Answer:
Ф = 28.9°
Explanation:
given:
radius (r) = 117m
velocity (v) = 25.1 m/s
required: angle Ф
Ф = inv tan (v² / (r * g)) we know that g = 9.8
Ф = inv tan (25.1² / (117 * 9.8))
Ф = 28.9°
A player is positioned 35 m[40 degrees W of S] of the net. He shoot the puck 25 m [E] to a teammate. What second displacement does the puck have to travel in order to make it to the net?
Answer:
x=22.57 m
Explanation:
Given that
35 m in W of S
angle = 40 degrees
25 m in east
From the diagram
The angle
[tex]\theta=90-40=50^o[/tex]
From the triangle OAB
[tex]cos40^o=\frac{35^2+25^2-x^2}{2\times 35\times 25}[/tex]
[tex]1340.57=35^2+25^2-x^2[/tex]
x=22.57 m
Therefore the answer of the above problem will be 22.57 m
Muons are elementary particles that are formed high in the atmosphere by the interactions of cosmic rays with atomic nuclei. Muons are radioactive and have average lifetimes of about two-millionths of a second. Even though they travel at almost the speed of light, they have so far to travel through the atmosphere that very few should be detected at sea level - at least according to classical physics. Laboratory measurements, however, show that muons in great number do reach the earth's surface. What is the explanation?
Answer:
Muons reach the earth in great amount due to the relativistic time dilation from an earthly frame of reference.
Explanation:
Muons travel at exceedingly high speed; close to the speed of light. At this speed, relativistic effect starts to take effect. The effect of this is that, when viewed from an earthly reference frame, their short half life of about two-millionth of a second is dilated. The dilated time, due to relativistic effects on time for travelling at speed close to the speed of light, gives the muons an extended relative travel time before their complete decay. So in reality, the muon do not have enough half-life to survive the distance from their point of production high up in the atmosphere to sea level, but relativistic effect due to their near-light speed, dilates their half-life; enough for them to be found in sufficient amount at sea level.
Two carts are connected by a loaded spring on a horizontal, frictionless surface. The spring is released and the carts push away from each other. Cart 1 has mass M and Cart 2 has mass M/3.
a) Is the momentum of Cart 1 conserved?
Yes
No
It depends on M
b) Is the momentum of Cart 2 conserved?
Yes
No
It depends on M
c) Is the total momentum of Carts 1 and 2 conserved?
Yes
No
It depends on M
d) Which cart ends up moving faster?
Cart 1
Cart 2
They move at the same speed
e) If M = 6 kg and Cart 1 moves with a speed of 16 m/s, what is the speed of Cart 2?
0 m/s
4.0 m/s
5.3 m/s
16 m/s
48 m/s
64 m/s
Answer:
a) yes
b) no
c) yes
d)Cart 2 with mass [tex]\frac{M}{3}[/tex] is expected to be more faster
e) u₂ = 48 m/s
Explanation:
a) the all out linear momentum of an arrangement of particles of Cart 1 not followed up on by external forces is constant.
b) the linear momentum of Cart 2 will be acted upon by external force by Cart 1 with mass M, thereby it's variable and the momentum is not conserved
c) yes, the momentum is conserved because no external force acted upon it and both Carts share the same velocity after the reaction
note: m₁u₁ + m₂u₂ = (m₁ + m₂)v
d) Cart 2 with mass [tex]\frac{M}{3}[/tex] will be faster than Cart 1 because Cart 2 is three times lighter than Cart 1.
e) Given
m₁= M
u₁ = 16m/s
m₂ =[tex]\frac{M}{3}[/tex]
u₂ = ?
from law of conservation of momentum
m₁u₁= m₂u₂
M× 16 = [tex]\frac{M}{3}[/tex] × u₂(multiply both sides by 3)
therefore, u₂ = [tex]\frac{3(M .16)}{M}[/tex] ("." means multiplication)
∴u₂ = 3×16 = 48 m/s
The density of a sample of metal was measured to be 8.91 g/cm3. An X-ray diffraction experiment measures the edge of a face-centered cubic cell as 352.4 pm. Part APart complete What is the atomic weight of the metal
Answer:
The atomic weight of the metal is 58.7 g/mol
Explanation:
Given;
density of the metal sample, ρ = 8.91 g/cm³
edge length of the face centered cubic cell, α = 352.4 pm = 352.4 x 10⁻¹⁰ cm
Volume of the unit cell of the metal;
V = α³
V = (352.4 x 10⁻¹⁰ cm)³
V = 4.376 x 10⁻²³ cm³
Mass of the metal in unit cell
mass = density x volume
mass = 8.91 g/cm³ x 4.376 x 10⁻²³ cm³
mass = 3.899 x 10⁻²² g
Atomic weight, based on 4 atoms per unit cell;
4 atoms = 3.899 x 10⁻²² g
6.022 x 10²³ atoms = ?
= (6.022 x 10²³atoms x 3.899 x 10⁻²² g) / (4 atoms)
= 58.699 g/mol
= 58.7 g/mol (this metal is Nickel)
Theerefore, the atomic weight of the metal is 58.7 g/mol
If, the limits of the visible spectrum are approximately 3000 A.U. and 5000 A.U. respectively. Determine the angular breadth of the first order visible spectrum produced by a plane diffraction grating having 12000 lines per inch when light is incident normally on the grating.
Answer:
θ₁ = 0.04º , θ₂ = 0.00118º
Explanation:
The equation that describes the diffraction pattern of a network is
d sin θ = m λ
where the diffraction order is, in this case they indicate that the order
m = 1
θ = sin⁻¹ (λ / d)
Trfuvsmod ls inrsd fr ll red s SI units
d = 12000 line / inc (1 inc / 2.54cm) = 4724 line / cm
the distance between two lines we can look for it with a direct proportions rule
If there are 4724 lines in a centimeter, the distance for two hundred is
d = 2 lines (1 cm / 4724 line) = 4.2337 10⁻⁴ cm
let's calculate the angles
λ = 300 10-9 m
θ₁ = sin⁻¹ (300 10-9 / 4,2337 10-4)
θ₁ = sin⁻¹ (7.08 10-4)
θ₁ = 0.04º
λ = 5000
θ₂ = sin-1 (500 10-9 / 4,2337 10-4)
θ₂ = 0.00118º
11. A seesaw sits in static equilibrium. A child with a mass of 30 kg sits 1 m away from a pivot point. Another child sits 0.75 m away from the pivot point on the opposite side. The second child's mass is _____ kg.
Answer:
40 kgExplanation:
Find the diagram relating to the question for proper explanation of the question below.
Using the principle of moment
Sum of clockwise moments = Sum of anticlockwise moments
Moment = Force * perpendicular distance
For anti-clockwise moment:
Since the 30 kg moves in the anticlockwise direction according to the diagram
ACW moment = 30 * 1 = 30 kgm
For clockwise moment
If another child sits 0.75 m away from the pivot point on the opposite side, moment of the child in clockwise direction = M * 0.75 = 0.75M (M is the mass of the unknown child).
Equating both moments we have;
0.75M = 30
M = 30/0.75
M = 40 kg
The second child's mass is 40 kg
What will be the volume and density of stone if mass of stone is 10 gram .please tell the answer fast it's very urgent I will mark as a brain me answer if you will answer it correct.
Answer:
[tex]\large \boxed{\text{3.3 cm}^{3}}[/tex]
Explanation:
Assume the stone consists of basalt, which has a density of 3.0 g/cm³.
[tex]\rho = \text{10 g}\times\dfrac{\text{1 cm}^{3}}{\text{3.0 g}} = \text{3.3 cm}^{3}\\\\\text{The volume of the stone is $\large \boxed{\textbf{3.3 cm}^{3}}$}[/tex]
can I get help please?
what is the preferred method of using percentage data by using a circle divided into sections
Answer:
A pie chart is a type of graph in which a circle is divided into sectors that each represents a proportion of the whole
Explanation:
pie charts are a useful way to organize data in order to see the size of components relative to the whole.
At t = 0, a battery is connected to a series arrangement of a resistor and an inductor. If the inductive time constant is 36.0 ms, at what time is the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field?
Answer:
The rate at which energy is dissipated in the resistor is equal to the rate at which energy is stored in the inductor's magnetic field in 24.95 ms.
Explanation:
The energy stored in the inductor is given as
E₁ = ½LI²
The rate at which energy is stored in the inductor is
(dE₁/dt) = (d/dt) (½LI²)
Since L is a constant
(dE₁/dt) = ½L × 2I (dI/dt) = LI (dI/dt)
(dE₁/dt) = LI (dI/dt)
Rate of Energy dissipated in a resistor = Power = I²R
(dE₂/dt) = I²R
When the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field
(dE₁/dt) = (dE₂/dt)
OK (dI/dt) = I²R
L (dI/dt) = IR
Current in a this kind of series setup of inductor and resistor at any time, t, is given as
I = (V/R) (1 - e⁻ᵏᵗ)
k = (1/time constant) = (R/L)
(dI/dt) = (kV/R) e⁻ᵏᵗ = (RV/RL) e⁻ᵏᵗ = (V/L) e⁻ᵏᵗ
L (dI/dt) = IR
L [(V/L) e⁻ᵏᵗ] = R [(V/R) (1 - e⁻ᵏᵗ)
V e⁻ᵏᵗ = V (1 - e⁻ᵏᵗ)
e⁻ᵏᵗ = 1 - e⁻ᵏᵗ
2 e⁻ᵏᵗ = 1
e⁻ᵏᵗ = (1/2) = 0.5
e⁻ᵏᵗ = 0.5
In e⁻ᵏᵗ = In 0.5 = -0.69315
- kt = -0.69315
kt = 0.69315
k = (1/time constant)
Time constant = 36.0 ms = 0.036 s
k = (1/0.036) = 27.78
27.78t = 0.69315
t = (0.69315/27.78) = 0.02495 = 24.95 ms
Hope this Helps!!!
In a series RC circuit, the resistor voltage is 124 V and the capacitor voltage is 167 V. What is the total voltage
Answer:
208 V
Explanation:
resistor voltage = Vr = 124 V
capacitor voltage Vc = 167 V
the total voltage in the RC circuit is the resultant voltage of the resistor and the capacitor
total voltage i= [tex]\sqrt{Vr^{2} + Vc^{2} }[/tex]
==> [tex]\sqrt{124^{2} + 167^{2} } =[/tex] 208 V
Alternating Current In Europe, the voltage of the alternating current coming through an electrical outlet can be modeled by the function V 230 sin (100t), where tis measured in seconds and Vin volts.What is the frequency of the voltage
Answer:
[tex]\frac{50}{\pi }[/tex]Hz
Explanation:
In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;
V(t) = V sin (ωt + Ф) -----------------(i)
Where;
V = amplitude value of the voltage
ω = angular frequency = 2 π f [f = cyclic frequency or simply, frequency]
Ф = phase difference between voltage and current.
Now,
From the question,
V(t) = 230 sin (100t) ---------------(ii)
By comparing equations (i) and (ii) the following holds;
V = 230
ω = 100
Ф = 0
But;
ω = 2 π f = 100
2 π f = 100 [divide both sides by 2]
π f = 50
f = [tex]\frac{50}{\pi }[/tex]Hz
Therefore, the frequency of the voltage is [tex]\frac{50}{\pi }[/tex]Hz
A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string horizontal. At the lowest point in its swing when it is moving horizontally, the ball collides elastically with a 2.0-kg block initially at rest on a horizontal frictionless surface. What is the speed of the block just after the collision
Answer:
[tex]v_{2}=3.5 m/s[/tex]
Explanation:
Using the conservation of energy we have:
[tex]\frac{1}{2}mv^{2}=mgh[/tex]
Let's solve it for v:
[tex]v=\sqrt{2gh}[/tex]
So the speed at the lowest point is [tex]v=7 m/s[/tex]
Now, using the conservation of momentum we have:
[tex]m_{1}v_{1}=m_{2}v_{2}[/tex]
[tex]v_{2}=\frac{1*7}{2}[/tex]
Therefore the speed of the block after the collision is [tex]v_{2}=3.5 m/s[/tex]
I hope it helps you!
2. A pair of narrow, parallel slits sep by 0.25 mm is illuminated by 546 nm green light. The interference pattern is observed on a screen situated at 1.3 m away from the slits. Calculate the distance from the central maximum to the
Answer:
for the first interference m = 1 y = 2,839 10-3 m
for the second interference m = 2 y = 5,678 10-3 m
Explanation:
The double slit interference phenomenon, for constructive interference is described by the expression
d sin θ = m λ
where d is the separation between the slits, λ the wavelength and m an integer that corresponds to the interference we see.
In these experiments in general the observation screen is L >> d, let's use trigonometry to find the angles
tan θ = y / L
with the angle it is small,
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
the distance between the central maximum and an interference line is
y = m λ L / d
let's reduce the magnitudes to the SI system
λ = 546 nm = 546 10⁻⁹ m
d = 0.25 mm = 0.25 10⁻³ m
let's substitute the values
y = m 546 10⁻⁹ 1.3 / 0.25 10⁻³
y = m 2,839 10⁻³
the explicit value for a line depends on the value of the integer m, for example
for the first interference m = 1
the distance from the central maximum to the first line is y = 2,839 10-3 m
for the second interference m = 2
the distance from the central maximum to the second line is y = 5,678 10-3 m
Newton’s first law says that if motion changes, then a force is exerted. Describe a collision in terms of the forces exerted on both objects.
Answer:
In collision between equal-mass objects, each object experiences the same acceleration, because of equal force exerted on both objects.
Explanation:
In a collision two objects, there is a force exerted on both objects that causes an acceleration of both objects. These forces that act on both objects are equal in magnitude and opposite in direction.
Thus, in collision between equal-mass objects, each object experiences the same acceleration, because of equal force exerted on both objects.
please help In a video game, a ball moving at 0.6 meter/second collides with a wall. After the collision, the velocity of the ball changed to -0.4 meter/second. The collision takes 0.2 seconds to occur. What’s the acceleration of the ball during the collision? Use . a= v-u/t
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]
How fast must a 2500-kg elephant move to have the same kinetic energy as a 67.0-kg sprinter running at 15.0 m/s
Answer:
2.45 m/s
Explanation:
kinetic energy = 1/2 * m * v^2
then, 0.5 * 2500 * x^2 = 0.5 * 67 * 15^2
by solving for x, X = 2.45 m/s
A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, the mass is released from rest at x = 10.0 cm. ( That is, the spring is stretched by 10.0 cm.) (a) Determine the frequency of the oscillations. (b) Determine the maximum speed of the mass. Where dos the maximum speed occur? (c) Determine the maximum acceleration of the mass. Where does the maximum acceleration occur? (d) Determine the total energy of teh oscillating system. (e) Express the displacement as a function of time.
Answer:
(a) f = 0.58Hz
(b) vmax = 0.364m/s
(c) amax = 1.32m/s^2
(d) E = 0.1J
(e) [tex]x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)[/tex]
Explanation:
(a) The frequency of the oscillation, in a spring-mass system, is calulated by using the following formula:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex] (1)
k: spring constant = 20.0N/m
m: mass = 1.5kg
you replace the values of m and k for getting f:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz[/tex]
The frequency of the oscillation is 0.58Hz
(b) The maximum speed is given by:
[tex]v_{max}=\omega A=2\pi f A[/tex] (2)
A: amplitude of the oscillations = 10.0cm = 0.10m
[tex]v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}[/tex]
The maximum speed of the mass is 0.364 m/s.
The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.
(c) The maximum acceleration is given by:
[tex]a_{max}=\omega^2A=(2\pi f)^2 A[/tex]
[tex]a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}[/tex]
The maximum acceleration is 1.32 m/s^2
The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.
(d) The total energy of the system is:
[tex]E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J[/tex]
The total energy is 0.1J
(e) The displacement as a function of time is:
[tex]x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)[/tex]
Please Help!!! I WILL GIVE BRAINLIEST!!!! An electron is in motion at 4.0 × 10^6 m/s horizontally when it enters a region of space between two parallel plates, starting at the negative plate. The electron deflects downwards and strikes the bottom plate. The magnitude of the electric field between the plates is 4.0 x 10^2 N/C and separation between the charged plates is 2.0 cm. a.) Determine the horizontal distance traveled by the electron when it hits the plate. b.)Determine the velocity of the electron as it strikes the plate.
Answer:
Explanation:
Given that
speed u=4*10^6 m/s
electric field E=4*10^3 N/c
distance b/w the plates d=2 cm
basing on the concept of the electrostatices
now we find the acceleration b/w the plates
acceleration a=qE/m=1.6*10^-19*4*10^3/9.1*10^-31=0.7*10^15 =7*10^14 m/s
now we find the horizantal distance travelled by electrons hit the plates
horizantal distance X=u[2y/a]^1/2
=4*10^6[2*2*10^-2/7*10^14]^1/2
=3*10^-2=3 cm
now we find the velocity f the electron strike the plate
v^2-(4*10^6)^2=2*7*10^14*2*10^-2
v^2=16*10^12+28*10^12
v^2=44*10^12
speed after hits =>V=6.6*10^6 m/s
_____________ friction is the interlocking of surfaces due to irregularities on the surfaces preventing those surfaces from moving/sliding against each other. For surfaces moving/sliding on each other, ___________ friction overwhelms kinetic friction to that movement/sliding. Kinetic friction is alway larger than ____________ friction. Kinetic friction is alway equal to _________ friction.
Answer:
STATIC, STATIC
KINETIC friction is less than static friction
Explanation:
In this exercise you are asked to complete the sentences with the correct words.
STATIC friction prevents the relative movement of two surfaces in contact.
For moving surfaces the friction is STATIC is greater than the kinetic friction.
For the last two sentences I think they are misspelled, the correct thing is
KINETIC friction is less than static friction
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.75 A out of the junction. How many electrons per second move past a point in wire 3?
Answer:
number of electrons = 2.18*10^18 e
Explanation:
In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.
Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:
[tex]i_1+i_3=i_2[/tex] (1)
i1 = 0.40 A
i2 = 0.75 A
you solve the equation i3 from the equation (1):
[tex]i_3=i_2-i_1=0.75A-0.40A=0.35A[/tex]
Next, you take into account that 1A = 1C/s = 6.24*10^18
Then, you have:
[tex]0.35A=0.35\frac{C}{s}=0.35*\frac{6.24*10^{18}e}{s}=2.18*10^{18}\frac{e}{s}[/tex]
The number of electrons that trough the wire 3 is 2.18*10^18 e/s
Two spaceships are observed from earth to be approaching each other along a straight line. Ship A moves at 0.40c relative to the earth observer, while ship B moves at 0.60c relative to the same observer. What speed does the captain of ship A report for the speed of ship B
Answer:
0.80 c
Explanation:
The computation of speed is shown below:-
Here, The speed of the captain ship A report for speed of the ship B which is
[tex]S = \frac{S_A + S_B}{1 + \frac{(S_AS_B)}{c^2} }[/tex]
where
[tex]S_A[/tex] indicates the speed of the ship A
[tex]S_B[/tex] indicates the speed of the ship B
and
C indicates the velocity of life
now we will Substitute 0.40c for A and 0.60 for B in the equation which is
[tex]S = \frac{0.40c + 0.60c}{1 + \frac{(0.40c)(0.60c)}{c^2} }[/tex]
after solving the above equation we will get
0.80 c
So, The correct answer is 0.80c
A machinist is required to manufacture a circular metal disk with area 1300 cm2. (a) What radius produces such a disk
Answer:
Radius r = 20.34 cm
The radius that can produces such a disk is 20.34 cm
Explanation:
Area of a circle;
A = πr^2
A = area
r = radius
Making r the subject of formula;
r = √(A/π) ........1
Given;
A = 1300 cm^2
Substituting into the equation 1;
r = √(1300/π)
r = 20.34214472564 cm
r = 20.34 cm
The radius that can produces such a disk is 20.34 cm
a 5.0 charge is placed at the 0 cm mark of a meterstick and a -4.0 charge is placed at the 50 cm mark. what is the electric field at the 30 cm mark
Answer:
-1748*10^N/C
Explanation:
See attached file
Briefly describe the relationship between an equipotential surface and an electric field, and use this to explain why we will plot equipotential lines.
Answer:
E = - dV/dx
Explanation:
Las superficies equipòtenciales son superficie donde el potencial eléctrico es constante por lo cual nos podemos desplazaren ella sin realizar nigun trabajo.
El campo electrico es el campo que existen algún punto en el espacio creado por alguna ddistribucion de carga.
De los antes expuesto las dos magnitudes están relacionadas
E = - dV/dx
por lo cual el potenical es el gradiente del potencial eléctrico.
Como el campo eléctrico sobre un superficie equipotenciales constante, podemos colocar una punta de prueba con un potencial dado y seguir la linea que de una diferencia de potencial constar, lo cual permite visualizar las forma de cada linea equipotencial
Consider a conducting rod of length 31 cm moving along a pair of rails, and a magnetic field pointing perpendicular to the plane of the rails. At what speed (in m /s) must the sliding rod move to produce an emf of 0.75 V in a 1.75 T field?
Answer:
The speed of the rod is 1.383 m/s
Explanation:
Given;
length of the conducting rod, L = 31 cm = 0.31 m
induced emf on the rod, emf = 0.75V
magnetic field around the rod, B = 1.75 T
Apply the following Faraday's equation for electromagnetic induction in a moving rod to determine the speed of the rod.
emef = BLv
where;
B is the magnetic field
L is length of the rod
v is the speed of the rod
v = emf / BL
v = (0.75) / (1.75 x 0.31)
v = 1.383 m/s
Therefore, the speed of the rod is 1.383 m/s
If, instead, the ball is revolved so that its speed is 3.7 m/s, what angle does the cord make with the vertical?
Complete Question:
A 0.50-kg ball that is tied to the end of a 1.5-m light cord is revolved in a horizontal plane, with the cord making a 30° angle with the vertical.
(a) Determine the ball’s speed. (b) If, instead, the ball is revolved so that its
speed is 3.7 m/s, what angle does the cord make with the vertical?
(Check attached image for the diagram.)
Answer:
(a) The ball’s speed, v = 2.06 m/s
(b) The angle the cord makes with the vertical is 50.40⁰
Explanation:
If the ball is revolved in a horizontal plane, it will form a circular trajectory,
the radius of the circle, R = Lsinθ
where;
L is length of the string
The force acting on the ball is given as;
F = mgtanθ
This above is also equal to centripetal force;
[tex]mgTan \theta = \frac{mv^2}{R} \\\\Recall, R = Lsin \theta\\\\mgTan \theta = \frac{mv^2}{Lsin \theta}\\\\v^2 = glTan \theta sin \theta\\\\v = \sqrt{glTan \theta sin \theta} \\\\v = \sqrt{(9.8)(1.5)(Tan30)(sin30)} \\\\v = 2.06 \ m/s[/tex]
(b) when the speed is 3.7 m/s
[tex]v = \sqrt{glTan \theta sin \theta} \ \ \ ;square \ both \ sides\\\\v^2 = glTan \theta sin \theta\\\\v^2 = gl(\frac{sin \theta}{cos \theta}) sin \theta\\\\v^2 = \frac{gl*sin^2 \theta}{cos \theta} \\\\v^2 = \frac{gl*(1- cos^2 \theta)}{cos \theta}\\\\gl*(1- cos^2 \theta) = v^2cos \theta\\\\(9.8*1.5)(1- cos^2 \theta) = (3.7^2)cos \theta\\\\14.7 - 14.7cos^2 \theta = 13.69cos \theta\\\\14.7cos^2 \theta + 13.69cos \theta - 14.7 = 0 \ \ \ ; this \ is \ quadratic \ equation\\\\[/tex]
[tex]Cos\theta = \frac{13.69\sqrt{13.69^2 -(-4*14.7*14.7)} }{14.7} \\\\Cos \theta = 0.6374\\\\\theta = Cos^{-1}(0.6374)\\\\\theta = 50.40 ^o[/tex]
Therefore, the angle the cord makes with the vertical is 50.40⁰
A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct
Complete question:
A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct?
A. the interior field points in a direction parallel to the exterior field
B. There is no electric field on the interior of the conducting sphere.
C. The interior field points in a direction perpendicular to the exterior field.
D. the interior field points in a direction opposite to the exterior field.
Answer:
B. There is no electric field on the interior of the conducting sphere.
Explanation:
Conductors are said to have free charges that move around easily. When the conductor is now placed in a static electric field, the free charges react to attain electrostatic equilibrium (steady state).
Here, a solid conducting sphere is placed in an external uniform electric field. Until the lines of the electric field are perpendicular to the surface, the free charges will move around the spherical conductor, causing polarization. There would be no electric field in the interior of the spherical conductor because there would be movement of free charges in the spherical conductor in response to any field until its neutralization.
Option B is correct.
There is no electric field on the interior of the conducting sphere.
Particle A has charge qA and particle B has charge qB. When they are separated by a distance ri, they experience an attractive force Fi. The particles are moved without altering their charges. Now they experience an attractive force with a magnitude of 36Fi. Find an expression for their new separation.
Answer:
[tex]r_f=\frac{1}{6}r_i[/tex]
Explanation:
To find the new separation of the charges, you first take into account the formula for the electric force, when the force are separated a distance of ri.
You use the following expression:
[tex]F_i=k\frac{q_Aq_B}{r_i^2}[/tex] (1)
k: Coulomb's constant
qA: charge of A particle
qB: charge of B particle
When the charges are separated to a new distance rf, the new force is 36Fi, if the charges have not changed, you have:
[tex]F_f=36F_i=k\frac{q_Aq_B}{r_f^2}[/tex] (2)
To find the new separation you replace the expression for Fi of the equation (1) into the equation (2) and solve for rf in terms of ri:
[tex]36F_i=36k\frac{q_Aq_B}{r_i^2}=k\frac{q_Aq_B}{r_f^2}\\\\\frac{36}{r_i^2}=\frac{1}{r_f^2}\\\\r_f=\frac{1}{6}r_i[/tex]
The new separation of the charges is 1/6 times of the initial separation
a fly undergoes a displacement of - 5.80 while accelerating at -1.33 m/s^2 for 4.22 s. what was the initial velocity of the fly?
Answer:
[tex]v_i = 1.44\frac{m}{s}[/tex]
Explanation:
The computation of the initial velocity of the fly is shown below:-
But before that we need to do the following calculations
For 4.22 seconds
[tex]\bar v = \frac{-5.80 m}{4.22 s}[/tex]
[tex]= -1.37\frac{m}{s}[/tex]
For uniform acceleration
[tex]\bar v = \frac{v_i +v_f}{2}[/tex]
[tex]= v_i + v_f[/tex]
[tex]= -2.74\frac{m}{s}[/tex]
With initial and final velocities
[tex]= -1.33\frac{m}{s^2}[/tex]
[tex]= \frac{v_i +v_f}{4.22s}[/tex]
[tex]= -v_i + v_f[/tex]
[tex]= -5.61\frac{m}{s}[/tex]
So, the initial velocity is
[tex]v_i = 1.44\frac{m}{s}[/tex]
We simply applied the above steps to reach at the final solution i.e initial velocity