Answer:
A) V = -136.36 V , B) V = 4.85 10³ V , C) V = 1.62 10⁴ V
Explanation:
To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is
V = k ∑ [tex]q_{i} / r_{i}[/tex]
where [tex]q_{i}[/tex] and [tex]r_{i}[/tex] are the loads and the point distances.
A) We apply this equation to our case
V = k (q₁ / r₁ + q₂ / r₂)
They ask us for the potential at the midpoint of separation
r = 3.30 / 2 = 1.65 m
this distance is much greater than the radius of the spheres
let's calculate
V = 9 10⁹ (1.1 10⁻⁸ / 1.65 + (-3.6 10⁻⁸) / 1.65)
V = 9 10¹ / 1.65 (1.10 - 3.60)
V = -136.36 V
B) The potential at the surface sphere A
r₂ is the distance of sphere B above the surface of sphere A
r₂ = 3.30 -0.02 = 3.28 m
r₁ = 0.02 m
we calculate
V = 9 10⁹ (1.1 10⁻⁸ / 0.02 - 3.6 10⁻⁸ / 3.28)
V = 9 10¹ (55 - 1,098)
V = 4.85 10³ V
C) The potential on the surface of sphere B
r₂ = 0.02 m
r₁ = 3.3 -0.02 = 3.28 m
V = 9 10⁹ (1.10 10⁻⁸ / 3.28 - 3.6 10⁻⁸ / 0.02)
V = 9 10¹ (0.335 - 180)
V = 1.62 10⁴ V
In a single-slit diffraction experiment, the width of the slit through which light passes is reduced. What happens to the width of the central bright fringe
Answer:
It becomes wider
Explanation:
Because The bigger the object the wave interacts with, the more spread there is in the interference pattern. Decreasing the size of the opening increases the spread in the pattern.
A 34.8 kg runner has a kinetic energy of 1.09 x 10³ J. What is the speed of the runner?
Answer:
7.91 m/s
Explanation:
Speed: This can be defined as the ratio of this to time. Speed can also be the magnitude of a velocity or speed is velocity without direction.
The S.I unit of speed is m/s.
From the question,
K.E = 1/2(mv²)................ Equation 1
Where K.E = Kinetic Energy, m = mass of the runner, v = velocity of the runner.
make v the subject of the equation
v = √(2K.E/m).................Equation 2
Given: K.E = 1.09×10³ J, m = 34.8 kg.
Substitute into equation 2
v = √(2× 1.09×10³/34.8)
v = √(62.64)
v = 7.91 m/s.
Hence the speed of the runner = 7.91 m/s
The angle of incidence of a light beam in air onto a reflecting surface is continuously variable. The reflected ray is found to be completely polarized when the angle of incidence is 64.2° .What is the index of refraction of the reflecting material?
Answer:
n = 2.0686
Explanation:
When an unpolarized ray of light is reflected on a surface, the reflected ray is partially polarized, complete polarization occurs when it is true that between the transmitted and reflected ray one has 90, the relationship is
n = so tea
let's calculate
n = tan 64.2
n = 2.0686
You are at the carnival with you your little brother and you decide to ride the bumper cars for fun. You each get in a different car and before you even get to drive your car, the little brat crashes into you at a speed of 3 m/s.
A. Knowing that the bumper cars each weigh 80 kg, while you and your brother weigh 60 and 30 kg,respectively, write down the equations you need to use to figure out how fast you and your brother are moving after the collision.
B. After the collision, your little brother reverses direction and moves at 0.36 m/s. How fast are you moving after the collision?
C. Assuming the collision lasted 0.05 seconds, what is the average force exerted on you during the collision?
D. Who undergoes the larger acceleration, you or your brother? Explain.
Answer:
a) The equation is [tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
b) Your velocity after collision is 2.64 m/s
c) The force you felt is 7392 N
d) you and your brother undergo an equal amount of acceleration
Explanation:
Your mass [tex]m_{y}[/tex] = 60 kg
your brother's mass [tex]m_{b}[/tex] = 30 kg
mass of the car [tex]m_{c}[/tex] = 80 kg
your initial speed [tex]u_{y}[/tex] = 0 m/s (since you've not started moving yet)
your brother's initial velocity [tex]u_{b}[/tex] = 3 m/s
your final speed [tex]v_{y}[/tex] after collision = ?
your brother's final speed [tex]v_{b}[/tex] after collision = ?
a) equations you need to use to figure out how fast you and your brother are moving after the collision is
[tex](m_{y}+m_{c} )u_{y} + (m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
but [tex]u_{y}[/tex] = 0 m/s
the equation reduces to
[tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
b) if your little brother reverses with velocity of 0.36 m/s it means
[tex]v_{b}[/tex] = -0.36 m/s (the reverse means it travels in the opposite direction)
then, imputing values into the equation, we'll have
[tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
(30 + 80)3 = (60 + 80)[tex]v_{y}[/tex] + (30 + 80)(-0.36)
330 = 140[tex]v_{y}[/tex] - 39.6
369.6 = 140[tex]v_{y}[/tex]
[tex]v_{y}[/tex] = 369.6/140 = 2.64 m/s
This means you will also reverse with a velocity of 2.64 m/s
c) your initial momentum = 0 since you started from rest
your final momentum = (total mass) x (final velocity)
==> (60 + 80) x 2.64 = 369.6 kg-m/s
If the collision lasted for 0.05 s,
then force exerted on you = (change in momentum) ÷ (time collision lasted)
force on you = ( 369.6 - 0) ÷ 0.05 = 7392 N
d) you changed velocity from 0 m/s to 2.64 m/s in 0.05 s
your acceleration is (2.64 - 0)/0.05 = 52.8 m/s^2
your brother changed velocity from 3 m/s to 0.36 m/s in 0.05 s
his deceleration is (3 - 0.36)/0.05 = 52.8 m/s
you and your brother undergo an equal amount of acceleration. This is because you gained the momentum your brother lost
Consider a long rod of mass, m, and length, l, which is thin enough that its width can be ignored compared to its length. The rod is connected at its end to frictionless pivot.
a) Find the angular frequency of small oscillations, w, for this physical pendulum.
b) Suppose at t=0 it pointing down (0 = 0) and has an angular velocity of 120 (that is '(t = 0) = 20) Note that 20 and w both have dimensions of time-1. Find an expression for maximum angular displacement for the pendulum during its oscillation (i.e. the amplitude of the oscillation) in terms of 20 and w assuming that the angular displacement is small.
Answer:
Explanation:
The rod will act as pendulum for small oscillation .
Time period of oscillation
[tex]T=2\pi\sqrt{\frac{l}{g} }[/tex]
angular frequency ω = 2π / T
= [tex]\omega=\sqrt{\frac{g}{l} }[/tex]
b )
ω = 20( given )
velocity = ω r = ω l
Let the maximum angular displacement in terms of degree be θ .
1/2 m v ² = mgl ( 1 - cosθ ) ,
[ l-lcosθ is loss of height . we have applied law of conservation of mechanical energy .]
.5 ( ω l )² = gl( 1 - cos θ )
.5 ω² l = g ( 1 - cosθ )
1 - cosθ = .5 ω² l /g
cosθ = 1 - .5 ω² l /g
θ can be calculated , if value of l is given .
In the far future, astronauts travel to the planet Saturn and land on Mimas, one of its 62 moons. Mimas is small compared with the Earth's moon, with mass Mm = 3.75 ✕ 1019 kg and radius Rm = 1.98 ✕ 105 m, giving it a free-fall acceleration of g = 0.0636 m/s2. One astronaut, being a baseball fan and having a strong arm, decides to see how high she can throw a ball in this reduced gravity. She throws the ball straight up from the surface of Mimas at a speed of 41 m/s (about 91.7 mph, the speed of a good major league fastball).
(a) Predict the maximum height of the ball assuming g is constant and using energy conservation. Mimas has no atmosphere, so there is no air resistance.
(b) Now calculate the maximum height using universal gravitation.
(c) How far off is your estimate of part (a)? Express your answer as a percent difference and indicate if the estimate is too high or too low.
Answer:
a) h = 13,205.4 m
b) r_f = 2.12 106 m
c) e% = 0.68%
Explanation:
a) This is an exercise we are asked to use energy conservation,
Starting point. On the surface of Mimas
Em₀ = K = ½ m v²
Final point. Where the ball stops
[tex]Em_{f}[/tex] = U = m g h
Em₀ = Em_{f}
½ m v² = m g h
h = ½ v² / g
let's calculate
h = ½ 41² / 0.0636
h = 13,205.4 m
b) For this part we are asked to use the law of universal gravitation, write the energy
starting point. Satellite surface
Em₀ = K + U = ½ m v² - GmM / r_o
final point. Where the ball stops
[tex]Em_{f}[/tex]= U = - G mM / r_f
Em₀ = Em_{f}
½ m v² - G m M / r_o = - G mM / r_f
In this case all distances are measured from the center of the satellite
1 / rf = 1 / GM (-½ v² + G M / r_o)
let's calculate
1 / rf = 1 / (6.67 10⁻¹¹ 3.75 10¹⁹) (- ½ 41 2 + 6.67 10⁻¹¹ 3.75 10¹⁹ / 1.98 105)
1 / r_f = 3,998 10⁻¹¹(-840.5 + 12.63 10³)
1 / r_f = 4,714 10⁻⁷
r_f = 1 / 4,715 10⁻⁷
r_f = 2.12 106 m
to measure this distance from the satellite surface
r_f ’= r_f - r_o
r_f ’= 2.12 106 - 1.98 105
r_f ’= 1,922 106 m
c) the percentage difference is
e% = 13 205.4 / 1,922 106 100
e% = 0.68%
The estimate of part a is a little low
What is the length of the x-component of the vector shown below?
9
380
х
A. 5.5
B. 30.0
O O
O C. 7.1
O D. 8.6
Answer:
x-component = 7.1
Explanation:
x-component = 9cos38 = 7.09 =7.1
Answer:
7.1
Explanation:
use cos
A container with volume 1.83 L is initially evacuated. Then it is filled with 0.246 g of N2. Assume that the pressure of the gas is low enough for the gas to obey the ideal-gas law to a high degree of accuracy. If the root-mean-square speed of the gas molecules is 192 m/s, what is the pressure of the gas?
Answer:
The pressure is [tex]P = 1652 \ Pa[/tex]
Explanation:
From the question we are told that
The volume of the container is [tex]V = 1.83 \ L = 1.83 *10^{-3 } \ m^3[/tex]
The mass of [tex]N_2[/tex] is [tex]m_n = 0.246 \ g = 0.246 *10^{-3} \ kg[/tex]
The root-mean-square velocity is [tex]v = 192 \ m/s[/tex]
The root -mean square velocity is mathematically represented as
[tex]v = \sqrt{ \frac{3 RT}{M_n } }[/tex]
Now the ideal gas law is mathematically represented as
[tex]PV = nRT[/tex]
=> [tex]RT = \frac{PV}{n }[/tex]
Where n is the number of moles which is mathematically represented as
[tex]n = \frac{ m_n }{M }[/tex]
Where M is the molar mass of [tex]N_2[/tex]
So
[tex]RT = \frac{PVM_n }{m _n }[/tex]
=> [tex]v = \sqrt{ \frac{3 \frac{P* V * M_n }{m_n } }{M_n } }[/tex]
=> [tex]v = \sqrt{ \frac{ 3 * P* V }{m_n } } }[/tex]
=> [tex]P = \frac{v^2 * m_n}{3 * V }[/tex]
substituting values
=> [tex]P = \frac{( 192)^2 * 0.246 *10^{-3}}{3 * 1.83 *10^{-3} }[/tex]
=> [tex]P = 1652 \ Pa[/tex]
If a 950 kg merry-go-round platform of radius 4.5 meters is driven by a mechanism located 2.0 meters from its center of rotation, how much force must the mechanism provide to get the platform moving at 5.5 revolutions per minute after 12 seconds if it were initially at rest
Answer:
F = 213.75 N
Explanation:
First we need to calculate the angular acceleration of merry-go-round. For that purpose we use 1st equation of motion in angular form.
ωf = ωi + αt
where,
ωf=final angular velocity=(5.5 rev/min)(2π rad/1 rev)(1 min/60 s)=0.58 rad/s
ωi =initial angular velocity = 0 rad/s
t = time = 12 s
α = angular acceleration = ?
Therefore,
0.58 rad/s = 0 rad/s + α(12 s)
α = (0.58 rad/s)/(12 s)
α = 0.05 rad/s²
Now, we shall find the linear acceleration of the merry-go-round:
a = rα
where,
a = linear acceleration = ?
r = radius = 4.5 m
Therefore,
a = (4.5 m)(0.05 rad/s²)
a = 0.225 m/s²
Now, the force is given by Newton;s 2nd Law:
F = ma
where,
F = Force = ?
m = mass pf merry-go-round = 950 kg
Therefore,
F = (950 kg)(0.225 m/s²)
F = 213.75 N
A truck accidentally rolls down a driveway for 8.0\,\text m8.0m8, point, 0, start text, m, end text while a person pushes against the truck with a force of 850\,\text N850N850, start text, N, end text to bring it to a stop. What is the change in kinetic energy for the truck?
Answer:
Explanation:
According to work energy theorem
change in kinetic energy of truck = work done against it
work done against it = force x displacement
= - 850 x 8 = 6800 J
change in kinetic energy of truck = - 6800 J .
energy will be reduced by 6800 J
Answer:-6800
Explanation:
wo parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area of 5.30 cm2 and plate separation of 2.65 mm. However, the first capacitor C1 is filled with air, while the second capacitor C2 is filled with a dielectric that has a dielectric constant of 2.10. (a) What is the charge stored on each capacitor
Complete Question
Two parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area of 5.30 cm2 and plate separation of 2.65 mm. However, the first capacitor C1 is filled with air, while the second capacitor C2 is filled with a dielectric that has a dielectric constant of 2.10.
(a) What is the charge stored on each capacitor
(b) What is the total charge stored in the parallel combination?
Answer:
a
i [tex]Q_1 = 2.124 *10^{-11} \ C[/tex]
ii [tex]Q_2 = 4.4604 *10^{-11} \ C[/tex]
b
[tex]Q_{eq} = 6.5844 *10^{-11} \ C[/tex]
Explanation:
From the question we are told that
The voltage of the battery is [tex]V = 12.0 \ V[/tex]
The plate area of each capacitor is [tex]A = 5.30 \ cm^2 = 5.30 *10^{-4} \ m^2[/tex]
The separation between the plates is [tex]d = 2.65 \ mm = 2.65 *10^{-3} \ m[/tex]
The permittivity of free space has a value [tex]\epsilon_o = 8.85 *10^{-12} \ F/m[/tex]
The dielectric constant of the other material is [tex]z = 2.10[/tex]
The capacitance of the first capacitor is mathematically represented as
[tex]C_1 = \frac{\epsilon * A }{d }[/tex]
substituting values
[tex]C_1 = \frac{8.85 *10^{-12 } * 5.30 *10^{-4} }{2.65 *10^{-3} }[/tex]
[tex]C_1 = 1.77 *10^{-12} \ F[/tex]
The charge stored in the first capacitor is
[tex]Q_1 = C_1 * V[/tex]
substituting values
[tex]Q_1 = 1.77 *10^{-12} * 12[/tex]
[tex]Q_1 = 2.124 *10^{-11} \ C[/tex]
The capacitance of the second capacitor is mathematically represented as
[tex]C_2 = \frac{ z * \epsilon * A }{d }[/tex]
substituting values
[tex]C_1 = \frac{ 2.10 *8.85 *10^{-12 } * 5.30 *10^{-4} }{2.65 *10^{-3} }[/tex]
[tex]C_1 = 3.717 *10^{-12} \ F[/tex]
The charge stored in the second capacitor is
[tex]Q_2 = C_2 * V[/tex]
substituting values
[tex]Q_2 = 3.717*10^{-12} * 12[/tex]
[tex]Q_2 = 4.4604 *10^{-11} \ C[/tex]
Now the total charge stored in the parallel combination is mathematically represented as
[tex]Q_{eq} = Q_1 + Q_2[/tex]
substituting values
[tex]Q_{eq} = 4.4604 *10^{-11} + 2.124*10^{-11}[/tex]
[tex]Q_{eq} = 6.5844 *10^{-11} \ C[/tex]
A tornado passes in front of a building, causing the pressure to drop there by 25% in 1 second. Part A If a door on the side of the building is 8.1 feet tall and 3 feet wide, what is the net force on the closed door.
Answer:
F_net = 264, 26 pound
Explanation:
For this exercise we use Newton's second law
F_net = F_int - F_outside
where the force can be found from the definition of pressure
P = F / A
F = P A
we substitute
F_net = P_inside A - P_outside A
F_net = A (P_inside - P_outside)
indicate that the pressure on the outside is 25% less than the pressure on the inside
P_outside = 0.25 P_inside
The area is
A = L W
we substitute
F_net = L W P_inside (1-0.25)
let's calculate
suppose the pressure inside is atmospheric pressure
P_inside= P_atmospheric = 1,013 10⁵ Pa = 14.7 PSI
F_net = 8.1 3 14.5 0.75
F_net = 264, 26 pound
A mass M = 4 kg attached to a string of length L = 2.0 m swings in a horizontal circle with a speed V. The string maintains a constant angle \theta\:=\:θ = 35.4 degrees with the vertical line through the pivot point as it swings. What is the speed V required to make this motion possible?
Answer:
The velocity is [tex]v = 2.84 1 \ m/s[/tex]
Explanation:
The diagram showing this set up is shown on the first uploaded image (reference Physics website )
From the question we are told that
The mass is m = 4 kg
The length of the string is [tex]L = 2.0 \ m[/tex]
The constant angle is [tex]\theta = 35.4 ^o[/tex]
Generally the vertical forces acting on the mass to keep it at equilibrium vertically is mathematically represented as
[tex]Tcos (\theta ) - mg = 0[/tex]
=> [tex]mg = Tcos (\theta )[/tex]
Now let the force acting on mass horizontally be k so from SOHCAHTOA rule
[tex]sin (\theta ) = \frac{k }{T}[/tex]
=> [tex]k = T sin \theta[/tex]
Now this k is also equivalent to the centripetal force acting on the mass which is mathematically represented as
[tex]F_v = \frac{m v^2}{r}[/tex]
So
[tex]k = F_v[/tex]
Which
=> [tex]T sin \theta= \frac{ m v^2}{ r }[/tex]
So
[tex]\frac{Tsin (\theta )}{Tcos (\theta )} = \frac{mg}{ \frac{mv^2}{r} }[/tex]
=> [tex]Tan (\theta ) = \frac{v^2}{ r * g }[/tex]
=> [tex]v = \sqrt{r * g * tan (\theta )}[/tex]
Now the radius is evaluated using SOHCAHTOA rule as
[tex]sin (\theta) = \frac{ r}{L}[/tex]
=> [tex]r = L sin (\theta)[/tex]
substituting values
[tex]r = 2 sin ( 35.4 )[/tex]
[tex]r = 1.1586 \ m[/tex]
So
[tex]v = \sqrt{1.1586* 9.8 * tan (35.4 )}[/tex]
[tex]v = 2.84 1 \ m/s[/tex]
A spherical balloon is inflated with gas at the rate of 500 cubic centimeters per minute. How fast is the radius of the balloon increasing at the instant the radius is 40 centimeters
Answer:
0.245cm/minExplanation:
The volume of the spherical balloon is expressed as V = 4/3πr³ where r is the radius of the spherical balloon. If the spherical balloon is inflated with gas at the rate of 500 cubic centimetres per minute then dV/dt = 500cm³.
Using chain rule to express dV/dt;
dV/dt = dV/dr*dr/dt
dr/dt is the rate at which the radius of the gallon is increasing.
From the formula, dV/dr = 3(4/3πr^3-1))
dV/dr = 4πr²
dV/dt = 4πr² *dr/dt
500 = 4πr² *dr/dt
If radius r = 40;
500 = 4π(40)² *dr/dt
500 = 6400π*dr/dt
dr/dt = 500/6400π
dr/dt = 5/64π
dr/dt = 0.245cm/min
Hence, the radius of the balloon is increasing at the rate of 0.245cm/min
Three cars (car 1, car 2, and car 3) are moving with the same velocity when the driver suddenly slams on the brakes, locking the wheels. The most massive car is car 1, the least massive is car 3, and all three cars have identical tires. For which car does friction do the largest amount of work in stopping the car
Answer:
Car 3
Explanation:
A coil of area 0.320 m2 is rotating at 100 rev/s with the axis of rotation perpendicular to a 0.430 T magnetic field. If the coil has 700 turns, what is the maximum emf generated in it
Answer:
The maximum emf generated in the coil is 60527.49 V
Explanation:
Given;
area of coil, A = 0.320 m²
angular frequency, f = 100 rev/s
magnetic field, B = 0.43 T
number of turns, N = 700 turns
The maximum emf generated in the coil is calculated as,
E = NBAω
where;
ω is the angular speed = 2πf
E = NBA(2πf)
Substitute in the given values and solve for E
E = 700 x 0.43 x 0.32 x 2π x 100
E = 60527.49 V
Therefore, the maximum emf generated in the coil is 60527.49 V
How much work is required to carry an electron from positive terminal of 12Volt battery to negative terminal?
Answer:
Explanation:
Work required = q x V
where q is charge on electron and V is potential difference
= 1.6 x 10⁻¹⁹ x 12
= 19.2 x 10⁻¹⁹ J
A sailor strikes the side of his ship just below the surface of the sea. He hears the echo of the wave reflected from the ocean floor directly below 2.5 ss later.
How deep is the ocean at this point? (Note: Use the bulk modulus method to determine the speed of sound in this fluid, rather than using a tabluated value.)
_____ m
Answer:
1248m
The time that wave moves from the wave source to the ocean floor is half the total travel time: t = 2.5/2 = 1.25s
The speed of sound in seawater is 1560 m/s
Therefore, s = vt = (1560 m/s)(1.25s) =1248 m = 1.2km
A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.
A. Assume that the collision is perfectly elastic, what will be the speed of the 0.300 kg object after the collision?
B. What will be the direction of the 0.300 kg object after the collision?
C. What will be the speed of the 0.900 kg object after the collision?
Answer:
a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.
Explanation:
a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:
Momentum
[tex]m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}[/tex]
Energy
[tex]\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})[/tex]
[tex]m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}[/tex]
Where:
[tex]m_{1}[/tex], [tex]m_{2}[/tex] - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.
[tex]v_{1,o}[/tex], [tex]v_{2,o}[/tex] - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.
If [tex]m_{1} = 0.400\,kg[/tex], [tex]m_{2} = 0.900\,kg[/tex], [tex]v_{1,o} = +5.86\,\frac{m}{s}[/tex], [tex]v_{2,o} = 0\,\frac{m}{s}[/tex], the system of equation is simplified as follows:
[tex]2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}[/tex]
[tex]13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}[/tex]
Let is clear [tex]v_{1,f}[/tex] in first equation:
[tex]0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}[/tex]
[tex]v_{1,f} = 5.86-2.25\cdot v_{2,f}[/tex]
Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:
[tex]13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}[/tex]
[tex]13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}[/tex]
[tex]13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}[/tex]
[tex]2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0[/tex]
[tex]2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0[/tex]
There are two solutions:
[tex]v_{2,f} = 0\,\frac{m}{s}[/tex] or [tex]v_{2,f} = 3.606\,\frac{m}{s}[/tex]
The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.
Now, the final speed of the 0.400-kg puck is: ([tex]v_{2,f} = 3.606\,\frac{m}{s}[/tex])
[tex]v_{1,f} = 5.86-2.25\cdot (3.606)[/tex]
[tex]v_{1,f} = -2.254\,\frac{m}{s}[/tex]
The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.
b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.
c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.
Two small plastic spheres are given positive electrical charges. When they are 20.0 cm apart, the repulsive force between them has magnitude 0.200 N.
1. What is the charge on each sphere if the two charges are equal? (C)
2. What is the charge on each sphere if one sphere has four times the charge of the other? (C)
Answer:
A. 2.97x 10^-6C
B. 1.48x10^ -6 C
Explanation:
Pls see attached file
Answer:
1) +9.4 x 10^-7 C
2) +4.72 x 10^-7 C and +1.9 x 10^-6 C
Explanation:
The two positive charges will repel each other
Repulsive force on charges = 0.200 N
distance apart = 20.0 cm = 0.2 m
charge on each sphere = ?
Electrical force on charged spheres at a distance is given as
F = [tex]\frac{kQq}{r^{2} }[/tex]
where F is the force on the spheres
k is the Coulomb's constant = 8.98 x 10^9 kg⋅m³⋅s⁻²⋅C⁻²
Q is the charge on of the spheres
q is the charge on the other sphere
r is their distance apart
since the charges are equal, i.e Q = q, the equation becomes
F = [tex]\frac{kQ^{2} }{r^{2} }[/tex]
making Q the subject of the formula
==> Q = [tex]\sqrt{\frac{Fr^{2} }{k} }[/tex]
imputing values into the equation, we have
Q = [tex]\sqrt{\frac{0.2*0.2^{2} }{8.98*10^{9} } }[/tex] = +9.4 x 10^-7 C
If one charge has four times the charge on the other, then
charge on one sphere = q
charge on the other sphere = 4q
product of both charges = [tex]4q^{2}[/tex]
we then have
F = [tex]\frac{4kq^{2} }{r^{2} }[/tex]
making q the subject of the formula
==> q = [tex]\sqrt{\frac{Fr^{2} }{4k} }[/tex]
imputing values into the equation, we have
q = [tex]\sqrt{\frac{0.2*0.2^{2} }{4*8.98*10^{9} } }[/tex] = +4.72 x 10^-7 C
charge on other sphere = 4q = 4 x 4.72 x 10^-7 = +1.9 x 10^-6 C
A computer has a mass of 3 kg. What is the weight of the computer?
A. 288 N.
B. 77.2 N
C. 3N
D. 29.4 N
Answer:
29.4 NOption D is the correct option.
Explanation:
Given,
Mass ( m ) = 3 kg
Acceleration due to gravity ( g ) = 9.8 m/s²
Weight ( w ) = ?
Now, let's find the weight :
[tex]w \: = \: m \times g[/tex]
plug the values
[tex] = 3 \times 9.8[/tex]
Multiply the numbers
[tex] = 29.4 \: [/tex] Newton
Hope this helps!!
best regards!!
An astronomy student, for her PhD, really needs to estimate the age of a cluster of stars. Which of the following would be part of the process she would follow?
A. plot an H-R diagram for the stars in the cluster
B. count the number of M type stars in the cluster
C. measure the Doppler shift of a number of the stars in the cluster
D. search for planets like Jupiter around the stars in the center of the cluster
E. search for x-rays coming from the center of the cluster
Answer:
A. plot an H-R diagram for the stars in the cluster.
Explanation:
A star cluster can be defined as a constellation of stars, due to gravitational force, which has the same origin.
The astronomy student would have to plot an H-R diagram for the stars in the cluster and determine the age of the cluster by observing the turn-off point. The turn-off is majorly as a result of gradual depletion of the source of energy of the star. Thus, it projects off the constellation.
g Power is _________________the force required to push something the work done by a system the speed of an object the rate that the energy of a system is transformed the energy of a system
Answer:
the rate that the energy of a system is transformed
Explanation:
We can define energy as the capacity or ability to do work. Power is defined as the rate of doing work or the rate at which energy is transformed. It can also be regarded as the time rate of energy transfer. In older physics literature, power is sometimes referred to as activity.
Power is given by energy/time. Its unit is watt which is defined as joule per second. Another popular unit of power is horsepower. 1 horsepower = 746 watts.
Very large magnitude of power is measured in killowats and megawatts.
Estimate the radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 6.5 cm from the center of the bulb. Assume that light is completely absorbed.
..........................................................
The radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 6.5 cm from the center of the bulb and the light is completely absorbed is 1.5707x10⁻⁶ N/m².
What is the Radiation pressure?Radiation pressure was defined as the mechanical pressure exerted upon any surface due to the exchange of momentum between the object and the electromagnetic field.
Radiation pressure always includes the Momentum of light or electromagnetic radiation of any wavelength that can be absorbed, reflected, or otherwise emitted by matter on any scale.
E.g: Black-body radiation
Given the values are,
Wattage of bulb = W = 25 W
distance = d = 6.5 cm = 0.065 m
To know the Radiation Pressure,
It can be given by
P = I/c
Where, c = 299792458 m/s is the speed of light,
I is the intensity of radiation and given by
I = W/4πd²
Where W is the Wattage of bulb and d is the distance
I = 25/4π*0.065²
I = 470.872 w/m²
so, the radiation pressure becomes
P = 470.872/299792458
P = 1.5707x10⁻⁶ N/m²
Therefore, the radiation pressure due to a 25 W bulb at a distance of 6.5 cm is 1.5707x10⁻⁶ N/m²
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A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 26 m/s when a 60 kg skydiver drops out by releasing his grip on the glider.
What is the glider's speed just after the skydiver lets go?
Answer:
The glider’s speed after the skydiver lets go is 26 m/s
Explanation:
To calculate the glider’s speed just after the skydiver lets go, we will need to use the conservation of momentum
Mathematically;
mv = mv + mv
so 680 * 26 = (680-60)v + 60 * 26
17680 = 620v + 1560
17680-1560 = 620v
16120 = 620v
v = 16120/620
v = 26 m/s
7. How many 1.00 µF capacitors must be connected in parallel to store a charge of 1.00 C with a potential of 110 V across the capacitors?
Answer:
q = C V charge on 1 capacitor
q = 1 * 10E-6 * 110 = 1.1 * 10E-4 C per capacitor
N = Q / q = 1 / 1.1 * 10E-4 = 9091 capacitors
9.09 × 10³ capacitors must be connected in parallel.
How to calculate the number of capacitors connected in parallel?
Given C = 1.00μF = 1 × 10⁻⁶ F
q = 1.00 C
V = 110 V
The equivalent capacitance is given by
Ceq = q/V
where q = total charge on all the capacitors
V = potential difference
For N number of identical capacitors in parallel,
Ceq = NC
Therefore,
NC = q/V
N = q/VC
Putting on the values in the above formula,
N = 1/ (110)(1 × 10⁻⁶)
= 1 / 110 × 10⁻⁶
= 9.09 × 10³
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At a certain instant the current flowing through a 5.0-H inductor is 3.0 A. If the energy in the inductor at this instant is increasing at a rate of 3.0 J/s, how fast is the current changing
Answer:
The current is changing at the rate of 0.20 A/s
Explanation:
Given;
inductance of the inductor, L = 5.0-H
current in the inductor, I = 3.0 A
Energy stored in the inductor at the given instant, E = 3.0 J/s
The energy stored in inductor is given as;
E = ¹/₂LI²
E = ¹/₂(5)(3)²
E = 22.5 J/s
This energy is increased by 3.0 J/s
E = 22.5 J/s + 3.0 J/s = 25.5 J/s
Determine the new current at this given energy;
25.5 = ¹/₂LI²
25.5 = ¹/₂(5)(I²)
25.5 = 2.5I²
I² = 25.5 / 2.5
I² = 10.2
I = √10.2
I = 3.194 A/s
The rate at which the current is changing is the difference between the final current and the initial current in the inductor.
= 3.194 A/s - 3.0 A/s
= 0.194 A/s
≅0.20 A/s
Therefore, the current is changing at the rate of 0.20 A/s.
The rate at which the current is changing is;
di/dt = 0.2 A/s
We are given;
Inductance; L = 5 H
Current; I = 3 A
Rate of Increase of energy; dE/dt = 3 J/s
Now, the formula for energy stored in inductor is given as;
E = ¹/₂LI²
Since we are looking for rate at which current is changing, then we differentiate both sides of the energy equation to get;
dE/dt = LI (di/dt)
Plugging in the relevant values gives;
3 = (5 × 3)(di/dt)
di/dt = 3/(5 × 3)
di/dt = 0.2 A/s
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At a fixed point, P, the electric and magnetic field vectors in an electromagnetic wave oscillate at angular frequency w . At what angular frequency does the Poynting vector oscillate at that point
Answer:
Poynting vector oscillate at a frequency of 2omega
Explanation:
This is because The poynting vector is proportional to the cross product of electric and magnetic field vectors. So Because both fields oscillate sinusoidally with frequency w, trigonometric identities show that their product is a sinusoidal function of frequency of 2w.
You should have observed that there are some frequencies where the output is stronger than the input. Discuss how that is even possible from a conservation of energy standpoint. Also, can you relate this behavior to the transient (natural) response of the circuit that you observed in the previous lab
Answer:
w = √ 1 / CL
This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence
Explanation:
This problem refers to electrical circuits, the circuits where this phenomenon occurs are series RLC circuits, where the resistor, the capacitor and the inductance are placed in series.
In these circuits the impedance is
X = √ (R² + ([tex]X_{C}[/tex] -[tex]X_{L}[/tex])² )
where Xc and XL is the capacitive and inductive impedance, respectively
X_{C} = 1 / wC
X_{L} = wL
From this expression we can see that for the resonance frequency
X_{C} = X_{L}
the impedance of the circuit is minimal, therefore the current and voltage are maximum and an increase in signal intensity is observed.
This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence
V = IR
Since the contribution of the two other components is canceled, this occurs for
X_{C} = X_{L}
1 / wC = w L
w = √ 1 / CL
A charged particle is moving with speed v perpendicular to a uniform magnetic field. A second identical charged particle is moving with speed 2v perpendicular to the same magnetic field. If the frequency of revolution of the first particle is f, the frequency of revolution of the second particle is
Answer:
the frequency of revolution of the second particle is f
Explanation:
centripetal force is balanced by the magnetic force for object under magnetic field is given as
Mv²/r= qvB
But v= omega x r
Omega= 2pi x f
f= qB/2pi x M
So since frequency does not depend on the velocity.therefore the frequency of revolution of the second particle remains the same and its equal to f