaster from point B to C is 264870 J is; v = 31.295 m/s
A) Formula for gravitational potential energy is;
PE = mgh
At point A;
mass; m = 900 kg
height; h = 80 m
Thus;
PE = 900 × 9.8 × 80
PE = 705600 J
B) Kinetic energy of the roller coaster at point C is given as;
KE = PE - W
We are given Workdone; W = 264870 J
Thus;
KE = 705600 - 264870
KE = 440730 J
Thus, velocity at point C is gotten from the formula of kinetic energy;
KE = ½mv²
v = √(2KE/m)
v = √(2 × 440730/900)
v = 31.295 m/s
Read more at; https://brainly.com/question/14295020
collision occurs betweena 2 kg particle traveling with velocity and a 4 kg particle traveling with velocity. what is the magnitude of their velocity
Answer:
metre per seconds
Explanation:
because velocity = distance ÷ time
A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through a potential difference of 2V0, what speed would it gain
Answer:
Explanation:
Let the charge on proton be q .
energy gain by proton in a field having potential difference of V₀
= V₀ q
Due to gain of energy , its kinetic energy becomes 1/2 m v₀²
where m is mass and v₀ is velocity of proton
V₀ q = 1/2 m v₀²
In the second case , gain of energy in electrical field
= 2 V₀q , if v be the velocity gained in the second case
2 V₀q = 1/2 m v²
1/2 m v² = 2 V₀q = 2 x 1/2 m v₀²
mv² = 2 m v₀²
v = √2 v₀
Two particles, of charges q1 and q2, are separated by a distance d on the x-axis with q1 at the origin and q2 in the positive direction. The net electric field due to the particles is zero at x = d/4. With V = 0 at infinity, locate (in terms of d) any point on the x-axis (other than infinity) at which the electric potential due to the two particles is zero.
Answer:
No point on the x-axis
Pls see attached file