Answer:
a) P = 10.27 kW
b) Pmax = 10.65 kW
c) E = 5.47 MJ
Explanation:
Mass of the loaded car, m = 950 kg
Angle of inclination of the shaft, θ = 28°
Acceleration due to gravity, g = 9.8 m/s²
The speed of the car, v = 2.35 m/s
Change in time, t = 14.0 s
a) The power that must be provided by the winch motor when the car is moving at constant speed.
P = Fv
The force exerted by the motor, F = mg sinθ
P = mgv sinθ
P = 950 * 9.8 *2.35* sin28°
P = 10,271.3 W
P = 10.27 kW
b) Maximum power that the motor must provide:
[tex]P = mv\frac{dv}{dt} + mgvsin \theta\\dv/dt = \frac{2.35 - 0}{14} \\dv/dt = 0.168 m/s^2\\P = (950*2.35*0.168) + (950*9.8*2.35* sin28)\\P = 374.74 + 10271.3\\P = 10646.04 W\\10.65 kW[/tex]
c) Total energy transferred:
Length of the track, d = 1250 m
[tex]E = 0.5 mv^2 + mgd sin \theta\\E = (0.5 * 950 * 2.35^2) + (950 * 9.8 * 1250 * sin 28)\\E = 2623.19 + 5463475.31\\E = 5466098.50 J\\E = 5.47 MJ[/tex]
When one person was talking in a small room, the sound intensity level was 60 dB everywhere within the room. Then, there were 14 people talking in similar manner simultaneously in the room, what was the resulting sound intensity level?
A. 64 dB
B. 60 dB
C. 69 dB
D. 79 dB
E. 71 dB
Answer:
E= 71dB
Explanation:
See attached file for step by step calculation
A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation between the plates is increased, what will happen to the charge on the capacitor and the electric potential across it
Answer:
The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)
Explanation:
The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. [tex]Q[/tex], the amount of charge stored in this capacitor, will stay the same.
The formula [tex]\displaystyle Q = C\, V[/tex] relates the electric potential across a capacitor to:
[tex]Q[/tex], the charge stored in the capacitor, and[tex]C[/tex], the capacitance of this capacitor.While [tex]Q[/tex] stays the same, moving the two plates apart could affect the potential [tex]V[/tex] by changing the capacitance [tex]C[/tex] of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:
[tex]\displaystyle C = \frac{\epsilon\, A}{d}[/tex],
where
[tex]\epsilon[/tex] is the permittivity of the material between the two plates.[tex]A[/tex] is the area of each of the two plates.[tex]d[/tex] is the distance between the two plates.Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of [tex]\epsilon[/tex]. Neither will that change the area of the two plates.
However, as [tex]d[/tex] (the distance between the two plates) increases, the value of [tex]\displaystyle C = \frac{\epsilon\, A}{d}[/tex] will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.
On the other hand, the formula [tex]\displaystyle Q = C\, V[/tex] can be rewritten as:
[tex]V = \displaystyle \frac{Q}{C}[/tex].
The value of [tex]Q[/tex] (charge stored in this capacitor) stays the same. As the value of [tex]C[/tex] becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.
Consider the Earth and the Moon as a two-particle system.
Find an expression for the gravitational field g of this two-particle system as a function of the distance r from the center of the Earth. (Do not worry about points inside either the Earth or the Moon. Assume the Moon lies on the +r-axis. Give the scalar component of the gravitational field. Do not substitute numerical values; use variables only. Use the following as necessary: G, Mm, Me, r, and d for the distance from the center of Earth to the center of the Moon.)"
sorry but I don't understand
Given a double slit apparatus with slit distance 1 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 400 nm on the slits
Answer:
The maximum number of bright spot is [tex]n_{max} =5001[/tex]
Explanation:
From the question we are told that
The slit distance is [tex]d = 1 \ mm = 0.001 \ m[/tex]
The wavelength is [tex]\lambda = 400 \ nm = 400*10^{-9 } \ m[/tex]
Generally the condition for interference is
[tex]n * \lambda = d * sin \theta[/tex]
Where n is the number of fringe(bright spots) for the number of bright spots to be maximum [tex]\theta = 90[/tex]
=> [tex]sin( 90 )= 1[/tex]
So
[tex]n = \frac{d }{\lambda }[/tex]
substituting values
[tex]n = \frac{ 1 *10^{-3} }{ 400 *10^{-9} }[/tex]
[tex]n = 2500[/tex]
given there are two sides when it comes to the double slit apparatus which implies that the fringe would appear on two sides so the maximum number of bright spots is mathematically evaluated as
[tex]n_{max} = 2 * n + 1[/tex]
The 1 here represented the central bright spot
So
[tex]n_{max} = 2 * 2500 + 1[/tex]
[tex]n_{max} =5001[/tex]
A car moving at a speed of 25 m/s enters a curve that traces a circular quarter turn of radius 129 m. The driver gently applies the brakes, slowing the car with a constant tangential acceleration of magnitude 1.2 m/s2.a) Just before emerging from the turn, what is the magnitudeof the car's acceleration?
b) At that same moment, what is the angle q between the velocity vector and theacceleration vector?
I am having trouble because this problem seems to have bothradial and tangential accleration. I tried finding the velocityusing V^2/R, but then that didnt take into account thedeceleration. Any help would be great.
Answer:
8.7 m/s^2
82.15°
Explanation:
Given:-
- The initial speed of the car, vi = 25 m/s
- The radius of track, r = 129 m
- Car makes a circular " quarter turn "
- The constant tangential acceleration, at = 1.2 m/s^2
Solution:-
- We will solve the problem using rotational kinematics. Determine the initial angular velocity of car ( wi ) as follows:
[tex]w_i = \frac{v_i}{r} \\\\w_i = \frac{25}{129}\\\\w_i = 0.19379 \frac{rad}{s}[/tex]
- Now use the constant tangential acceleration ( at ) and determine the constant angular acceleration ( α ) for the rotational motion as follows:
at = r*α
α = ( 1.2 / 129 )
α = 0.00930 rad/s^2
- We know that the angular displacement from the initial entry to the exit of the turn is quarter of a turn. The angular displacement would be ( θ = π/2 ).
- Now we will use the third rotational kinematic equation of motion to determine the angular velocity at the exit of the turn (wf) as follows:
[tex]w_f^2 = w_i^2 + 2\alpha*theta\\\\w_f = \sqrt{0.19379^2 + 0.00930\pi } \\\\w_f = 0.25840 \frac{rad}{s}[/tex]
- We will use the evaluated final velocity ( wf ) and determine the corresponding velocity ( vf ) as follows:
[tex]v_f = r*w_f\\\\v_f = 129*0.2584\\\\v_f = 33.33380 \frac{x}{y}[/tex]
- Now use the formulation to determine the centripetal acceleration ( ac ) at this point as follows:
[tex]a_c = \frac{v_f^2}{r} \\\\a_c = \frac{33.3338^2}{129} \\\\a_c = 8.6135 \frac{m}{s^2}[/tex]
- To determine the magnitude of acceleration we will use find the resultant of the constant tangential acceleration ( at ) and the calculated centripetal acceleration at the exit of turn ( ac ) as follows:
[tex]|a| = \sqrt{a^2_t + a_c^2} \\\\|a| = \sqrt{1.2^2 + 8.6135^2} \\\\|a| = 8.7 \frac{m}{s^2}[/tex]
- To determine the angle between the velocity vector and the acceleration vector. We need to recall that the velocity vector only has one component and always tangential to the curved path. Hence, the velocity vector is parallel to the tangential acceleration vector ( at ). We can use the tangential acceleration ( at ) component of acceleration ( a ) and the centripetal acceleration ( ac ) component of the acceleration and apply trigonometric ratio as follows:
[tex]q = arctan \frac{a_c}{a_t} = arctan \frac{8.7}{1.2} \\\\q = 82.15 ^.[/tex]
Answer: The angle ( q ) between acceleration vector ( a ) and the velocity vector ( v ) at the exit of the turn is 82.15° .
Charge of uniform surface density (0.20 nC/m2) is distributed over the entire xy plane. Determine the magnitude of the electric field at any point having z
The question is not complete, the value of z is not given.
Assuming the value of z = 4.0m
Answer:
the magnitude of the electric field at any point having z(4.0 m) =
E = 5.65 N/C
Explanation:
given
σ(surface density) = 0.20 nC/m² = 0.20 × 10⁻⁹C/m²
z = 4.0 m
Recall
E =F/q (coulumb's law)
E = kQ/r²
σ = Q/A
A = 4πr²
∴ The electric field at point z =
E = σ/zε₀
E = 0.20 × 10⁻⁹C/m²/(4 × 8.85 × 10⁻¹²C²/N.m²)
E = 5.65 N/C
How much work will it take to lift a 2-kg pair of hiking boots 2 meters off the
ground and onto a shelf in your closet?
O A. 2.45 J
OB. 4J
C. 39.2 J
D. 20 J
Answer:
Option C - 39.2 J
Explanation:
We are given that;
Mass; m = 2 kg.
Distance moved off the floor;d = 10 m.
Acceleration due to gravity;g = 9.8 m/s².
We want to find the work done.
Now, the Formula for work done is given by;
Work = Force × displacement.
In this case, it's force of gravity to lift up the boots, thus;
Formula for this force is;
Force = mass x acceleration due to gravity
Force = 2 × 9.8 = 19.2 N
∴ Work done = 19.6 × 2
Work done = 39.2 J.
Hence, the Work done to life the boot of 2 kg to a height of 2 m is 39.2 J.
Answer:39.2J
Explanation: I just answered this question and this was the correct answer. 4J is the wrong answer.
a certain volume of dry air at NTP is allowed to expand five times of it original volume under adiabatic condition.calculate the final pressure.(air=1.4)
Answer:
Final pressure 0.105atm
Explanation:
Let V1 represent the initial volume of dry air at NTP.
under adiabatic condition: no heat is lost or gained by the system. This does not implies that the constant temperature throughout the system , but rather that no heat gained or loss by the system.
Adiabatic expansion:
[tex]\frac{T_1}{T_2} =(\frac{V_1}{V_2} )^{\gamma -1}[/tex]
273/T2=(5V1/V1)^(1.4−1)
273/T2=5^0.4
Final temperature T2=143.41 K
Also
P1/P2=(V2/V1)^γ
1/P2=(5V1/V1)^1.4
Final pressure P2=0.105atm
Two people play tug of war. The 100-kg person on the left pulls with 1,000 N, and the 70-kg person on the right pulls with 830 N. Assume that neither person releases their grip on the rope with either hand at any time, assume that the rope is always taut, and assume that the rope does not stretch. What is the magnitude of the tension in the rope in Newtons
Answer:
The tension on the rope is T = 900 N
Explanation:
From the question we are told that
The mass of the person on the left is [tex]m_l = 100 \ kg[/tex]
The force of the person on the left is [tex]F_l = 1000 \ N[/tex]
The mass of the person on the right is [tex]m_r = 70 \ kg[/tex]
The force of the person on the right is [tex]F_r = 830 \ N[/tex]
Generally the net force is mathematically represented as
[tex]F_{Net} = F_l - F_r[/tex]
substituting values
[tex]F_{Net} = 1000-830[/tex]
[tex]F_{Net} = 170 \ N[/tex]
Now the acceleration net acceleration of the rope is mathematically evaluated as
[tex]a = \frac{F_{net}}{m_I + m_r }[/tex]
substituting values
[tex]a = \frac{170}{100 + 70 }[/tex]
[tex]a = 1 \ m/s ^2[/tex]
The force [tex]m_i * a[/tex]) of the person on the left that caused the rope to accelerate by a is mathematically represented as
[tex]m_l * a = F_r -T[/tex]
Where T is the tension on the rope
substituting values
[tex]100 * 1 = 1000 - T[/tex]
=> T = 900 N
Two cannonballs are dropped from a second-floor physics lab at height h above the ground. Ball B has four times the mass of ball A. When the balls pass the bottom of a first-floor window at height above the ground, the relation between their kinetic energies, KA and KB, is
Answer:
1:4
Explanation:
The formula for calculating kinetic energy is:
[tex]KE=\dfrac{1}{2}mv^2[/tex]
If the mass is multiplied by 4, then, the kinetic energy must be increased by 4 as well. Since they will be travelling at the same speed when they are at the same point, the relation between KA and KB must be 1:4 or 1/4. Hope this helps!
The relation between the kinetic energies of the freely falling balls A and B is obtained as [tex]\frac{KE_{A}}{KE_{B}} =\frac{1}{4}[/tex].
Kinetic EnergyThe kinetic energy of an object depends on the mass and velocity with which it moves.
While under free-fall, the mass of an object does not affect the velocity with which it falls.
So, the velocities of both the balls are the same.
Let the mass of ball A is 'm'
So, the mass of ball B is '4m'
The kinetic energy of ball A is given by;
[tex]KE_{A}=\frac{1}{2} mv^2[/tex]
The kinetic energy of ball B is given by;
[tex]KE_{B}=\frac{1}{2} 4mv^2 = 2mv^2[/tex]
Therefore, the ratio of kinetic energies of A and B is,
[tex]\frac{KE_{A}}{KE_{B}} =\frac{1}{4}[/tex]
Learn more about kinetic energy here:
https://brainly.com/question/11580018
A cowboy fires a silver bullet with a muzzle speed of 200 m/s into the pine wall of a saloon. Assume all the internal energy generated by the impact remains with the bullet. What is the temperature change of the bullet?
Explanation:
KE = q
½ mv² = mCΔT
ΔT = v² / (2C)
ΔT = (200 m/s)² / (2 × 236 J/kg/°C)
ΔT = 84.7°C
This question involves the concepts of the law of conservation of energy.
The temperature change of the bullet is "84.38°C".
What is the Law of Conservation of Energy?According to the law of conservation of energy, total energy of the system must remain constant. Therefore, in this situation.
[tex]Kinetic\ energy\ of\ bullet\ before\ impact=heat\ absorbed\ in\ bullet\\\\\frac{1}{2}mv^2=mC\Delta T\\\\\Delta T = \frac{v^2}{2C}[/tex]
where,
ΔT = change in temperature of the bullet = ?C = specific heat capacity of silver = 237 J/kg°Cv = speed of bullet = 200 m/sTherefore,
[tex]\Delta T = \frac{(200\ m/s)^2}{2(237\ J/kg.^oC)}[/tex]
ΔT = 84.38°C
Learn more about the law of conservation of energy here:
https://brainly.com/question/20971995
#SPJ2
Rope BCA passes through a pulley at point C and supports a crate at point A. Rope segment CD supports the pulley and is attached to an eye anchor embedded in a wall. Rope segment BC creates an angle of ϕ = 51.0 ∘ with the floor and rope segment CD creates an angle θ with the horizontal. If both ropes BCA and CD can support a maximum tensile force Tmax = 120 lb , what is the maximum weight Wmax of the crate that the system can support? What is the
Answer:
Wmax = 63.65 ≈ 64 lb
Explanation:
Light in vacuum is incident on the surface of a glass slab. In the vacuum the beam makes an angle of 38.0° with the normal to the surface, while in the glass it makes an angle of 26.0° with the normal. What is the index of refraction of the glass?
Answer:
n_glass = 1.404
Explanation:
In order to calculate the index of refraction of the light you use the Snell's law, which is given by the following formula:
[tex]n_1sin\theta_1=n_2sin\theta_2[/tex] (1)
n1: index of refraction of vacuum = 1.00
θ1: angle of the incident light respect to normal of the surface = 38.0°
n2: index of refraction of glass = ?
θ2: angle of the refracted light in the glass respect to normal = 26.0°
You solve the equation (1) for n2 and replace the values of all parameters:
[tex]n_2=n_1\frac{sin\theta_1}{sin\theta_2}=(1.00)\frac{sin(38.0\°)}{sin(26.0\°)}\\\\n_2=1.404[/tex]
The index of refraction of the glass is 1.404
When a certain capacitor carries charge of magnitude Q on each of its plates, it stores energy Ep. In order to store twice as much energy, how much charge should it have on its plates
2Q
Explanation:
When a capacitor carries some certain charge, the energy stored in the capacitor is its electric potential energy E. The magnitude of this potential energy is given by;
E = [tex]\frac{1}{2}qV[/tex] ------------(i)
Where;
q = charge between the plates of the capacitor
V = potential difference between the plates of the capacitor
From the question;
q = Q
E = Ep
Therefore, equation (i) becomes;
Ep = [tex]\frac{1}{2} QV[/tex] ----------------(ii)
Make V subject of the formula in equation (ii)
V = [tex]\frac{2E_{p}}{Q}[/tex]
Now, when the energy is doubled i.e E = 2Ep, equation (i) becomes;
2Ep = [tex]\frac{1}{2}qV[/tex]
Substitute the value of V into the equation above;
2Ep = [tex]\frac{1}{2}[/tex]([tex]q *\frac{2E_{p}}{Q}[/tex])
Solve for q;
[tex]2E_{p}[/tex] = [tex]\frac{2qE_p}{2Q}[/tex]
[tex]2E_{p}[/tex] = [tex]\frac{qE_p}{Q}[/tex]
[tex]q = 2Q[/tex]
Therefore, the charge, when the energy stored is twice the originally stored energy, is twice the original charge. i.e 2Q
1) A net force of 75.5 N is applied horizontally to slide a 225 kg crate across the floor.
a. Compute the acceleration of the crate?
Answer:
The acceleration of the crate is [tex]0.3356\,\frac{m}{s^2}[/tex]
Explanation:
Recall the formula that relates force,mass and acceleration from newton's second law;
[tex]F=m\,a[/tex]
Then in our case, we know the force applied and we know the mass of the crate, so we can solve for the acceleration as shown below:
[tex]F=m\,a\\75.5\,N=225\,\,kg\,\,a\\a=\frac{75.5}{225} \,\frac{m}{s^2} \\a=0.3356\,\,\frac{m}{s^2}[/tex]
If 2 balls had the same volume but ball a has twice as much mass as babil which one will have the greater density
An inquisitive physics student and mountain climber climbs a 47.0-m-high cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.12 m/s.
(a) How long after release of the first stone do the two stones hit the water?
(b) What initial velocity must the second stone have if the two stones are to hit the water simultaneously?
magnitude =
(c) What is the speed of each stone at the instant the two stones hit the water?
first stone =
second stone =
Answer:
a) Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds, b) The initial velocity of the second stone is -16.038 meters per second, c) The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.
Explanation:
a) The time after the release after the release of the first stone can be get from the following kinematic formula for the first rock:
[tex]y_{1} = y_{1,o} + v_{1,o} \cdot t +\frac{1}{2}\cdot g \cdot t^{2}[/tex]
Where:
[tex]y_{1}[/tex] - Final height of the first stone, measured in meters.
[tex]y_{1,o}[/tex] - Initial height of the first stone, measured in meters.
[tex]v_{1,o}[/tex] - Initial speed of the first stone, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]g[/tex] - Gravity constant, measured in meters per square second.
Given that [tex]y_{1,o} = 47\,m[/tex], [tex]y_{1} = 0\,m[/tex], [tex]v_{1,o} = -2.12\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the following second-order polynomial is built:
[tex]-4.984\cdot t^{2} - 2.12\cdot t + 47 = 0[/tex]
Roots of the polynomial are, respectively:
[tex]t_{1} \approx 2.866\,s[/tex] and [tex]t_{2}\approx -3.291\,s[/tex]
Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds.
b) As the second stone is thrown a second later than first one, its height is represented by the following kinematic expression:
[tex]y_{2} = y_{2,o} + v_{2,o}\cdot (t-t_{o}) + \frac{1}{2}\cdot g \cdot (t-t_{o})^{2}[/tex]
[tex]y_{2}[/tex] - Final height of the second stone, measured in meters.
[tex]y_{2,o}[/tex] - Initial height of the second stone, measured in meters.
[tex]v_{2,o}[/tex] - Initial speed of the second stone, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]t_{o}[/tex] - Initial absolute time, measured in seconds.
[tex]g[/tex] - Gravity constant, measured in meters per square second.
Given that [tex]y_{2,o} = 47\,m[/tex], [tex]y_{2} = 0\,m[/tex], [tex]t_{o} = 1\,s[/tex], [tex]t = 2.866\,s[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the following expression is constructed and the initial speed of the second stone is:
[tex]1.866\cdot v_{2,o}+29.926 = 0[/tex]
[tex]v_{2,o} = -16.038\,\frac{m}{s}[/tex]
The initial velocity of the second stone is -16.038 meters per second.
c) The final speed of each stone is determined by the following expressions:
First stone
[tex]v_{1} = v_{1,o} + g \cdot t[/tex]
Second stone
[tex]v_{2} = v_{2,o} + g\cdot (t-t_{o})[/tex]
Where:
[tex]v_{1,o}, v_{1}[/tex] - Initial and final velocities of the first stone, measured in meters per second.
[tex]v_{2,o}, v_{2}[/tex] - Initial and final velocities of the second stone, measured in meters per second.
If [tex]v_{1,o} = -2.12\,\frac{m}{s}[/tex] and [tex]v_{2,o} = -16.038\,\frac{m}{s}[/tex], the final speeds of both stones are:
First stone
[tex]v_{1} = -2.12\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (2.866\,s)[/tex]
[tex]v_{1} = -30.227\,\frac{m}{s}[/tex]
Second stone
[tex]v_{2} = -16.038\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (2.866\,s-1\,s)[/tex]
[tex]v_{2} = -34.338\,\frac{m}{s}[/tex]
The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.
A particle with charge q is to be brought from far away to a point near an electric dipole. Net nonzero work is done if the final position of the particle is on:__________
A) any point on the line through the charges of the dipole, excluding the midpoint between the two charges.
B) any point on a line that is a perpendicular bisector to the line that separates the two charges.
C) a line that makes an angle of 30 ∘ with the dipole moment.
D) a line that makes an angle of 45 ∘with the dipole moment.
Answer:
Net nonzero work is done if the final position of the particle is on options A, C and D
Explanation:
non zero work is done if following will be the final position of the charges :
A) Any point on the line through the charges of the dipole , excluding the midpoint between the two charges.
C) A line that makes an angle 30° with the dipole moment.
D) A line that makes an angle 45° with the dipole moment.
An interference pattern is produced by light with a wavelength 550 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.500 mm .
a. If the slits are very narrow, what would be the angular position of the second- order, two-slit interference maxima?
b. Let the slits have a width 0.300 mm. In terms of the intensity lo at the center of the central maximum, what is the intensity at the angular position in part "a"?
Answer:
a
[tex]\theta = 0.0022 rad[/tex]
b
[tex]I = 0.000304 I_o[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 550 \ nm = 550 *10^{-9} \ m[/tex]
The distance of the slit separation is [tex]d = 0.500 \ mm = 5.0 *10^{-4} \ m[/tex]
Generally the condition for two slit interference is
[tex]dsin \theta = m \lambda[/tex]
Where m is the order which is given from the question as m = 2
=> [tex]\theta = sin ^{-1} [\frac{m \lambda}{d} ][/tex]
substituting values
[tex]\theta = 0.0022 rad[/tex]
Now on the second question
The distance of separation of the slit is
[tex]d = 0.300 \ mm = 3.0 *10^{-4} \ m[/tex]
The intensity at the the angular position in part "a" is mathematically evaluated as
[tex]I = I_o [\frac{sin \beta}{\beta} ]^2[/tex]
Where [tex]\beta[/tex] is mathematically evaluated as
[tex]\beta = \frac{\pi * d * sin(\theta )}{\lambda }[/tex]
substituting values
[tex]\beta = \frac{3.142 * 3*10^{-4} * sin(0.0022 )}{550 *10^{-9} }[/tex]
[tex]\beta = 0.06581[/tex]
So the intensity is
[tex]I = I_o [\frac{sin (0.06581)}{0.06581} ]^2[/tex]
[tex]I = 0.000304 I_o[/tex]
How do I find an apparent weight in N for a metal connected to a string submerged in water if a scale shows the mass 29.52 g when it is submerged ? Also how do I measure its density
The Tension of the string is going to be less when submerged in water by a value called the buoyancy force, so below in the attached file is explanation on how to calculate the apparent weight and density of the submerged object
Three solid, uniform, cylindrical flywheels, each of mass 65.0 kg and radius 1.47 m, rotate independently around a common axis through their centers. Two of the flywheels rotate in one direction at 8.94 rad/s, but the other one rotates in the opposite direction at 3.42 rad/s.
Required:
Calculate the magnitude of the net angular momentum of the system.
Answer:
the angular momentum is 1015.52 kg m²/s
Explanation:
given data
mass of each flywheel, m = 65 kg
radius of flywheel, r = 1.47 m
ω1 = 8.94 rad/s
ω2 = - 3.42 rad/s
to find out
magnitude of the net angular momentum
solution
we get here Moment of inertia that is express as
I = 0.5 m r² .................1
put here value and we get
I = 0.5 × 65 × 1.47 × 1.47
I = 70.23 kg m²
and
now we get here Angular momentum that is express as
L = I × ω ...........................2
and Net angular momentum will be
L = 2 × I x ω1 - I × ω2
put here value and we get
L = 2 × 70.23 × 8.94 - 70.23 × 3.42
L = 1015.52 kg m²/s
so
the angular momentum is 1015.52 kg m²/s
The magnitude of the net angular momentum of the system will be "1015.52 kg.m²/s".
MomentumAccording to the question,
Flywheel's mass, m = 65 kg
Flywheel's radius, r = 1.47 m
ω₁ = 8.94 rad/s
ω₂ = 3.42 rad/s
We know,
The moment of inertia (I),
= 0.5 m r²
By substituting the values,
= 0.5 × 65 × 1.47 × 1.47
= 70.23 kg.m²
hence, The angular momentum be:
→ L = I × ω or,
= 2 × I × ω₁ - l × ω₂
= 2 × 70.23 × 8.94 - 70.23 × 3.42
= 1015.52 kg.m²/s
Thus the above answer is correct.
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Four point charges have the same magnitude of 2.4×10^−12C and are fixed to the corners of a square that is 4.0 cm on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.
Answer:
7.2N/C
Explanation:
Pls see attached file
The ball tends to come back to the centerline of the flow when it is pushed by an external disturbance. Explain this phenomenon using the curvature of streamlines.
Answer is given below
Explanation:
given data
we will consider here
Ping-Pong ball weighs = 3.1 g
diameter = 4.2 cm
solution
Whenever the ball is pushed, the length of the airflow along the outer edge increases and it accelerates. According to Bernoulli's equation. As the speed increases, the pressure decreases, so the pressure at the outer end is reduced. As the pressure at the outer edge is low, the extra air jet pushes it back to the center line.
The first Leyden jar was probably discovered by a German clerk named E. Georg von Kleist. Because von Kleist was not a scientist and did not keep good records, the credit for the discovery of the Leyden jar usually goes to physicist Pieter Musschenbroek from Leyden, Holland. Musschenbroek accidentally discovered the Leyden jar when he tried to charge a jar of water and shocked himself by touching the wire on the inside of the jar while holding the jar on the outside. He said that the shock was no ordinary shock and his body shook violently as though he had been hit by lightning. The energy from the jar that passed through his body was probably around 1 J, and his jar probably had a capacitance of about 1 nF.A) Estimate the charge that passed through Musschenbroek's body.
B) What was the potential difference between the inside and outside of the Leyden jar before Musschenbroek discharged it?
Answer:
a) q = 4.47 10⁻⁵ C
b) ΔV = 4.47 10⁴ V
Explanation:
A Leyden bottle works as a condenser that accumulates electrical charge, so we can use the formula of the energy stored in a capacitor
U = Q² / 2C
Q = √ (2UC)
let's reduce the magnitudes to the SI system
c = 1 nF = 1 10⁻⁹ F
let's calculate
q = √ (2 1 10⁻⁹-9)
q = 0.447 10⁻⁴ C
q = 4.47 10⁻⁵ C
b) for the potential difference we use
C = Q / ΔV
ΔV = Q / C
ΔV = 4.47 10⁻⁵ / 1 10⁻⁹
ΔV = 4.47 10⁴ V
Two small identical speakers are connected (in phase) to the same source. The speakers are 3 m apart and at ear level. An observer stands at X, 4 m in front of one speaker. If the amplitudes are not changed, the sound he hears will be least intense if the wavelength is:
a. 1 m
b. 2 m
c. 3 m
d. 4 m
e. 5 m
Answer:
b. 2 m
Explanation:
Given that:
the identical speakers are connected in phases ;
Let assume ; we have speaker A and speaker B which are = 3 meter apart
An observer stands at X = 4m in front of one speaker.
If the amplitudes are not changed, the sound he hears will be least intense if the wavelength is:
From above; the distance between speaker A and speaker B can be expressed as:
[tex]\sqrt{3^2 + 4^2 } \\ \\ = \sqrt{9+16 } \\ \\ = \sqrt{25} \\ \\ = 5 \ m[/tex]
The path length difference will now be:
= 5 m - 4 m
= 1 m
Since , we are to determine the least intense sound; the destructive interference for that path length will be half the wavelength; which is
= [tex]\dfrac{1}{2}*4 \ m[/tex]
= 2 m
The sound will be heard with least intensity if the wavelength is 2 m. Hence, option (b) is correct.
Given data:
The distance between the speakers is, d = 3 m.
The distance between the observer and speaker is, s = 4 m.
The amplitude of sound wave is the vertical distance from the base to peak of wave. Since sound amplitudes are not changed in the given problem. Then the distance between speaker A and speaker B can be expressed as:
[tex]=\sqrt{3^{2}+4^{2}}\\\\=\sqrt{25}\\\\=5\;\rm m[/tex]
And the path length difference is,
= 5 m - 4 m
= 1 m
Since , we are to determine the least intense sound; the destructive
interference for that path length will be half the wavelength; which is
[tex]=\dfrac{1}{2} \times s\\\\=\dfrac{1}{2} \times 4[/tex]
= 2 m
Thus, we can conclude that the sound will be heard with least intensity if the wavelength is 2 m.
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A goalie kicks a soccer ball straight vertically into the air. It takes 5.00 s for the ball to reach its maximum height and come back down to the level of the crossbar. Assume the crossbar of a soccer goal is 2.44 m above the ground. (a) How fast was the ball originally moving when it was kicked. (b) How much longer would it take the ball to reach the ground?
Answer:
(a) vo = 24.98m/s
(b) t = 5.09 s
Explanation:
(a) In order to calculate the the initial speed of the ball, you use the following formula:
[tex]y=y_o+v_ot-\frac{1}{2}gt^2[/tex] (1)
y: vertical position of the ball = 2.44m
yo: initial vertical position = 0m
vo: initial speed of the ball = ?
g: gravitational acceleration = 9.8m/s²
t: time on which the ball is at 2.44m above the ground = 5.00s
You solve the equation (1) for vo and replace the values of the other parameters:
[tex]v_o=\frac{y-y_o+1/2gt^2}{t}[/tex]
[tex]v_o=\frac{2.44m-0.00m+1/2(9.8m/s^2)(5.00s)^2}{5.00s}\\\\v_o=24.98\frac{m}{s}[/tex]
The initial speed of the ball is 24.98m/s
(b) To find the time the ball takes to arrive to the ground you use the equation (1) for y = 0m (ground) and solve for t:
[tex]0=24.98t-\frac{1}{2}(9.8)t^2\\\\t=5.09s[/tex]
The time that the ball takes to arrive to the ground is 5.09s
We have that for the Question, it can be said that the speed of ball and How much longer would it take the ball to reach the ground is
u=25.13m/sX=0.095sec
From the question we are told
A goalie kicks a soccer ball straight vertically into the air. It takes 5.00 s for the ball to reach its maximum height and come back down to the level of the crossbar. Assume the crossbar of a soccer goal is 2.44 m above the ground.
(a) How fast was the ball originally moving when it was kicked.
(b) How much longer would it take the ball to reach the ground?
a)
Generally the Newton equation for the Motion is mathematically given as
[tex]S=ut+1/2at^2\\\\Therefore\\\\2.44=ut+1/2(9.8)(5)^2\\\\u=25.13m/s\\\\[/tex]
b)
Generally the Newton equation for the Motion is mathematically given as
[tex]S=ut+1/2at^2\\\\Therefore\\\\t=\frac{-24}{a}\\\\t=\frac{-2*25.013}{9.81}\\\\t=5.095sec\\\\[/tex]
Therefore
[tex]X=5.095-5[/tex]
X=0.095sec
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at the temperature at which we live, earth's core is solid or liquid?
Explanation:
The Earth has a solid inner core
Doubling the potential across a given capacitor causes the energy stored in that capacitor to reduce to:_______
a. one-half.
b. double.
c. reduce to one-fourth.
d. quadruple.
Answer:
D. quadrupleExplanation:
The stored energy varies with the square of the electric charge stored in the capacitor. If you double the charge, the stored energy in the capacitor will quadruple or increase by a factor of 4.
Doubling the potential across a given capacitor causes the energy stored in that capacitor to reduce to :
D. Quadruple
"Energy"Doubling the potential across a given capacitor causes the energy stored in that capacitor to reduce to Quadruple.
The stored energy shifts with the square of the electric charge put away within the capacitor.
In case you twofold the charge, the put away vitality within the capacitor will fourfold or increment by a calculate of 4.
Thus, the correct answer is D.
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An experiment is set up to test the angular resolution of an optical device when red light (wavelength ????r ) shines on an aperture of diameter D . Which aperture diameter gives the best resolution? D=(1/2)????r D=????r D=2????r
Explanation:
As per Rayleigh criterion, the angular resolution is given as follows:
[tex]\theta=\frac{1.22 \lambda}{D}[/tex]
From this expression larger the size of aperture, smaller will be the value of angular resolution and hence, better will be the device i.e. precision for distinguishing two points at very high angular difference is higher.
According to the model in which active galactic nuclei are powered by supermassive black holes, the high luminosity of an active galactic nucleus primarily consists of
Answer:
the high luminosity of an active galactic nucleus primarily consists of light emitted by hot gas in an accretion disk that swirls around the black hole