Answer:
the acceleration during the collision is: - 5 [tex]\frac{m}{s^2}[/tex]
Explanation:
Using the formula:
[tex]a=\frac{\Delta\,v}{\Delta\,t}[/tex]
we get:
[tex]a=\frac{-0.4-0.6}{0.2} \,\frac{m}{s^2} =\frac{-1}{0.2} \,\frac{m}{s^2} =-5\,\,\frac{m}{s^2}[/tex]
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
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
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.
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.
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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
A hungry 177 kg lion running northward at 81.8 km/hr attacks and holds onto a 32.0 kg Thomson's gazelle running eastward at 59.0 km/hr. Find the final speed of the lion–gazelle system immediately after the attack.
Answer:
The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.
Explanation:
Let suppose that lion and Thomson's gazelle are running at constant speed before and after collision and that collision is entirely inelastic. Given the absence of external force, the Principle of Momentum Conservation is applied such that:
[tex]\vec p_{L} + \vec p_{G} = \vec p_{F}[/tex]
Where:
[tex]\vec p_{L}[/tex] - Linear momentum of the lion, measured in kilograms-meters per second.
[tex]\vec p_{G}[/tex] - Linear momentum of the Thomson's gazelle, measured in kilograms-meters per second.
[tex]\vec p_{F}[/tex] - Linear momentum of the lion-Thomson's gazelle, measured in kilograms-meters per second.
After using the definition of momentum, the system is expanded:
[tex]m_{L}\cdot \vec v_{L} + m_{G}\cdot \vec v_{G} = (m_{L} + m_{G})\cdot \vec v_{F}[/tex]
Vectorially speaking, the final velocity of the lion-gazelle system is:
[tex]\vec v_{F} = \frac{m_{L}}{m_{L}+m_{G}}\cdot \vec v_{L} + \frac{m_{G}}{m_{L}+m_{G}}\cdot \vec v_{G}[/tex]
Where:
[tex]m_{L}[/tex], [tex]m_{G}[/tex] - Masses of the lion and the Thomson's gazelle, respectively. Measured in kilograms.
[tex]\vec v_{L}[/tex], [tex]\vec v_{G}[/tex], [tex]\vec v_{F}[/tex] - Velocities of the lion, Thomson's gazelle and the lion-gazelle system. respectively. Measured in meters per second.
If [tex]m_{L} = 177\,kg[/tex], [tex]m_{G} = 32\,kg[/tex], [tex]\vec v_{L} = 81.8\cdot j\,\left[\frac{km}{h} \right][/tex] and [tex]\vec v_{G} = 59.0\cdot i\,\left[\frac{km}{h} \right][/tex], the final velocity of the lion-gazelle system is:
[tex]\vec v_{F} = \frac{177\,kg}{177\,kg+32\,kg}\cdot \left(81.8\cdot j\right)\,\left[\frac{km}{h} \right] + \frac{32\,kg}{177\,kg+32\,kg}\cdot \left(59.0\cdot i\right)\,\left[\frac{km}{h} \right][/tex]
[tex]\vec v_{F} = 9.033\cdot i + 69.276\cdot j\,\left[\frac{km}{h} \right][/tex]
The speed of the system is the magnitude of the velocity vector, which can be found by means of the Pythagorean theorem:
[tex]\|\vec v_{F}\| = \sqrt{\left(9.033\frac{km}{h} \right)^{2}+\left(69.276\frac{km}{h} \right)^{2}}[/tex]
[tex]\|\vec v_{F}\| \approx 69.862\,\frac{km}{h}[/tex]
The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.
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.
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Zuckerman’s test for sensation seeking measures which of the following characteristics?
dangerousness, antisocial traits, “letting loose,’ and intolerance for boredom
thrill and adventure seeking, experience seeking, disinhibition, and susceptibility to boredom
adventurousness, physical prowess, creative morality, and charisma
dangerousness, adventurousness, creativity, and thrill and adventure seeking
The correct answer is B. thrill and adventure seeking, experience seeking, disinhibition, and susceptibility to boredom
Explanation:
Marvin Zuckerman was an important American Psychologists mainly known for his research about personality and the creation of a model to study this aspect of human psychology. This model purposes five factors define personality, these are the thrill and adventure-seeking that involves seeking for adventures and danger; experience seeking that implies a strong interest in participating in new activities; disinhibition that implies being open and extrovert; and susceptibility to boredom that implies avoiding boredom or repetition. Thus, option B correctly describes the characteristics used in Zuckerman's test.
A spherical balloon contains a charge +Q uniformly distributed over its surface. When it has a diameter D, the electric field at its surface has magnitude E. If the balloon is not blown up to thrice this diameter without changing the charge, the electric field at its surface is?
Answer:
E = 1/9 E₀
Explanation:
In this exercise we are told that the electric field is Eo when the diameter of the balloon is D, the expression
we are asked to shorten the electric field when the diameter is 3D with the same eclectic charge
For this we can use the gauss law to find the field in the new diameter, for this we create a Gaussian surface in the form of a sphere
Ф = ∫ E. dA = [tex]q_{int}[/tex] /ε₀
In this case the lines of the electric field and the radii of the sphere are parallel, therefore the scalar product is reduced to the algebraic product and the charge inside the sphere is the initial charge Q
A = 4π r²
E 4π r² = Q /ε₀
E = 1 /4πε₀ Q / r²
the value of the indicated distance is 3 times the initial diamete
r = 3 D / 2
we substitute
E = 1/4 πε₀ Q (2/ 3D)²
for the initial conditions
E₀ = 1 / 4πε₀ Q (2/D)²
subtitled in the equation above
E = 1/9 E₀
A horizontal spring with spring constant 290 N/m is compressed by 10 cm and then used to launch a 300 g box across the floor. The coefficient of kinetic friction between the box and the floor is 0.23. What is the box's launch speed?
Answer:
Explanation:
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A copper transmission cable 180 km long and 11.0 cm in diameter carries a current of 135 A.
Required:
a. What is the potential drop across the cable?
b. How much electrical energy is dissipated as thermal energy every hour?
Answer:
a) 43.98 V
b) E = 21.37 MJ
Explanation:
Parameters given:
Length of cable = 180 km = 180000 m
Diameter of cable = 11 cm = 0.11 m
Radius = 0.11 / 2 = 0.055 m
Current, I = 135 A
a) To find the potential drop, we have to find the voltage across the wire:
V = IR
=> V = IρL / A
where R = resistance
L = length of cable
A = cross-sectional area
ρ = resistivity of the copper wire = 1.72 * 10^(-8) Ωm
Therefore:
V = (135 * 1.72 * 10^(-8) * 180000) / (π * 0.055^2)
V = 43.98 V
The potential drop across the cable is 43.98 V
b) Electrical energy is given as:
E = IVt
where t = time taken = 1 hour = 3600 s
Therefore, the energy dissipated per hour is:
E = 135 * 43.98 * 3600
E = 21.37 MJ (mega joules, 10^6)
If two radio telescope dishes are wired together in the right way, the "D" used in determining the angular resolution is determined by
Answer:
D is determined by distance between the telescopes.
Explanation:
A 54.0 kg ice skater is moving at 3.98 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.802 m around the pole.
(a) Determine the force exerted by the horizontal rope on her arms.N
(b) What is the ratio of this force to her weight?(force from part a / her weight)
Answer:
(a) force is 1066.56N
Explanation:
(a) MV²/R
When separated by distance d, identically charged point-like objects A and B exert a force of magnitude F on each other. If you reduce the charge of A to one-fourth its original value, and the charge of B to one-fourth, and reduce the distance between the objects by half, what will be the new force that they exert on each other in terms of force F
Answer:
F ’= F 0.25
Explanation:
This problem refers to the electric force, which is described by Coulomb's law
F = k q₁ q₂ / r²
where k is the Coulomb constant, q the charges and r the separation between them.
The initial conditions are
F = k q_A q_B / d²
they indicate that the loads are reduced to ¼ q and the distance is reduced to ½ d
F ’= k (q / 4 q / 4) / (0.5 d)²
F ’= k q / 16 / 0.25 d²
F ’= k q² / d² 0.0625 / 0.25
F ’= F 0.25
Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. If the charges are reduced to one-fourth of their original values and the distance is halved, the new force will be one-fourth of the original force.
Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. This can be explained through Coulomb's law.
What is Coulomb's law?Coulomb's law is a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.
[tex]F = k \frac{q_Aq_B}{d^{2} } = k \frac{q^{2} }{d^{2} } [/tex]
where,
[tex]q_A [/tex] and [tex]q_B[/tex] are the charges of A and B (and equal to q).k is the Coulomb's constant.If you reduce the charge of A to one-fourth its original value, and the charge of B to one-fourth, and reduce the distance between the objects by half, the new force will be:
[tex]F_2 = k \frac{(0.25q_A)(0.25q_B)}{(0.5d)^{2} } = 0.25k\frac{q^{2} }{d^{2} } = 0.25 F[/tex]
Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. If the charges are reduced to one-fourth of their original values and the distance is halved, the new force will be one-fourth of the original force.
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Samantha is refinishing her rusty wheelbarrow. She moves her sandpaper back and forth 45 times over a rusty area, each time moving with a total distance of 0.12 m. Samantha pushes the sandpaper against the surface with a normal force of 2.6 N. The coefficient of friction for the metal/sandpaper interface is 0.92. How much work is done by the normal force during the sanding process
Answer:
W = 12.96 J
Explanation:
The force acting in the direction of motion of the sand paper is the frictional force. So, we first calculate the frictional force:
F = μR
where,
F = Friction Force = ?
μ = 0.92
R = Normal Force = 2.6 N
Therefore,
F = (0.92)(2.6 N)
F = 2.4 N
Now, the displacement is given as:
d = (0.12 m)(45)
d = 5.4 m
So, the work done will be:
W = F d
W = (2.4 N)(5.4 m)
W = 12.96 J
Two blocks of masses m1 and m2 are placed in contact with each other on a smooth, horizontal surface. Block m1 is on the left of block m2 . A constant horizontal force F to the right is applied to m1 . What is the horizontal force acting on m2?
Answer:
The horizontal force acting on m2 is F + 9.8m1
Explanation:
Given;
Block m1 on left of block m2
Make a sketch of this problem;
F →→→→→→→→→→→-------m1--------m2
Apply Newton's second law of motion;
F = ma
where;
m is the total mass of the body
a is the acceleration of the body
The horizontal force acting on block m2 is the force applied to block m1 and force due to weight of block m1
F₂ = F + W1
F₂ = F + m1g
F₂ = F + 9.8m1
Therefore, the horizontal force acting on m2 is F + 9.8m1
The force acting on the block of mass m₂ is [tex]\frac{m_2F}{m_1+m_2}[/tex]
Force acting on the block:Given that there are two blocks of mass m₁ and m₂.
m₁ is on the left of block m₂. They are in contact with each other.
A force F is applied on m₁ to the right.
According to Newton's laws of motion:
The equation of motion of the blocks can be written as:
F = (m₁ + m₂)a
here, a is the acceleration.
so, acceleration:
a = F / (m₁ + m₂)
Now, the force acting on the block of mass m₂ is:
f = m₂a
[tex]f = \frac{m_2F}{m_1+m_2}[/tex]
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A motorcyclist changes his speed from 20 km / h to 100 km / h in 3 seconds, maintaining a constant acceleration in that time interval. If the mass of the motorcycle is 200 kg and that of its rider is 80 kg, what is the value of the net force to accelerate the motorcycle? Help!
Answer:
2000 N
Explanation:
20 km/h = 5.56 m/s
100 km/h = 27.78 m/s
F = ma
F = m Δv/Δt
F = (200 kg + 80 kg) (27.78 m/s − 5.56 m/s) / (3 s)
F = 2074 N
Rounded to one significant figure, the force is 2000 N.
A student in her physics lab measures the standing-wave modes of a tube. The lowest frequency that makes a resonance is 30 Hz. As the frequency is increased, the next resonance is at 90 Hz.
What will be the next resonance after this?
Answer:
The next resonance will be 150 Hz.
Explanation:
The frequency of the sound produced by a tube, both open and closed, is directly proportional to the speed of propagation. Hence, to produce the different harmonics of a tube, the wave propagation speed must be increased.
The frequency of the sound produced by a tube, both open and closed, is inversely proportional to the length of the tube. The greater the length of the tube, the frequency is lower.
Frecuency of the standing sound wave modes in a open-closed tube is:
fₙ=n*f₁ where m is an integer and f₁ is the first frecuency (30 Hz)
The next resonance is at 90 Hz. This means that it occurs when n = 3:
f₃=3*30 Hz= 90 Hz
This means that the next resonance occurs when n = 5:
f₅=5*30 Hz= 150 Hz
The next resonance will be 150 Hz.
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 25-kilogram object is placed on a compression spring, and it creates a displacement of 0.15 meters. What is the weight of an object that creates a displacement of 0.23 m on the same spring? Enter your answer as a number rounded to the nearest tenth, such as: 42.5
Answer:
I hope it is correct ✌️
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 car has a mass of 1200 kg and an acceleration of 4 m/s^2. If the friction on the car is 200 N, how much force is the thrust providing?
Answer:
5000N
Explanation:
According to Newton's second law of motion, the net force (∑F) acting on a body is the product of the mass (m) of the body and the acceleration (a) of the body caused by the force. i.e
∑F = m x a -------------(i)
From the question, the net force is the combined effect of the thrust (F) and the friction force (Fₓ). i.e
∑F = F + Fₓ -------------(ii)
Where;
Fₓ = -200N [negative sign because the friction force opposes motion]
Combine equations(i) and (ii) together to get;
F + Fₓ = m x a
F = ma - Fₓ -------------(iii)
Where;
m = mass of car = 1200kg
a = acceleration of the car = 4m/s²
Now substitute the values of m, a and Fₓ into equation (iii) as follows;
F = (1200 x 4) - (-200)
F = 4800 + 200
F = 5000N
Therefore, the force the thrust is providing is 5000N
What is the role of the part in the diagram labeled Y?
modulate, amplify, and send out waves
O capture, amplify, and demodulate waves
change the amplitude and frequency of waves
O change the pulse and phase of waves
Question is incomplete and image is not attached ti the question. The required image is attached below, so the complete question is:
The diagram shows a device that uses radio waves.
What is the role of the part in the diagram labeled Y?
modulate, amplify, and send out waves capture, amplify, and demodulate waves change the amplitude and frequency of waves change the pulse and phase of wavesAnswer:
2. capture, amplify, and demodulate waves
Explanation:
The part Y labeled in the diagram refers to radio receiver which capture, amplify and demodulate the radio waves.
The radio receiver seperates required radio frequency signals through antenna and consist of an amplifier that amplify or increase the power of receiving signal. At the end, demodulators present in receivers recover the information from the modulated wave.
Hence, the correct option is 2.
Answer:
B
Explanation:
edge 2020
Which characteristic gives the most information about what kind of element an atom is ?
Answer:
The atomic number
Explanation:
A student slides her 80.0-kg desk across the level floor of her dormitory room a distance 4.40 m at constant speed. If the coefficient of kinetic friction between the desk and the floor is 0.400, how much work did she do
Answer:
F = umg where u is coefficient of dynamic friction
Explanation:
F = 0.4 x 80 x 9.81 = 313.92 N
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:
If a water wave completes one cycle in 2 seconds, what is
the period of the wave?
0.5 seconds
O4 seconds
2 seconds
0.2 seconds
Done
The period of a wave is the time it takes the wave to complete one cycle (at a fixed location).
So if a wave completes one cycle in 2 seconds, then that is its period.
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
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.
A cyclotron operates with a given magnetic field and at a given frequency. If R denotes the radius of the final orbit, then the final particle energy is proportional to which of the following?
A. 1/RB. RC. R^2D. R^3E. R^4
Answer:
C. R^2
Explanation:
A cyclotron is a particle accelerator which employs the use of electric and magnetic fields for its functioning. It consists of two D shaped region called dees and the magnetic field present in the dee is responsible for making sure the charges follow the half-circle and then to a gap in between the dees.
R is denoted as the radius of the final orbit then the final particle energy is proportional to the radius of the two dees. This however translates to the energy being proportional to R^2.