Explanation:
The revolutions of the Earth (also called translation movement), consist of the elliptical orbit that describes the Earth around the Sun.
In this sense, a complete revolution around the Sun occurs every 365 days, 5 hours, 48 minutes and about 46 seconds. It is thanks to this movement and that the Earth's axis is tilted with respect to the plane of its orbit about [tex]23\º[/tex], that the four seasons of the year exist.
For this reason, some regions receive different amounts of sunlight according to the seasons of the year. These variations are more evident near the poles and softer or imperceptible in the tropics (near the equator). Because near the equator the temperature tends to be more stable, with only two seasons: rain and drought.
The Earth's revolution around the Sun, taking approximately 365.24 days, determines the timing of seasons and the length of the year, and when viewed from the North Pole, the revolution is counterclockwise.
The revolution of the Earth around the Sun characterizes the Earth's journey through space as it orbits the Sun. This movement takes approximately 365.24 days, which equals one year.
The revolution is responsible for the timing of seasons and the length of the year. When observed from above the North Pole, the Earth's revolution around the Sun occurs counterclockwise, which is different from the rotation of the Earth on its own axis, the latter causing day and night cycles. Therefore, the correct answer to the student's question about what characterizes the Earth's revolution is (c) it determines the timing of seasons and the length of the year.
An infinitely long straight wire has a uniform linear charge density of Derive the 4. equation for the electric field a distance R away from the wire using Gauss's Law for Electrostatics.
Answer:
[tex]E = \frac{\lambda}{2\pi \epsilon_0 r}[/tex]
Explanation:
Let the linear charge density of the charged wire is given as
[tex]\frac{q}{L} = \lambda[/tex]
here we can use Gauss law to find the electric field at a distance r from wire
so here we will assume a Gaussian surface of cylinder shape around the wire
so we have
[tex]\int E. dA = \frac{q}{\epsilon_0}[/tex]
here we have
[tex]E \int dA = \frac{\lambda L}{\epsilon_0}[/tex]
[tex]E. 2\pi r L = \frac{\lambda L}{\epsilon_0}[/tex]
so we have
[tex]E = \frac{\lambda}{2\pi \epsilon_0 r}[/tex]
A bullet of mass 12 g strikes a ballistic pendulum of mass 2.2 kg. The center of mass of the pendulum rises a vertical distance of 10 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.
Answer:
The bullet's initial speed 258.06 m/s.
Explanation:
It is given that,
It is given that,
Mass of the bullet, m₁ = 12 g = 0.012 kg
Mass of the pendulum, m₂ = 2.2 kg
The center of mass of the pendulum rises a vertical distance i.e. h = 10 cm = 0.1 m
We need to find the initial speed of the bullet. Here, the kinetic energy of the bullet gets converted to potential energy of the system as :
[tex]\dfrac{1}{2}(m_1+m_2)V^2=(m_1+m_2)gh[/tex]
[tex]V=\sqrt{2gh}[/tex]....................(1)
Where
V is the speed of bullet and pendulum at the time of collision.
Let v be the initial speed of the bullet. Using conservation of momentum as :
[tex]m_1v=(m_1+m_2)V[/tex]
[tex]V=(\dfrac{m_1+m_2}{m_1})\sqrt{2gh}[/tex]
[tex]V=(\dfrac{0.012\ kg+2.2\ kg}{0.012 kg})\sqrt{2\times 9.8\ m.s^2\times 0.1\ m}[/tex]
V = 258.06 m/s
So, the initial speed of the bullet is 258.06 m/s. Hence, this is the required solution.
An unknown material, with a mass of 54 gm, was submerged in water. The object occupied 14 gm of water. What is the density of the material?
Answer:
Density of unknown material = 3.86 g/cm³
Explanation:
Mass of unknown material, M = 54 gm
The object occupied 14 gm of water.
Volume of unknown material = Volume of 14 gm of water.
Density of water = 1 gm/cm³
Volume of 14 gm of water
[tex]V=\frac{14}{1}=14cm^3[/tex]
Volume of unknown material, V = Volume of 14 gm of water = 14 cm³
Density of unknown material
[tex]\rho =\frac{M}{V}=\frac{54}{14}=3.86g/cm^3[/tex]
g The “size” of the atom in Rutherford’s model is about 8 × 10−11 m. Determine the attractive electrostatics force between a electron and a proton separated by this distance. Answer in units of N.
Answer:
[tex]3.6\cdot 10^{-8} N[/tex]
Explanation:
The electrostatic force between the proton and the electron is given by:
[tex]F=k\frac{q_p q_e}{r^2}[/tex]
where
[tex]k=9.00\cdot 10^9 Nm^2 C^{-2}[/tex] is the Coulomb constant
[tex]q_p = 1.6\cdot 10^{-19} C[/tex] is the magnitude of the charge of the proton
[tex]q_e = 1.6\cdot 10^{-19}C[/tex] is the magnitude of the charge of the electron
[tex]r=8\cdot 10^{-11}m[/tex] is the distance between the proton and the electrons
Substituting the values into the formula, we find
[tex]F=(9\cdot 10^9 ) \frac{(1.6\cdot 10^{-19})^2}{(8\cdot 10^{-11})^2}=3.6\cdot 10^{-8} N[/tex]
Points A and B lie above an infinite plane of negative charge that produces a uniform electric field of strength E=10 N/C. What is the potential difference ΔV = VB - VA between points A and B if the points are separated by a distance of 3m as illustrated below?
There is no illustration of the problem provided but I'll attempt to provide an answer.
The relationship between the electric potential difference between two points and the average strength of the electric field between those two points is given by:
║E║ = ΔV/d
║E║ is the magnitude of the average electric field, ΔV is the potential difference between A and B, and d is the distance between A and B.
We are given the following values:
║E║= 10N/C
d = 3m
Plug these values in and solve for ΔV
10 = ΔV/3
ΔV = 30V
If a pendulum has a period of 4 seconds at the north pole with gravity force 9.83 ms^-2 and a 3.97m pendulum length, how could the same pendulum be adjusted to have the same period as the equator with 9.78ms^-2 gravity force?
Answer:
The same pendulum could be adjusted to have the same period, in the equator must have a length of 3.949m.
Explanation:
Tnp= 4 sec
gnp= 9.83 m/sec²
Lnp= 3.97m
Tequ= 4 sec
gequ= 9.78 m/sec²
Lequ=?
Lequ= (Lnp* gequ) / gnp
Lequ= 3.949 m
A 16.0 m long, thin, uniform metal rod slides north at a speed of 21.0 m/s. The length of the rod maintains an east-west orientation while sliding. The vertical component of the Earth's magnetic field at this location has a magnitude of 42.0 µT. What is the magnitude of the induced emf between the ends of the rod (in mV)?
Answer:
14.112 mV
Explanation:
L = 16 m, v = 21 m/s, B = 42 μ T = 42 x 10^-6 T
The formula for the induced emf is given by
e = B x v x L
e = 42 x 10^-6 x 21 x 16 = 14.112 x 10^-3 V = 14.112 mV
Thus, the induce emf is 14.112 mV.
Determine the capacitive reactance for a 20 uF capacitor that is across a 20 volt, 60 Hz source
Answer:
Capacitive reactance is 132.6 Ω.
Explanation:
It is given that,
Capacitance, [tex]C=20\ \mu F=20\times 10^{-6}\ F=2\times 10^{-5}\ F[/tex]
Voltage source, V = 20 volt
Frequency of source, f = 60 Hz
We need to find the capacitive reactance. It is defined as the reactance for a capacitor. It is given by :
[tex]X_C=\dfrac{1}{2\pi fC}[/tex]
[tex]X_C=\dfrac{1}{2\pi \times 60\ Hz\times 2\times 10^{-5}\ F}[/tex]
[tex]X_C=132.6\ \Omega[/tex]
So, the capacitive reactance of the capacitor is 132.6 Ω. Hence, this is the required solution.
A ski tow operates on a slope of angle 15.6 ∘ of length 340 m . The rope moves at a speed of 12.7 km/h and provides power for 53 riders at one time, with an average mass per rider of 69.0 kg .
Estimate the power required to operate the tow
Answer:
P = 34034.2 Watt
Explanation:
As we know that the slope angle is given as
[tex]\theta = 15.6^0[/tex]
now the weight of the rider along the slope is given as
[tex]W = mgsin\theta[/tex]
[tex]W = 69(9.81)sin15.6[/tex]
[tex]W = 182 N[/tex]
now total weight of all 53 riders along the slope is given as
[tex]F = 53 W[/tex]
[tex]F = 9647.5 N[/tex]
now the speed of the rope is given as
[tex]v = 12.7 km/h = 3.53 m/s[/tex]
now the power required is given as
[tex]P = F.v[/tex]
[tex]P = 9647.5(3.53) [/tex]
[tex]P = 34034.2 Watt[/tex]
The power required to operate the tow is 34020.45 W.
Weight of a single riderThe weight of a single rider is calculated as follows;
[tex]W = mgsin(\theta)\\\\W = (69 \times 9.8) sin(15.6)\\\\W = 181.84 \ N[/tex]
Weight of the 53 ridersW = 53 x 181.84
W = 9637.52 N
Power required to operate the towThe power required to operate the tow is calculated as follows;
P = Fv
where;
F is the upward force due to weight of the riders = 9637.52 Nv is speed of the rope = 12.7 km/h = 3.53 m/sP = 9637.52 x 3.53 = 34020.45 W
Learn more about power here: https://brainly.com/question/25263760
Give a calculate answer to show that the two values (English system and metric system) for the Planck Constant are equivalent.
h=4.14*10-15 (eV-sec) or 6.63*10-34( Joule-sec)
Answer:
Given values of Planck Constant are equivalent in English system and metric system.
Explanation:
Value of Planck's constant is given in English system as 4.14 x 10⁻¹⁵eV s.
Converting this in to metric system .
We have 1 eV = 1.6 x 10⁻¹⁹ J
Converting
4.14 x 10⁻¹⁵eV s = 4.14 x 10⁻¹⁵x 1.6 x 10⁻¹⁹ = 6.63 x 10⁻³⁴ Joule s
So Given values of Planck Constant are equivalent in English system and metric system.
A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.2 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall? (That is, find the angle's rate of change when the bottom of the ladder is 8 ft from the wall.) rad/s
Answer:0.2 rad/s
Explanation:
Given data
Velocity of the bottom point of the ladder=1.2Ft/s
Length of ladder=10ft
distance of the bottom most point of ladder from origin=8ft
From the data the angle θ with ladder makes with horizontal surface is
Cosθ=[tex]\frac{8}{10}[/tex]
θ=36.86≈37°
We have to find rate of change of θ
From figure we can say that
[tex]x^{2}[/tex]+[tex]y^{2}[/tex]=[tex]AB^{2}[/tex]
Differentiating above equation we get
[tex]\frac{dx}{dt}[/tex]=-[tex]\frac{dy}{dt}[/tex]
i.e [tex]{V_A}=-{V_B}=1.2ft/s[/tex]
[tex]{at\theta}={37}[/tex]
[tex]Y=6ft[/tex]
[tex]and\ about\ Instantaneous\ centre\ of\ rotation[/tex]
[tex]{\omega r_A}={V_A}[/tex]
[tex]{\omega=\frac{1.2}{6}[/tex]
ω=0.2rad/s
i.e.Rate of change of angle=0.2 rad/s
A person exerts a force of 50 N on the end of a door 71 cm wide. (a)What is the magnitude of the torque if the force is exerted perpendicular to the door? (b) What is the magnitude of the torque if the force is exerted at a 45 degree angle to the face of the door?
Answer:
a) 35.5
b)25.1
Explanation:
Torque=distance* (perpendicular component of the force to the distance)
a) force=50 N distance=0,71m(we need the distance in meters to have the same units with the Newton)
T=50*0.75
b) only one component of the force makes torque, so we multiply all the 50N with sin(45) to find the perpendicular component to the distance
T=50*sin(45)*0.71
Find the heat that flows in 1.0 s through a lead brick 13 cm long if the temperature difference between the ends of the brick is 8.0°C. The cross-sectional area of the brick is 13 cm^2. Express your answer using two significant figures.
Final answer:
The heat flow through the lead brick can be calculated using Fourier's law of heat conduction. The thermal conductivity of lead is used along with the provided dimensions and temperature difference to find the heat flow, which is 2.2 W to two significant figures.
Explanation:
To calculate the heat flow through a lead brick, we can use Fourier's law of heat conduction, which states that the heat transfer rate (Q) through a material is proportional to the thermal conductivity of the material (k), the area (A) through which the heat is being transferred, the temperature difference (ΔT) across the material, and inversely proportional to the thickness (d) of the material.
The formula to calculate heat flow is: Q = k × A × ΔT × t / d
First, we need to convert all measurements to SI units: the length from cm to m (13 cm = 0.13 m), the area from cm² to m² (13 cm² = 0.0013 m²), and the time from seconds to hours (1.0 s). The thermal conductivity of lead is approximately 35 W/m°C. Plugging in the values: Q = 35 W/m°C × 0.0013 m² × 8.0°C × 1.0 s / 0.13 m
By simplifying the equation, we find the heat flow through the lead brick:
Q = (35 × 0.0013 × 8.0 × 1.0) / 0.13 W = 2.215 W
To two significant figures, the heat flow is 2.2 W.
In a two-slit experiment, monochromatic coherent light of wavelength 500 nm passes through a pair of slits separated by 1.30 x 10^-5 m. At what angle away from the centerline does the first bright fringe occur? O 1.56° O 2.20° O 3.85° O 2.73° O 4.40°
Answer:
2.20°
Explanation:
λ = wavelength of the coherent light = 500 x 10⁻⁹ m
d = slits separation = 1.30 x 10⁻⁵ m
n = order of the fringe = 1
θ = angle made by the first bright fringe with the center line = ?
For first bright fringe, Using the equation
d Sinθ = n λ
(1.30 x 10⁻⁵) Sinθ = (1) (500 x 10⁻⁹)
[tex]Sin\theta =\frac{500\times 10^{-9}}{1.30\times 10^{-5}}[/tex]
Sinθ = 0.0385
θ = 2.20°
Final answer:
To find the angle of the first bright fringe in a double-slit experiment, the formula for constructive interference is used. By substituting the provided measurements into the formula and calculating the inverse sine, the angle is found to be approximately 2.20°.
Explanation:
The question refers to a Young's double-slit experiment, where monochromatic light of a known wavelength is used to produce an interference pattern on a screen. To find the angle at which the first bright fringe occurs, we can use the formula for constructive interference in a double-slit setup, which is given by:
d sin(θ) = mλ
Where d is the separation between the slits, θ is the angle of the fringe from the centerline, m is the order number of the fringe (m=1 for the first bright fringe), and λ is the wavelength of the light.
Plugging in the values provided:
d = 1.30 x 10-5 m
λ = 500 x 10-9 m
m = 1
Now we solve for θ:
sin(θ) = mλ / d
sin(θ) = (1)(500 x 10-9 m) / (1.30 x 10-5 m)
sin(θ) ≈ 0.038462
Using a calculator to find the inverse sine, we get:
θ ≈ 2.20°
Liquid sodium can be used as a heat transfer fluid in some nuclear reactors due to its high thermal conductivity and low neutron absorptivity. Its vapor pressure is 40.0 torr at 633 degree C and an enthalpy of vaporization of 1.00x10^2 kJ/mol. To what temperature can it be heated if a maximum of 500 torr pressure is allowed in the system?
Answer:
1119.1 K
Explanation:
From Clausius-Clapeyron equation:
[tex]\frac{dP}{dT}=[/tex]Δ[tex]\frac{h_{v} }{R} (\frac{1}{T^{2} } )dT[/tex]
The equation may be integrated considering the enthalpy of vaporization constant, and its result is:
[tex]ln(\frac{P_{2} }{P_{1} } )=-[/tex]Δ[tex]\frac{h_{v} }{R}*(\frac{1}{T_{2} }-\frac{1}{T_{1} })[/tex]
Isolating the temperature [tex]T_{2}[/tex]
[tex]T_{2}=\frac{1}{\frac{-R}{dhv}*ln(\frac{P_{2} }{P_{1}}) +\frac{1}{T_{1}} }[/tex]
[tex]T_{2}=\frac{1}{\frac{-8.314}{1.00*10^5}*ln(\frac{500}{40}) +\frac{1}{906.15}}[/tex]
[tex]T_{2}=1119.1K[/tex]
Note: Remember to change the units of the enthalpy vaporization to J/mol; and the temperatures must be in Kelvin units.
There is a format mistake with the enthalpy of vaporization, each 'Δ' correspond to that.
To make a bounce pass, a player throws a 0.60-kg basketball toward the floor. The ball hits the floor with a speed of 5.4 m/s at an angle of 65° to the vertical. If the ball rebounds with the same speed and angle, what was the impulse delivered to it by the floor?
The impulse delivered to the ball by the floor after rebounding with the same speed and angle is; Δp = 2.74 N.s
We are given;
Mass of basketball; m = 0.6 kg
Speed; v = 5.4 m/s
Angle to the vertical; θ = 65°
We want to find the impulse if the ball rebounds with the same speed and angle above.
Now, as the x-component of the momentum remains constant, the impulse would be equal to the change in the y-component of the balls' momentum.
Thus;
Δp = m[(v_f * cos θ) - (v_i * cos θ)]
Earlier we saw that we were given the rebound speed to be the same with the initial speed but however, as the y-axis is pointing upwards, it means that the initial velocity will be negative.
Thus;
v_f = 5.4 m/s
v_i = -5.4 m/s
Plugging in the relevant values into the impulse equation gives;
Δp = 0.6[(5.4 * cos 65) - (-5.4 * cos 65)]
Δp = (0.6 × cos 65) × (5.4 + 5.4)
Δp = 2.74 N.s
Read more at; https://brainly.com/question/6815023
The impulse delivered to the ball by the floor after the rebound is 5.87 Ns.
The given parameters;
mass of the ball, m = 0.6 kgspeed of the ball, v = 5.4 m/sdirection of the ball, = 65° to the vertical.The impulse delivered to the ball is the change in the linear momentum of the ball in vertical direction.
The magnitude of the impulse delivered to the ball by the floor is calculated as;
[tex]J = \Delta P = m(v_f - v_0)\\\\J = m(vsin\theta - vsin\theta)\\\\the \ final \ speed \ occured \ in \ opposite \ direction;\\\\J = m(vsin\theta + vsin\theta)\\\\J = m(2vsin\theta)\\\\J = 2mv \times sin(\theta)\\\\J = 2\times 0.6 \times 5.4 \times sin(65)\\\\J = 5.87 \ Ns[/tex]
Thus, the impulse delivered to the ball by the floor after the rebound is 5.87 Ns.
Learn more here:https://brainly.com/question/11483875
Near Earth, the density of protons in the solar wind (a stream of particles from the Sun) is 11.6 cm-3, and their speed is 563 km/s. (a) Find the current density of these protons. (b) If Earth's magnetic field did not deflect the protons, what total current Earth receive?
Answer:
a) 10462341.6×10⁻¹³ A/m²
b) 133709238.907 A
Explanation:
n = Density of the protons in the solar wind = 11.6 cm⁻³ = 11.6×10⁶ m⁻³
v = Velocity of the protons = 563 km/s = 563000 m/s
e = Charge of a proton = 1.602×10⁻¹⁹ coulombs
R = Radius of Earth = 6.3781×10⁶ m
A = Area of Earth = πR² = π(6.3781×10⁶)²=127.8×10¹² m²
a) Current density
J = nev
⇒J = 11.6×10⁶×1.602×10⁻¹⁹×563000
⇒J = 10462341.6×10⁻¹³ A/m²
∴ Current density of these protons is 10462341.6×10⁻¹³ A/m²
b) Current
I = JA
⇒I = 10462341.6×10⁻¹³×127.8×10¹²
⇒I = 1337092389.07×10⁻¹
⇒I = 133709238.907 A
∴ Total current Earth receives is 133709238.907 A
When turned on, a fan requires 5.0 seconds to get up to its final operating rotational speed of 1200 rpm. a) How large is the final operating angular speed, in rad/s? b) What was the angular acceleration of the fan, in rad/s^2?
Answer:
a)
125.6 rad/s
b)
25.12 rad/s²
Explanation:
a)
t = time required by the fan to get up to final operating speed = 5 sec
w = final operating rotational speed = 1200 rpm
we know that :
1 revolution = 2π rad
1 min = 60 sec
w = [tex]1200\frac{rev}{min}\frac{2\pi rad}{1 rev}\frac{1 min}{60 sec}[/tex]
w = [tex]\frac{1200\times 2\pi }{60}\frac{rad}{s}[/tex]
w = 125.6 rad/s
b)
w₀ = initial angular speed = 0 rad/s
α = angular acceleration
using the equation
w = w₀ + α t
125.6 = 0 + α (5)
α = 25.12 rad/s²
The _____ is used to express absolute temperatures in the English system of measurement. A. Fahrenheit scale B. Kelvin scale C. Celsius scale D. Rankine scale
Answer:
B). Kelvin Scale
Explanation:
Kelvin scale is the absolute scale which is used to express temperature in English System
We have different temperature scales
1) Fahrenheit Scale
Generally use in British system of units
2) Celcius Scale
It is used to given temperature of different scales and its relation with kelvin
[tex]^0 C = K - 273[/tex]
3) Rankine Scale
It is used in thermodynamic scales with large temperature range
A large sphere of material is placed in a dissolving liquid. (a) When the radius is 1 m, it is observed to be changing at 8 cm/hour. At that instant, what is the rate at which the sphere's volume is being dissolved with respect to time?
Final answer:
To determine the rate at which the sphere's volume is dissolving, differentiate the volume formula with respect to time and substitute the given rate of the radius change. The dissolution rate at the instant when the radius is 1 meter is 4π*0.08 m^3/hour.
Explanation:
To find the rate at which the sphere's volume is being dissolved with respect to time, we can apply the formula for the volume of a sphere, V = &frac43;πr^3, and differentiate it with respect to time (t). Since the radius (r) of the sphere is changing, we have:
dV/dt = d(&frac43;πr^3)/dt = 4πr^2*(dr/dt)
Given that the radius is shrinking at a rate of 8 cm/hour (which we need to convert to meters since our volume will be in cubic meters), the radius change in m/hour is:
dr/dt = 0.08 m/hour
At the instant when r is 1 meter, plugging the values into the equation gives us:
dV/dt = 4π(1 m)^2*(0.08 m/hour) = 4π*0.08 m^3/hour
Therefore, the rate at which the sphere's volume is being dissolved is 4π*0.08 m^3/hour.
The Eiffel tower is made of iron and is about 300 m tall. Due to thermal expansion, its height changes between winter and summer. Estimate the height change the tower undergoes if the difference in temperature between the seasons is 23º C. Enter your answer in meters but do not include units. Provide your answer to the hundredths place.
Answer:
Change in length is 8.3 cm
Explanation:
Coefficient of thermal expansion for iron is given as
[tex]\alpha = 12 \times 10^{-6} per ^0C[/tex]
now as we know that the change in the length due to thermal expansion depends of change in temperature and initial length of the object
so here we will have
[tex]\Delta L = L_o \alpha \Delta T[/tex]
here we know that
[tex]L_o = 300 m[/tex]
[tex]\Delta T = 23^o C[/tex]
now we will have
[tex]\Delta L = (300)(12 \times 10^{-6})(23)[/tex]
[tex]\Delta L = 0.083 m[/tex]
so change in length is approx 8.3 cm
If a heat engine pulls 4390.0 J of heat from the hot reservoir and exhausts 3582.2 J of heat to the cold reservoir, what was the work done by the engine? What is the efficiency of this engine in percent? (do not enter the percent sign)
Answer:
18.4
Explanation:
Q1 = 4390 J
Q2 = 3582.2 J
The efficiency of heat engine is given by
n = 1 - Q2 / Q1
n = 1 - 3582.2 / 4390
n = 0.184
n = 18.4 %
An ac source is connected to a resistor R = 75Ω, an inductor L = 0.01 H, and a capacitor C = 4 μF. What is the phase difference φ between the voltage of the source and the current in the circuit when the frequency of the source is equal to half the resonance frequency of the circuit?
Answer:
135°.
Explanation:
R = 75 ohm, L = 0.01 H, C = 4 micro F = 4 x 10^-6 F
Frequency is equal to the half of resonant frequency.
Let f0 be the resonant frequency.
[tex]f_{0}=\frac{1}{2\pi \sqrt{LC}}[/tex]
[tex]f_{0}=\frac{1}{2\times 3.14 \sqrt{0.01\times 4\times 10^{-6}}}[/tex]
f0 = 796.2 Hz
f = f0 / 2 = 398.1 Hz
So, XL = 2 x 3.14 x f x L = 2 x 3.14 x 398.1 x 0.01 = 25 ohm
[tex]X_{c}=\frac{1}{2\pi fC}[/tex]
Xc = 100 ohm
[tex]tan\phi = \frac{X_{L}-_{X_{C}}}{R}[/tex]
tan Ф = (25 - 100) / 75 = - 1
Ф = 135°
Thus, the phase difference is 135°.
An alpha particle, which has charge 3.20 ✕ 10−19 C, is moved from point A, where the electric potential is 2.40 ✕ 103 J/C, to point B, where the electric potential is 4.95 ✕ 103 J/C. Calculate the work in electron volts done by the electric field on the alpha particle.
Answer:
5100 eV
Explanation:
q = 3.2 x 10^-19 C
Va = 2.4 x 10^3 J/C
Vb = 4.95 x 10^3 J/C
Work done = q (Vb - Va)
W = 3.2 x 10^-19 x (4.95 - 2.4 ) x 10^3 = 8.16 x 10^-16 J
A we know that 1 eV = 1.6 x 10^-19 J
So, W = (8.16 x 10^-16) / (1.6 x 10^-19) = 5100 eV
An electron moving with a velocity = 5.0 × 107 m/s enters a region of space where perpendicular electric and a magnetic fields are present. The electric field is = - 104 . What magnetic field will allow the electron to go through, undeflected?
Answer:
magnetic field will allow the electron to go through 2 x [tex]10^{4}[/tex] T k
Explanation:
Given data in question
velocity = 5.0 × [tex]10^{7}[/tex]
electric filed = [tex]10^{4}[/tex]
To find out
what magnetic field will allow the electron to go through, undeflected
solution
we know if electron move without deflection i.e. net force is zero on electron and we can say both electric and magnetic force equal in magnitude and opposite in directions
so we can also say
F(net) = Fe + Fb i.e. = 0
q V B + q E = 0
q will be cancel out
[tex]10^{4}j[/tex] + 5e + 7i × B = 0
B = 2 x [tex]10^{4}[/tex] T k
In a hydraulic garage lift, the small piston has a radius of 5.0 cm and the large piston has a radius of 1.50 m. What force must be to the small piston in order to lift a Zamboni weighing 200,000 N on the large piston?
Answer:
222.22 N
Explanation:
r = 5 cm = 0.05 m, R = 1.5 m, f = ? , F = 200,000 N
Use pascal's law
F / A = f / a
F / 3.14 x R^2 = f / 3.14 x r^2
F / R^2 = f / r^2
f = F x r^2 / R^2
f = 200000 x 0.05 x 0.05 / (1.5 x 1.5)
f = 222.22 N
What is the capacitance of a parallel place capacitor with plates of area 1.0 cm2 that are separated by a distance of 0.1 mm? Assume that strontium titanate is the dielectric.
C = 0.274μF.
The capacitance when a dielectric is present is given by the equation:
C = kε₀(A/d)
Where k is the dielectric constant of the material, ε₀ is a universal constant, A is the area of the capacitor's plates, and d is the distance that separate the plates.
A parallel plates capacitor with plates of area 1.0cm² that are separated by a distance of 0.1mm, the dielectric constant of the stronium titanate is 310 at a temperature of 20 ° C.
K = 310.00, ε₀ = 8.85x10⁻¹²F/m, A = 1.00x10⁻²m², and d = 0.10x10⁻³m. Substituing the values in the equation C = kε₀(A/d):
C = (310.00)(8.85x10⁻¹²F/m)(1.00x10⁻²m²/0.10x10⁻³m)
C = (2.74x10⁻⁹F/m)(100m)
C = 2.74x10⁻⁷F
C = 0.274x10⁻⁶ = 0.274μF
A 22g bullet traveling 210 m/s penetrates a 2.0kg block of wood and emerges going 150m/s. If the block were stationary on a frictionless plane before the collision, what is the velocity of the block after the bullet passes through? What is the kinetic energy loss of the system?
Answer:
speed of block is 0.665 m/s
Kinetic energy loss is 237.16 J
Explanation:
Here we can use momentum conservation as there is no external force on the system in horizontal direction
so here we will have
[tex]m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]
[tex]0.022(210) + 2(0) = 0.022(150) + 2 v[/tex]
[tex]4.62 = 3.3 = 2 v[/tex]
[tex]v = \frac{4.63 - 3.3}{2}[/tex]
[tex]v = 0.665 m/s[/tex]
Now kinetic energy loss in the system is given as
[tex]Loss = KE_i - KE_f[/tex]
[tex]Loss = \frac{1}{2}(0.022)(210^2) - (\frac{1}{2}(0.022)(150^2) + \frac{1}{2}(2)(0.665)^2)[/tex]
[tex]Loss = 237.16 J[/tex]
The block's final velocity after the bullet passes through it is about 1.68 m/s. The kinetic energy loss of the system is about 602 Joules. These conclusions are drawn using the conservation of momentum and energy.
Explanation:This problem can be approached using the principles of conservation of momentum and energy. Let's first focus on momentum. Before the bullet hits, the momentum of the system is only contributed by the bullet. After the bullet passes through the block, both the bullet and the block contribute to the total momentum, which should be the same as before. This gives us the equation: 0.022 kg * 210 m/s (bullet's initial momentum) = 0.022 kg * 150 m/s (bullet's final momentum) + 2.0 kg * v (final momentum of the block). Solving this for the block's final velocity, v, will yield about 1.68 m/s.
Next, let's examine the kinetic energy loss. Before the collision, kinetic energy is only contributed by the bullet: KE_initial = 0.5 * 0.022 kg * (210 m/s)². After, it comes from both the bullet and the block: KE_final = 0.5 * 0.022 kg * (150 m/s)² + 0.5 * 2.0 kg * (1.68 m/s)². The kinetic energy loss is KE_initial - KE_final, which should be about 602 Joules. It's worth mentioning that this energy loss is transferred to other forms like heat and sound, illustrating the law of conservation of energy.
Learn more about conservation of momentum and energy here:https://brainly.com/question/35479656
#SPJ3
A 50-g ice cube at 0ºC is submerged into a container of liquid nitrogen. How many kg of nitrogen evaporates if it is at its boiling point of 77K and has a latent heat of vaporization of 200 kJ/kg? The specific heat of ice is 2100 J/(kg ºC).
Answer:
0.1029 Kg
Explanation:
[tex]Given\ data[/tex]
[tex]mass\of \ice\ cube =50gm[/tex]
[tex]Latent\ heat \of\ vapourization\ of nitrogen(L) =200kJ/kg[/tex]
[tex]specific\ heat\ of\ ice=2100 J/(kgc)[/tex]
[tex]boiling\ point\ of\ nitrogen=77K\approx-169^{\circ}c[/tex]
[tex]now\ Equating\ heat \ absorb\ by \ ice\ from\ liquid\ nitrogen [/tex]
[tex]{m_{ice}}c[{\Delta T}]={m_{nitrogen}}L[/tex]
[tex]{m_{ice}}2100[0-(-196)}=m_{nitrogen}{200\times 1000}[/tex]
[tex]m_{nitrogen}=0.1029Kg[/tex]
A satellite with a mass of 2,000 ㎏ is inserted into an orbit that is twice the Earth' s radius. W hat is the force of gravity on the satellite? The radius of the Earth is 6380 km, and the mass of the Earth is 5.98 X 1024 kg a. 2,000 N h. 4,900 N 19,600 N d 24,500 N e 36.200 N l Space Station which orbits Earth, astronauts experience apparenst
Answer:
option (b) 4900 N
Explanation:
m = 2000 kg, R = 6380 km = 6380 x 10^3 m, Me = 5.98 x 10^24 kg, h = R
F = G Me x m / (R + h)^2
F = G Me x m / 2R^2
F = 6.67 x 10^-11 x 5.98 x 10^24 x 2000 / (2 x 6380 x 10^3)^2
F = 4900 N