Answer:
the image is virtual and erect and the lens divergent; therefore the correct answer is C
Explanation:
In a thin lens the magnification given by
m = h '/ h = - q / p
where h ’is the height of the image, h is the height of the object, q is the distance to the image and p is the distance to the object.
It indicates that the object is straight and is placed at a distance p> f
analyze the situation tells us that the magnification is positive so the distance to the image must be negative, that is, that the image is on the same side as the object.
Consequently the lens must be divergent
The magnification value is
0.4 = h ’/ h
h ’= 0.4 h
therefore the erect images
therefore the image is virtual and erect and the lens divergent; therefore the correct answer is C
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
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]
While running, a person dissipates about 0.60 J of mechanical energy per step per kilogram of body mass. If a 52-kg person develops a power of 80 W during a race, how fast is the person running
Answer:
The person is running at a speed of 2.564 m/s
Explanation:
Given;
mechanical energy dissipated per kilogram per step, E/kg/S = 0.6 J/kg/S
mass of the person, m = 52 kg
power developed by the person, P = 80 W
mechanical energy of the person per step, E = 0.6 J/kg x 52 kg
[tex]E_{step}[/tex] = 31.2 J
mechanical energy for the total step, [tex]E_{total}[/tex] = 31.2 J x S
P = E / t
[tex]P_{avg} = \frac{E_{total}}{t} \\\\P_{avg} = \frac{E_{step}*S}{t}\\\\\frac{P_{avg}}{E_{step}} = \frac{S}{t} \\\\\frac{S}{t} = \frac{80}{31.2} \\\\\frac{S}{t} = 2.564 \ m/s[/tex]
Therefore, the person is running at a speed of 2.564 m/s
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 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
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
Inside a 30.2 cm internal diameter stainless steel pan on a gas stove water is being boiled at 1 atm pressure. If the water level in the pan drops by 1.45 cm in 18.6 min, determine the rate of heat transfer to the pan in watts. (Give your answer in 3 significant digits.)
Answer:
Q = 20.22 x 10³ W = 20.22 KW
Explanation:
First we need to find the volume of water dropped.
Volume = V = πr²h
where,
r = radius of pan = 30.2 cm/2 = 15.1 cm = 0.151 m
h = height drop = 1.45 cm = 0.0145 m
Therefore,
V = π(0.151 m)²(0.0145 m)
V = 1.038 x 10⁻³ m³
Now, we find the mass of the water that is vaporized.
m = ρV
where,
m = mass = ?
ρ = density of water = 1000 kg/m³
Therefore,
m = (1000 kg/m³)(1.038 x 10⁻³ m³)
m = 1.038 kg
Now, we calculate the heat required to vaporize this amount of water.
q = mH
where,
H = Heat of vaporization of water = 22.6 x 10⁵ J/kg
Therefore,
q = (1.038 kg)(22.6 x 10⁵ J/kg)
q = 23.46 x 10⁵ J
Now, for the rate of heat transfer:
Rate of Heat Transfer = Q = q/t
where,
t = time = (18.6 min)(60 s/1 min) = 1116 s
Therefore,
Q = (23.46 x 10⁵ J)/1116 s
Q = 20.22 x 10³ W = 20.22 KW
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.
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.
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].
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:
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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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 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
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]
How many components do vectors have, and what are they?
Answer:
There are two components for a two-dimensional coordinate system/vector.
Explanation:
For two-dimensional vectors, such as velocity, acceleraton, etc, there are two components, the x- and y-components.
These components could be rotated or translated, depending on the coordinate system.
Instead of rectangular cartesian system, the components could also be in the form of polar coordinates, such as radius and theta (angle).
For three-dimensional vectors, such as velocity in space, there are three components, in various coordinate systems.
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.
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]
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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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).
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]
Which characteristic gives the most information about what kind of element an atom is ?
Answer:
The atomic number
Explanation:
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
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.
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
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:
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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
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 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
Two parallel plates 0.800 cm apart are equally and oppositely charged. An electron is released from rest at the surface of the negative plate and simultaneously a proton is released from rest at the surface of the positive plate.
How far from the negative plate is the point at which the electron and proton pass each other?
Express your answer with the appropriate units.
Answer:
0.79 cm
Explanation:
The computation is shown below:-
Particle acceleration is
[tex]a = \frac{qE}{m}[/tex]
We will take d which indicates distance as from the negative plate, so the travel by proton is 0.800 cm - d at the same time
[tex]d = \frac{1}{2} a_et^2\\\\0.800 cm - d = \frac{1}{2} a_pt^2\\\\\frac{d}{0.800 cm - d} = \frac{a_e}{a_p} \\\\\frac{d}{0.800 cm - d} = \frac{m_p}{m_e} \\\\\frac{d}{0.800 cm - d} = \frac{1836m_e}{m_e}[/tex]
After solving the equation we will get 0.79 cm from the negative plate.
Therefore it is 0.79 cm far from the negative pate i.e the point at which the electron and proton pass each other
The point at which the electron and proton pass each other will be 0.79 cm.
What is the charge?When the matter is put in an electromagnetic field, it has an electric charge, which causes it to experience a force. A positive or negative electric charge can exist.
The given data in the problem is;
d' is the distance between the two parallel plates= 0.800 cm
The acceleration is given as;
[tex]\rm a= \frac{qE}{m} \\\\[/tex]
The distance from Newton's law is found as;
[tex]d = ut+\frac{1}{2} at^2 \\\\ u=0 \\\\ d= \frac{1}{2} at^2 \\\\ d-d' = \frac{1}{2} a_pt^2 \\\\ 0.800-d= \frac{1}{2} a_pt^2 \\\\\ \frac{d}{0.800-d} =\frac{a}{a_p} \\\\ \frac{d}{0.800-d} =\frac{m_p}{m} \\\\ \frac{d}{0.800-d} =\frac{1836m_e}{m_e} \\\\ d=0.79 \ cm[/tex]
Hence the point at which the electron and proton pass each other will be 0.79 cm.
To learn more about the charge refer to the link;
https://brainly.com/question/24391667