Answer:
b. sees the bow (front) light flash before the aft (rear) light.
Explanation:
Galactic Explorer is the mission in which we explore about the universe. The spaceship signal has set its bright signal lights off at the end of two ends of the spaceship. The General Kay has concluded that when lights flashed are in her reference frame it will be midpoint of ship. This can happen when Captain Ray, sees bow light flash before aft light at the center of Galactic Explorer.
A parallel-plate air capacitor is made from two plates 0.200 m square, spaced 0.900 cm apart. It is connected to a 140 V battery.
A. What is the capacitance?
B. What is the charge on each plate?
C. What is the electric field between the plates?
D. What is the energy stored in the capacitor?
E. If the battery is disconnected and then the plates are pulled apart to a separation of 1.40 cm , what are the answers to parts A, B, C, and D?
Enter your answer as four numbers corresponding to C, Q, E, U. Please enter the answer in the given order and in the same units as in parts A, B, C, and D.
Answer:
See detailed solution below
Explanation:
a) From C= εoεrA/d
Where;
C= capacitance of the capacitor
εo= permittivity of free space
εr= relative permittivity
A= cross sectional area
d= distance between the plates
Since the relative permittivity of air=1 and permittivity of free space = 8.85 × 10^−12 Fm−1
Then;
C= 8.85 × 10^−12 Fm−1 × 0.2m^2/0.009 m
C= 196.67 × 10^-12 F or 1.967 ×10^-10 F
b) Q= CV = 1.967 ×10^-10 F × 140 V = 2.75 × 10^-8 C
c) E= V/d = 140 V/0.009m = 15.56 Vm-1
d) W= 1/2 CV^2 = 1/2 × 1.967 ×10^-10 F × (140)^2 =1.93×10^-6J
Part II
When the distance is now 0.014 m
a) C= 8.85 × 10^−12 Fm−1 × 0.2m^2/0.014 m = 1.26×10^-10 F
b) W= 1/2 Q^2/C = 1/2 × ( 2.75 × 10^-8 C)^2 / 1.26×10^-10= 3×10^-6 J
Note that the voltage changes when the distance is changed but the charge remains the same
A speed skater moving to the left across frictionless ice at 8.6 m/s hits a 4.7-m-wide patch of rough ice. She slows steadily, then continues on at 5.3 m/s.
What is her acceleration on the rough ice?
Answer:
a = 4.88 m / s²
Explanation:
We can solve this exercise using the expressions of kinematics in one dimension
v² = v₀² - 2a x
where v is the velocity, v₀ is the initial velocity, at acceleration and ax is the distance, the negative sign is because the velocity decreases because
a = (v₀² - v²) / 2x
let's calculate
a = (8.6² - 5.3²) / (2 4.7)
a = 4.88 m / s²
Match each term to the best description.
1. Composed of numerous narrowly spaced parallel slits or grooves
2. Having the same wavelength, frequency, and in-phase Interaction of waves where they meet in space
3. The bending of waves near a boundary or as a wave passes through an opening
4. The zeroth order direct reflection fringe
a. Coherent
b. Diffraction
c. Grating
d. Specular dot
e. Interference
Answer:
1. Grating
2. Interference
3. Diffraction
4. Specular dot
Explanation:
1. Composed of numerous narrowly spaced slits and grooves ........ Grating
2. Having the same wavelength, frequency, and in-phase Interaction of waves where they meet in space ....... Interference
3. The bending of waves near a boundary or as a wave passes through an opening ...... Diffraction
4. The zeroth order direct reflection fringe ...... Specular dot
The time constant of a simple RL circuit is defined as _______. We say that R is the resistance of the circuit and L is the inductance of the circuit.
Answer:
The correct answer will be "[tex]\tau =\frac{L}{R}[/tex]".
Explanation:
The time it would take again for current or electricity flows throughout the circuit including its LR modules can be connected its full steady-state condition is equal to approximately 5[tex]\tau[/tex] as well as five-time constants.
It would be calculated in seconds by:
⇒ [tex]\tau=\frac{L}{R}[/tex]
, where
R seems to be the resistor function in ohms.L seems to be the inductor function in Henries.Earth orbiting the Sun The Earth is 1.5 ⋅ 10 8 km from the Sun (on average). How fast is the Earth orbiting the Sun in kilometers per second (on average)? You can assume the orbit of the Earth is a circle and that the circumference of a circle is equal to C = 2 π R where R is the radius of a circle (the distance between the center and the edge. Note that for our purposes, it is perfectly fine to assume π = 3 which allows for a pretty good approximation C = 6 R . Your answer does not need to be put into scientific notation, but if you choose to do so it will be marked correct! kilometers per second
Answer:
1 yr = 24 * 3600 * 365 = 3.2 * 10E7 sec
C = 6 R = 1.5 * 10E8 * 6 = 9 * 10E8 km circumference of orbit
v = C / t = 9 * 10E8 km / 3 * 10E7 sec = 30 km / sec = 18 mi/sec
g If the interaction of a particle with its environment restricts the particle to a finite region of space, the result is the quantization of ____ of the particle.
Answer:
the result is the quantization of __Energy__ of the particle
Explanation:
What value of D is required to make vt = 42.7 m/s the terminal velocity of a skydiver of mass 85.0 kg . Express your answer using two significant figures.
Complete Question
For a human body falling through air in a spread edge position , the numerical value of the constant D is about [tex]D = 0.2500 kg/m[/tex]
What value of D is required to make vt = 42.7 m/s the terminal velocity of a skydiver of mass 85.0 kg . Express your answer using two significant figures?
Answer:
The value of D is [tex]D = 0.457 \ kg/m[/tex]
Explanation:
From the question we are told that
The terminal velocity is [tex]v_t = 42.7 \ m/s[/tex]
The mass of the skydiver is [tex]m = 85.0 \ kg[/tex]
The numerical value of D is [tex]D = 0.2500 kg/m[/tex]
From the unit of D in the question we can evaluate D as
[tex]D = \frac{m * g }{v^2}[/tex]
substituting values
[tex]D = \frac{85 * 9.8 }{(42.7)^2}[/tex]
[tex]D = 0.457 \ kg/m[/tex]
A 0.20-kg block rests on a frictionless level surface and is attached to a horizontally aligned spring with a spring constant of 40 N/m. The block is initially displaced 4.0 cm from the equilibrium point and then released to set up a simple harmonic motion. A frictional force of 0.3 N exists between the block and surface. What is the speed of the block when it passes through the equilibrium point after being released from the 4.0-cm displacement point
Answer:
Approximately [tex]0.45\; \rm m \cdot s^{-1}[/tex].
Explanation:
The mechanical energy of an object is the sum of its potential energy and kinetic energy. Consider this question from the energy point of view:
Mechanical energy of the block [tex]0.04\; \rm m[/tex] away from the equilibrium position:
Elastic potential energy: [tex]\displaystyle \frac{1}{2} \, k\, x^2 = \frac{1}{2}\times \left(0.04\; \rm m\right)^2 \times 40\; \rm N \cdot m^{-1} = 0.032\; \rm J[/tex].Kinetic energy: [tex]0\; \rm J[/tex].While the block moves back to the equilibrium position, it keeps losing (mechanical) energy due to friction:
[tex]\begin{aligned}& \text{Work done by friction} = (-0.3\; \rm N) \times (0.04 \; \rm m) = -0.012\; \rm J\end{aligned}[/tex].
The opposite ([tex]0.012\; \rm N[/tex]) of that value would be the amount of energy lost to friction. Since there's no other form of energy loss, the mechanical energy of the block at the equilibrium position would be [tex]0.032\; \rm N - 0.012\; \rm N = 0.020\; \rm N[/tex].
The elastic potential energy of the block at the equilibrium position is zero. As a result, all that [tex]0.020\; \rm N[/tex] of mechanical energy would all be in the form of the kinetic energy of that block.
Elastic potential energy: [tex]0\; \rm J[/tex].Kinetic energy: [tex]0.020\; \rm J[/tex].Given that the mass of this block is [tex]0.020\; \rm kg[/tex], calculate its speed:
[tex]\begin{aligned}v &= \sqrt{\frac{2\, \mathrm{KE}}{m}} \\ &= \sqrt{\frac{2 \times 0.020\; \rm J}{0.20\; \rm kg}} \approx 0.45\; \rm m\cdot s^{-1}\end{aligned}[/tex].
3. A particle of charge +7.5 µC is released from rest at the point x = 60 cm on an x-axis. The particle begins to move due to the presence of a charge ???? that remains fixed at the origin. What is the kinetic energy of the particle at the instant it has moved 40 cm if a) ???? = +20 µC and b) ???? = −20 µC?
Answer:
HSBC keen vs kg get it yyyyyuuy
Explanation:
hgccccxfcffgbbbbbbbbbbghhyhhhgdghcjyddhhyfdghhhfdgbxbbndgnncvbhcxgnjffccggshgdggjhddh
nnnbvvvvvggfxrugdfutdfjhyfggigftffghhjjhhjyhrdffddfvvvvvvvvvvvbbbbbbbbbvvcxccghhyhhhjjjhjnnnnnnnnnnnnnbhbfgjgfhhccccccvvjjfdbngxvncnccbnxcvbchvxxghfdgvvhhihbvhbbhhvxcgbbbcxzxvbjhcxvvbnnxvnn
Consider two soap bubbles with radius r1 and r2 (r1 <r2) connected via a valve. What happens if we open the valve
Complete Question
The complete question is shown on the first uploaded image
Answer:
The pressure difference of the first bubble is [tex]\Delta P _1 =10 J/m^3[/tex]
The pressure difference of the second bubble is [tex]\Delta P _2 =20 J/m^3[/tex]
The pressure difference on the second bubble is higher than that of the first bubble so when the valve is opened pressure from second bubble will cause air to flow toward the first bubble making is bigger
Explanation:
From the question we are told that
The radius of the first bubble is [tex]r_1 = 10 \ mm=0.01 \ m[/tex]
The radius of the second bubble is [tex]r_2 = 5 \ mm = 0.005 \ m[/tex]
The surface tension of the soap solution is [tex]s = 25 \ mJ/m^2 = 25*10^{-3} J/m^2[/tex]
Generally according to the Laplace's Law for a spherical membrane the pressure difference is mathematically represented as
[tex]\Delta P = \frac{4 s}{R}[/tex]
Now the pressure difference for the first bubble is mathematically evaluated as
[tex]\Delta P _1 = \frac{4 s}{r_1}[/tex]
substituting values
[tex]\Delta P _1 = \frac{4 *25 *10^{-3}}{0.01}[/tex]
[tex]\Delta P _1 =10 J/m^3[/tex]
Now the pressure difference for the second bubble is mathematically evaluated as
[tex]\Delta P _2 = \frac{4 s}{r_1}[/tex]
[tex]\Delta P _2 = \frac{4 *25 *10^{-3}}{0.005}[/tex]
[tex]\Delta P _2 =20 J/m^3[/tex]
distributed uniformly over the surface of a metal sphere with a radius 24.0 cm. If the potential is zero at a point at infinity, find the value of the pote my jobntA total electric charge of 3.50 nC is distributed uniformly over the surface of a metal sphere with a radius 24.0 cm. If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) 48.0 cm (b) 2ial at the following distances from the center of the sphere: (a) 48.0 cm (b) 24.0 cm (c) 12.0 cm
Answer:
(a) V = 65.625 Volts
(b) V = 131.25 Volts
(c) V = 131.25 Volts
Explanation:
Recall that:
1) in a metal sphere the charges distribute uniformly around the surface, and the electric field inside the sphere is zero, and the potential is constant equal to:
[tex]V=k\frac{Q}{R}[/tex]
2) the electric potential outside of a charged metal sphere is the same as that of a charge of the same value located at the sphere's center:
[tex]V=k\frac{Q}{r}[/tex]
where k is the Coulomb constant ( [tex]9\,\,10^9\,\,\frac{N\,m^2}{C^2}[/tex] ), Q is the total charge of the sphere, R is the sphere's radius (0.24 m), and r is the distance at which the potential is calculated measured from the sphere's center.
Then, at a distance of:
(a) 48 cm = 0.48 m, the electric potential is:
[tex]V=k\frac{Q}{r}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.48} =65.625\,\,V[/tex]
(b) 24 cm = 0.24 m, - notice we are exactly at the sphere's surface - the electric potential is:
[tex]V=k\frac{Q}{r}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.24} =131.25\,\,V[/tex]
(c) 12 cm (notice we are inside the sphere, and therefore the potential is constant and the same as we calculated for the sphere's surface:
[tex]V=k\frac{Q}{R}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.24} =131.25\,\,V[/tex]
Answer:
c) a difference in electric potential
Explanation:
my insta: priscillamarquezz
A street light is at the top of a pole that has a height of 17 ft . A woman 5 ft tall walks away from the pole with a speed of 8 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40ft from the pole?
Answer:
8 ft/s
Explanation:
This is a straight forward question without much ado.
It is given from the question that she walks with a speed of 8 ft/s
A square copper plate, with sides of 50 cm, has no net charge and is placed in a region where there is a uniform 80 kN / C electric field directed perpendicular to the plate. Find a) the charge density of each side of the plate and b) the total load on each side.
Answer:
a) ±7.08×10⁻⁷ C/m²
b) 1.77×10⁻⁷ C
Explanation:
For a conductor,
σ = ±Eε₀,
where σ is the charge density,
E is the electric field,
and ε₀ is the permittivity of space.
a)
σ = ±Eε₀
σ = ±(8×10⁴ N/C) (8.85×10⁻¹² F/m)
σ = ±7.08×10⁻⁷ C/m²
b)
σ = q/A
7.08×10⁻⁷ C/m² = q / (0.5 m)²
q = 1.77×10⁻⁷ C
The relationship between the Period (T) caused by the oscillation of the mass on the end of a hanging spring and the mass (m) is:
Answer:
T= 2p√m/k
Explanation:
This is because the period of oscillation of the mass of spring system is directly proportional to the square root of the mass and it is inversely proportional to the square root of the spring constant.
The period of a mass on a spring is given by the equation
T=2π√m/k.
Where T is the period,
M is mass
K is spring constant.
An increase in mass in a spring increases the period of oscillation and decrease in mass decrease period of oscillation.
When there is the relationship between the Period (T) caused by the oscillation of the mass should be considered as the T= 2p√m/k.
Oscillation of the mass:The mass of the spring system with respect to period of oscillation should be directly proportional to the square root of the mass and it is inversely proportional to the square root of the spring constant.
So the following equation should be considered
T=2π√m/k.
Here,
T is the period,
M is mass
K is spring constant.
An increase in mass in a spring rises the period of oscillation and reduce in mass decrease period of oscillation.
Learn more about mass here: https://brainly.com/question/21860379
You have a 160-Ω resistor and a 0.430-H inductor. Suppose you take the resistor and inductor and make a series circuit with a voltage source that has a voltage amplitude of 30.0 V and an angular frequency of 220 rad/s .
Part A: What is the impedance of the circuit? ( Answer: Z = ? Ω )
Part B: What is the current amplitude? ( Answer: I = ? A )
Part C: What is the voltage amplitude across the resistor? ( Answer: VR = ? V )
Part D: What is the voltage amplitudes across the inductor? ( Answer: VL = ? V )
Part E: What is the phase angle ϕ of the source voltage with respect to the current? ( Answer: ϕ = ? degrees )
Part F: Does the source voltage lag or lead the current? ( Answer: the voltage lags the current OR the voltage leads the current )
Answer:
A. Z = 185.87Ω
B. I = 0.16A
C. V = 1mV
D. VL = 68.8V
E. Ф = 30.59°
Explanation:
A. The impedance of a RL circuit is given by the following formula:
[tex]Z=\sqrt{R^2+\omega^2L^2}[/tex] (1)
R: resistance of the circuit = 160-Ω
w: angular frequency = 220 rad/s
L: inductance of the circuit = 0.430H
You replace in the equation (1):
[tex]Z=\sqrt{(160\Omega)^2+(220rad/s)^2(0.430H)^2}=185.87\Omega[/tex]
The impedance of the circuit is 185.87Ω
B. The current amplitude is:
[tex]I=\frac{V}{Z}[/tex] (2)
V: voltage amplitude = 30.0V
[tex]I=\frac{30.0V}{185.87\Omega}=0.16A[/tex]
The current amplitude is 0.16A
C. The current I is the same for each component of the circuit. Then, the voltage in the resistor is:
[tex]V=\frac{I}{R}=\frac{0.16A}{160\Omega}=1*10^{-3}V=1mV[/tex] (3)
D. The voltage across the inductor is:
[tex]V_L=L\frac{dI}{dt}=L\frac{d(Icos(\omega t))}{dt}=-LIsin(\omega t)\\\\V_L=-(0.430H)(160\Omega)sin(220 t)=68.8sin(220t)\\\\V_L_{max}=68.8V[/tex]
E. The phase difference is given by:
[tex]\phi=tan^{-1}(\frac{\omega L}{R})=tan^{-1}(\frac{(220rad/s)(0.430H)}{160\Omega})\\\\\phi=30.59\°[/tex]
A student applies a constant horizontal 20 N force to a 12 kg box that is initially at rest. The student moves the box a distance of 3.0 m. What is the speed of the box at the end of the motion
Answer:
u = 10.02m/s
Explanation:
a = f/m
a = 20/12 = 1.67m/s²
U =2aS
u = 2 x 1.67 x 3
U = 10.02m/s
The gravitational energy of a swimmer on a driving board at different heights is shown in the table below. What is the driver's gravitational energy at 5m high? (A) 5500 J (B) 2750 J (C) 8800 J (D) 3300 J
Answer:
E = 2750 J at h = 5 m
Explanation:
The gravitational potential energy is given by :
[tex]E=mgh[/tex]
In this case, m is the mass of swimmer is constant at every heights. So,
At h = 1 m, E = 550 J
[tex]550=m\times 10\times 1\\\\m=55\ kg[/tex]
So, at h = 5 m, gravitational potential energy is given by :
[tex]E=55\times 10\times 5\\\\E=2750\ J[/tex]
So, the correct option is (B).
A hair dryer draws 1200 W, a curling iron draws 800 W, and an electric light fixture draws 500 W. If all three of these appliances are operating in parallel on a 120-V circuit, what is the total current drawn
Answer:
The Total current drawn is 20.83 Ampere.
Explanation:
https://brainly.com/question/15048481
the density of gold is 19 300kg/m cube. what is the mass of gold cube with the length 0.2015m?
Answer:
157.9 kg
Explanation:
Density: This can be defined as the ratio of the mass of a body and it's volume.
The S.I unit of density is kg/m³.
From the question,
Density = Mass/volume
D = m/v............................ Equation 1
Where D = Density of gold, m = mass of gold, v = volume of gold.
make m the subject of the equation
m = Dv.................... Equation 2
Since the gold is a cube,
v = l³................... Equation 3
Where l = length of the gold cube.
Substitute equation 3 into equation 2
m = Dl³............... Equation 4
Given: D = 19300 kg/m³, l = 0.2015 m
Substitute into equation 4
m = 19300(0.2015)³
m = 157.9 kg.
A car is traveling down a highway. It was moving with a velocity of 50m/s when the driver reads the speed limit and has to decelerate with an acceleration of -5m/s for 2 seconds. What is the momentum of this 500kg car after it decelerates?
Answer:
20,000 kg m/s
Explanation:
Given:
v₀ = 50 m/s
a = -5 m/s²
t = 2 s
Find: v
v = at + v₀
v = (-5 m/s²) (2 s) + (50 m/s)
v = 40 m/s
p = mv
p = (500 kg) (40 m/s)
p = 20,000 kg m/s
To practice Problem-Solving Strategy 15.2 Standing Waves.In lab, your instructor generates a standing wave using a thin string of length L = 2.25 m fixed at both ends. You are told that the standing wave is produced by the superposition of traveling and reflected waves, where the incident traveling waves propagate in the +x direction with an amplitude A = 2.45 mm and a speed vx = 14.5 m/s . The first antinode of the standing wave is a distance of x = 37.5 cm from the left end of the string, while a light bead is placed a distance of 18.8 cm to the right of the first antinode. What is the maximum transverse speed vy of the bead? Make sure to use consistent distance units in your calculations.To check your equation for the standing wave's transverse velocity, find the maximum transverse velocity at x = 75.0 cm .
Answer:
Explanation:
Let the equation of standing wave be as follows
y = A sinωt cos kx
A = 2.45 mm
y = 2.45 cosωt sin kx
given
[tex]\frac{\omega}{k}[/tex] = velocity = 14.5
Position of first antinode = 37.5 cm
kx = π / 2
k x 37.5 = π / 2
k = π / 75
ω / k = 14.5
ω = 14.5 x π / 75
= .607 rad /s
Maximum transverse speed
= ω A
= .607 x 2.45
= 1.49 mm / s
y = A sinωt cos kx
Transverse velocity
v = dy / dt
= ω A cosωt cos kx
Maximum transverse velocity at any x = ω A .
The characteristics of standing waves allows find the results for the questions about the speed of the rope are:
The transverse oscillatory velocity is: vy = 0.298 m / sGiven parameters
The amplitude of the wave A = 2.45 mm = 2.45 10-3 m Chord length L = 2.25 m Wave velocity v = 14.5 m / d The first antinode x = 37.5 cmTo find
Maximum rope swing speed.
The movement in a string is formed by two movements, a movement in the direction of the string with constant speed and a transverse movement where the speed varies as in a simple harmonic movement.
The standing wave is formed from the sum of the incident wave and the reflected wave.
y = A cos (kx- wt)
y = A cos (kx + wt)
resulting
y = A sin wt cos kx
the speed that the wave is given by
v = w / k
They indicate the position of the first antinode at this point the cosine function must be maximum.
kx = π
k = π/x
k = [tex]\frac{\pi }{0.375}[/tex]
k = 8.38 m⁻¹
let's find the angular velocity.
w = v k
w = 14.5 8.38
w = 121.5 rad / s
The expression for displacement in simple harmonic motion is:
x = A cos wt
The speed is defined by the variation of the position with respect to time.
v = [tex]\frac{dx}{dt}[/tex] =
v = - A w sin wt
To calculate the maximum speed we make the sine equal to 1.
[tex]v_{y \ max}[/tex] = w A
[tex]v_{y \ max}[/tex] = 121.5 2.45 10⁻³
[tex]v_{y \ max}[/tex] = 0.298 m / s
For point x = 75 cm = 0.750 m
We seek the value of
kx = 8.38 0.750
kx = 6.285 = 2π
therefore this point is also an antinode and the results do not change.
In conclusion, using the characteristics of standing waves, we can find the results for the questions about the speed of the rope are:
The transverse oscillatory velocity is: [tex]v_y[/tex] = 0.298 m / s
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.
1. How does the arrangement of a pulley system affect its ideal mechanical advantage
Answer: In a pulley, the ideal mechanical advantage is equal to the number of rope segments pulling up on the object. The more rope segments that are supporting to do the lifting work, the less pressure that is needed for the job.
Explanation:
A parallel-plate capacitor has 3.5 cm × 3.5 cm electrodes with surface charge densities ±1.0 × 106 C/m2. A proton traveling parallel to the electrodes at 1.9 × 106 m/s enters the center of the gap between them. By what distance has the proton been deflected sideways when it reaches the far edge of the capacitor? Assume the field is uniform inside the capacitor and zero outside the capacitor.
Answer:
x = 1.1*10^-7 m
Explanation:
You can consider that the electron travels in the horizontal positive x direction. Furthermore, you can consider that the electric field between the plates is totally vertical and points upward (which means that the negative plates is above the line of motion of the proton).
In order to calculate the vertical distance traveled by the electron when it has passed the region between the plates, you take into account the acceleration experienced by the proton, due to the electric field between the plates.
You first calculate the electric field between the plates, by using the following formula:
[tex]E=\frac{\sigma}{\epsilon_o}[/tex] (1)
σ: surface charge density of each plate = ±1.0*10^6 C/m^2
εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2
You replace the values of the parameters in the equation (1):
[tex]E=\frac{1.0*10^6C/m^2}{8.85*10^{-12}C^2/Nm}=1.12*10^{17}\frac{N}{C}[/tex]
Next, you calculate the acceleration of the electron produced by the electric force. You use the second Newton law for the electric force:
[tex]F_e=qE=ma[/tex] (2)
q: charge of the electron = 1.6*10^-19C
m: mass of the proton = 1.67*10^-27 kg
You solve the equation (2) for a:
[tex]a=\frac{qE}{m}=\frac{(1.6*10^{-19})(1.12*10^{17}N/C)}{1.67*10^{-27}kg}\\\\a=1.07*10^{25}\frac{m}{s^2}[/tex]
Next, you calculate the time that proton takes to reach the negative capacitor. You use the following formula:
[tex]y=v_{oy}t+\frac{1}{2}at^2[/tex] (3)
y: vertical distance from the center of the capacitor to the negative plate = 3.5cm/2 = 1.75cm = 0.0175m
voy: vertical initial speed of the proton = 0m/s (the proton enter to the region of electric field horizontally)
You solve the equation (3) for t:
[tex]y=\frac{1}{2}at^2\\\\t=\sqrt{\frac{2y}{a}}=\sqrt{\frac{2(0.0175m)}{1.05*10^{25}m/s^2}}=5.77*10^{-14}s[/tex]
Next, you calculate the horizontal distance traveled by the proton in the time calculated. You use the following formula:
[tex]x=v_ot[/tex] (4)
x: length of the capacitor = 3.5cm = 0.035m
vo: initial speed of the proton = 1.9*10^6 m/s
[tex]x=(1.9*10^6m/s)(5.77*10^{-14}s)=1.1*10^{-7}m[/tex]
The horizontal distance traveld by the electron when it impacts the negative plate of teh capacitor is 1.1*10^-7m
What is the length of the shadow cast on the vertical screen by your 10.0 cm hand if it is held at an angle of θ=30.0∘ above horizontal? Express your answer in centimeters to three significant figures. View Available Hint(s)
Answer:
The length is [tex]D = 5 \ cm[/tex]
Explanation:
From the question we are told that
The length of the hand is [tex]l = 10.0 \ cm[/tex]
The angle at the hand is held is [tex]\theta = 30 ^o[/tex]
Generally resolving the length the length of the hand to it vertical component we obtain that the length of the shadow on the vertical wall is mathematically evaluated as
[tex]D = l * sin(\theta )[/tex]
substituting values
[tex]D = 10 * sin (30)[/tex]
[tex]D = 5 \ cm[/tex]
A scientist claims to be a solid-state physicist. She is working on electromagnetics and trying to see if she can create an electromagnetic field by causing different solutions to flow through a tube. Is this scientist a solid-state physicist?
no, because she is working in electromagnetics, which is not a solid-state field
no, because she is not working on the atomic structures of a solid
yes, because she is working in electromagnetics, which is a solid-state field
yes, because she is causing solutions to flow through a tube
The correct answer is B. No, because she is not working on the atomic structures of a solid
Explanation:
Solid-state physics is a sub-discipline of physics that focuses on studying solids, this includes analyzing solids structures, features, and other phenomena that occur in substances in this state of the matter. This means a solid-state physics will not study or gases.
In this context, the fact the scientist is trying to create an electromagnetic field by using solutions and the flow of these show the scientists is not working with solids but with liquids or gases as solids do not flow. Also, her focus is not solids, and therefore she is not a solid-state physicist. Thus, it can be concluded she is not a solid-state physicists because she is not working on the structures of solids.
An object of mass 2 kg has a speed of 6 m/s and moves a distance of 8 m. What is its kinetic energy in joules?
Answer:
36 JoulesExplanation:
Mass ( m ) = 2 kg
Speed of the object (v) = 6 metre per second
Kinetic energy =?
Now,
We have,
Kinetic Energy = [tex] \frac{1}{2} \times m \times {v}^{2} [/tex]
Plugging the values,
[tex] = \frac{1}{2} \times 2 \times {(6)}^{2} [/tex]
Reduce the numbers with Greatest Common Factor 2
[tex] = {(6)}^{2} [/tex]
Calculate
[tex] = 36 \: joule[/tex]
Hope this helps...
Good luck on your assignment...
The Kinetic energy of the object will be "36 joules".
Kinetic energyThe excess energy of moving can be observed as that of the movement of an object, component, as well as the group of components. There would never be a negative (-) amount of kinetic energy.
According to the question,
Mass of object, m = 2 kg
Speed of object, v = 6 m/s
As we know the formula,
→ Kinetic energy (K.E),
= [tex]\frac{1}{2}[/tex] × m × v²
By substituting the values, we get
= [tex]\frac{1}{2}[/tex] × 2 × (6)²
= [tex]\frac{1}{2}[/tex] × 2 × 36
= 36 joule
Thus the above answer is appropriate.
Find out more information about Kinetic energy here:
https://brainly.com/question/25959744
A child is sitting on the seat of a swing with ropes 10 m long. Their father pulls the swing back until the ropes make a 37o angle with the vertical and then releases the swing. If air resistance is neglected, what is the speed of the child at the bottom of the arc of the swing when the ropes are vertical
Answer:
Explanation:
We shall apply conservation of mechanical energy law to solve the problem .
loss of height = L ( 1 - cos 37 ) where L is length of rope
loss of potential energy at the bottom = gain of kinetic energy .
mg L ( 1 - cos 37 ) = 1/2 m v² where v is velocity at the bottom
v² = 2 L g ( 1 - cos 37 )
= 2 x 10 x 9.8 ( 1 - cos 37 )
= 39.46
v = 6.28 m /s
n electromagnetic wave in vacuum has an electric field amplitude of 611 V/m. Calculate the amplitude of the corresponding magnetic field.
Answer:
The corresponding magnetic field is
Explanation:
From the question we are told that
The electric field amplitude is [tex]E_o = 611\ V/m[/tex]
Generally the magnetic field amplitude is mathematically represented as
[tex]B_o = \frac{E_o }{c }[/tex]
Where c is the speed of light with a constant value
[tex]c = 3.0 *0^{8} \ m/s[/tex]
So
[tex]B_o = \frac{611 }{3.0*10^{8}}[/tex]
[tex]B_o = 2.0 4 *10^{-6} \ Vm^{-2} s[/tex]
Since 1 T is equivalent to [tex]V m^{-2} \cdot s[/tex]
[tex]B_o = 2.0 4 *10^{-6} \ T[/tex]
When you release the mass, what do you observe about the energy?
Explanation:
Mass and energy are closely related. Due to mass–energy equivalence, any object that has mass when stationary (called rest mass) also has an equivalent amount of energy whose form is called rest energy, and any additional energy (of any form) acquired by the object above that rest energy will increase the object's total mass just as it increases its total energy. For example, after heating an object, its increase in energy could be measured as a small increase in mass, with a sensitive enough scale.
Two small charged spheres are 7.59 cmcm apart. They are moved, and the force on each of them is found to have been tripled. How far apart are they now?
Answer:
The two small charged spheres are now 4.382 cm apart
Explanation:
Given;
distance between the two small charged sphere, r = 7.59 cm
The force on each of the charged sphere can be calculated by applying Coulomb's law;
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
where;
F is the force on each sphere
q₁ and q₂ are the charges of the spheres
r is the distance between the spheres
[tex]F = \frac{kq_1q_2}{r^2} \\\\kq_1q_2 = Fr^2 \ \ (keep \ kq_1q_2 \ constant)\\\\F_1r_1^2 = F_2r_2^2\\\\r_2^2 = \frac{F_1r_1^2}{F_2} \\\\r_2 = \sqrt{\frac{F_1r_1^2}{F_2}} \\\\r_2 = r_1\sqrt{\frac{F_1}{F_2}}\\\\(r_1 = 7.59 \ cm, \ F_2 = 3F_1)\\\\r_2 = 7.59cm\sqrt{\frac{F_1}{3F_1}}\\\\r_2 = 7.59cm\sqrt{\frac{1}{3}}\\\\r_2 = 7.59cm *0.5773\\\\r_2 = 4.382 \ cm[/tex]
Therefore, the two small charged spheres are now 4.382 cm apart.