Answer:
The process of solving a circular motion problem is much like any other problem in physics class. The process involves a careful reading of the problem, the identification of the known and required information in variable form, the selection of the relevant equation(s), substitution of known values into the equation, and finally algebraic manipulation of the equation to determine the answer. Consider the application of this process to the following two circular motion problems.
Sample Problem #1
A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car.
The solution of this problem begins with the identification of the known and requested information.
Known Information:
m = 900 kg
v = 10.0 m/s
R = 25.0 m
Requested Information:
a = ????
Fnet = ????
To determine the acceleration of the car, use the equation a = v2 / R. The solution is as follows:
a = v2 / R
a = (10.0 m/s)2 / (25.0 m)
a = (100 m2/s2) / (25.0 m)
a = 4 m/s2
To determine the net force acting upon the car, use the equation Fnet = m•a. The solution is as follows.
Fnet = m • a
Fnet = (900 kg) • (4 m/s2)
Fnet = 3600 N
Sample Problem #2
A 95-kg halfback makes a turn on the football field. The halfback sweeps out a path that is a portion of a circle with a radius of 12-meters. The halfback makes a quarter of a turn around the circle in 2.1 seconds. Determine the speed, acceleration and net force acting upon the halfback.
The solution of this problem begins with the identification of the known and requested information.
Known Information:
m = 95.0 kg
R = 12.0 m
Traveled 1/4-th of the circumference in 2.1 s
Requested Information:
v = ????
a = ????
Fnet = ????
To determine the speed of the halfback, use the equation v = d / t where the d is one-fourth of the circumference and the time is 2.1 s. The solution is as follows:
v = d / t
v = (0.25 • 2 • pi • R) / t
v = (0.25 • 2 • 3.14 • 12.0 m) / (2.1 s)
v = 8.97 m/s
To determine the acceleration of the halfback, use the equation a = v2 / R. The solution is as follows:
a = v2 / R
a = (8.97 m/s)2 / (12.0 m)
a = (80.5 m2/s2) / (12.0 m)
a = 6.71 m/s2
To determine the net force acting upon the halfback, use the equation Fnet = m•a. The solution is as follows.
Fnet = m*a
Fnet = (95.0 kg)*(6.71 m/s2)
Fnet = 637 N
In Lesson 2 of this unit, circular motion principles and the above mathematical equations will be combined to explain and analyze a variety of real-world motion scenarios including amusement park rides and circular-type motions in athletics.
The charge that passes a cross-sectional area A=10-4 m2 varies with time according
to the relation
Q=4 + 2t + t2, where Q is in coulombs and t is in seconds. (a) Find the relation
that gives the instantaneous current at time t =2 s is
Answer:
Current = dQ/dt
or I = dQ/dt
Where I represents current.
Which is the rate of flow of charge.
Q=4 + 2t + t²
dQ/dt = 2 + 2t --- This is the relation that gives the instantaneous current.
At time t=2sec
dQ/dt = I = 2 + 2t
= 2 + 2(2)
=2 + 4
= 6A.
Why can’t we see the molecules moving in a solid or liquid?
Answer:
Because molecules are to small for us people to see with the bare eye, unless you use a telescope.
Explanation:
Explanation:
Because in solid form, the molecules stood in place, not moving. In liquid state, the molecules move slowly but not as fast as air molecules.
if a glass rod rubbed with Silk piece is taken to a ball pen rubeed with wool the ball pen_____
Answer:
The glass rod remains charged due to electrification by the silk
Which of the following items is an example of a symbol Bernice bobs her hair
what? Is there a picture you can show us with this question
A 3.7-kg object is acted on by two forces. One of the forces is 11 N acting toward the
east. What is the other force if the acceleration of the object is 1.0 m/s2 toward the east?
Answer:
F₂ = -7.3 N
Explanation:
Given that,
The mass of an object, m₁ = 3.7 kg
First force, F₁ = 11 N
The net acceleration of the object is 1 m/s².
We know that,
F₁+F₂ = ma
11+F₂ = (3.7)(1)
F₂ = 3.7-11
F₂ = -7.3 N
so, the other force is 7.3 N and it is acting in west direction.
Give an example of a physical entity that is quantized. State specifically what the entity is and what the limits are on its values.
Answer:
A charge is a physical entity that has been quantized. The limits on its values are the value of a charged particle quantized in the state where 'n'...
Explanation:
One example of a physical entity that is quantized is:
The amount of money in your pocket.
The amount can't have any fraction of 1 cent.
Its value must be an integer-multiple of cents, or 0.01 dollar.
When it increases or decreases, it jumps from one integer number of cents to the next integer number. It doesn't "slide" from one to the next. It can never have a value between two integer numbers of cents.
A spring with a 10-kg mass and a damping constant 15 can be held stretched 2 meters beyond its natural length by a force of 6 newtons. Suppose the spring is stretched 4 meters beyond its natural length and then released with zero velocity. Suppose the spring is stretched 4 meters beyond its natural length and then released with zero velocity.
Required:
Find the position of the mass at any time t.
Answer:
Explanation:
Given that:
mass = 10kg
damping constant C = 15 kg/s
length = 2 m
Force F = 6N
Using the Hooke's law:
F = kx
6 = 15x
k = 6 N /2 m
spring constant k = 3 N/m
For the critical damping
C² - 4k*m= 0
m = C²/4k
m = (15)²/4(3) kg
m = 225/12 kg
m = 18.75 kg
Some radar systems detect the size and shape of objects such as aircraft and geological terrain. What is the frequency of such a system which can detect objects as small as 19.1 cm?
Answer:
[tex]f=1.57\times 10^9\ Hz[/tex]
Explanation:
Given that,
A system can detect objects as small as 19.1 cm i.e. 0.191 m. It is the wavelength.
We know that,
Frequency, [tex]f=\dfrac{c}{\lambda}[/tex]
So,
[tex]f=\dfrac{3\times 10^8}{0.191}\\\\=1.57\times 10^9\ Hz[/tex]
So, the frequency of such a system is equal to[tex]1.57\times 10^9\ Hz[/tex].
Two resistors, A and B, are connected in parallel across a 6.0-V battery. The current through B is found to be 2.0 A. When the two resistors are connected in series to the 6.0- V battery, a voltmeter connected across resistor A measures a voltage of 4.0 V. Find the resistances of A and B
Answer:
The resistance of A is 6 ohms and the resistance of B is 3 ohms
Explanation:
Step 1: For the first connection (parallel connection), the resistance of B will be calculated.
Note: in a parallel connection, the voltage through each resistor is the same.
[tex]V = I_AR_A = I_BR_B\\\\R_B = \frac{V}{I_B} = \frac{6}{2} = 3 \ ohms[/tex]
Step 2: The resistance of A will be calculated from the second connection (series connection)
Note: in series connection, the current flowing in each resistor is the same
[tex]V = V_A + V_B\\\\V = IR_A + IR_B\\\\The \ voltage \ drop \ in \ B; \ V_B = V- V_A\\\\V_B = 6 - 4 = 2 \ V\\\\IR_B = 2\ V\\\\I = \frac{2 \ V}{R_B}= \frac{2}{3} \ A\\\\The \ resistance \ of \ A \ is \ calculated \ as ;\\\\IR_A = 4 \ V\\\\R_A = \frac{4}{I} = \frac{4 \times 3}{2} = 6 \ ohms[/tex]
An LC circuit is built with a 20 mH inductor and an 8.0 PF capacitor. The capacitor voltage has its maximum value of 25 V at t = 0 s.(a)How much time does it take until the capacitor is fully discharged for the first time? (b)What is the inductor current at that time?
Answer:
a) the required time is 0.6283 μs
b) the inductor current is 0.5 mA
Explanation:
Given the data in the question;
The capacitor voltage has its maximum value of 25 V at t = 0
i.e V[tex]_m[/tex] = V₀ = 25 V
we determine the angular velocity;
ω = 1 / √( LC )
ω = 1 / √( ( 20 × 10⁻³ H ) × ( 8.0 × 10⁻¹² F) )
ω = 1 / √( 1.6 × 10⁻¹³ )
ω = 1 / 0.0000004
ω = 2.5 × 10⁶ s⁻¹
a) How much time does it take until the capacitor is fully discharged for the first time?
V[tex]_m[/tex] = V₀sin( ωt )
we substitute
25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )
25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )
divide both sides by 25 V
sin( 2.5 × 10⁶ × t ) = 1
( 2.5 × 10⁶ × t ) = π/2
t = 1.570796 / (2.5 × 10⁶)
t = 0.6283 × 10⁻⁶ s
t = 0.6283 μs
Therefore, the required time is 0.6283 μs
b) What is the inductor current at that time?
[tex]I[/tex](t) = V₀√(C/L) sin(ωt)
{ sin(ωt) = 1 )
[tex]I[/tex](t) = V₀√(C/L)
we substitute
[tex]I[/tex](t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )
[tex]I[/tex](t) = 25 × 0.00002
[tex]I[/tex](t) = 0.0005 A
[tex]I[/tex](t) = 0.5 mA
Therefore, the inductor current is 0.5 mA
The time taken for the capacitor to fully discharge is 6.28 x 10⁻⁷ s.
The current in the inductor at the given time is 0.0005 A.
Angular velocity of circuitThe angular velocity of the circuit is calculated as follows;
[tex]\omega = \frac{1}{\sqrt{LC} } \\\\\omega = \frac{1}{\sqrt{20 \times 10^{-3} \times 8 \times 10^{-12} } } \\\\\omega = 2.5 \times 10^6 \ rad/s[/tex]
Time for the capacitor to fully dischargeV = V₀(sinωt)
sinωt = V/V₀
sinωt = = 25/25
sin(ωt) = 1
ωt = sin⁻¹ (1)
ωt = π/2
2.5 x 10⁶ t = π/2
t = 1.57 / (2.5 × 10⁶)
t = 6.28 x 10⁻⁷ s
Inductor current at the given timeThe current in the inductor at the given time is calculated as follows
[tex]I(t) = V_0 \sqrt{\frac{C}{L} } \\\\I = 25 \times \sqrt{\frac{8\times 10^{-12}}{20 \times 10^{-3}} } \\\\I = 0.0005 \ A[/tex]
Learn more about inductor current here: https://brainly.com/question/4425414
Lab: Energy Transfer Instructions Click the links to open the resources below. These resources will help you complete the assignment. Once you have created your file(s) and are ready to upload your assignment, click the Add Files button below and select each file from your desktop or network folder. Upload each file separately. Your work will not be submitted to your teacher until you click Submit. Documents Descriptive Lab Report Guide Descriptive Lab Report Rubric
Answer:
The second one says cant open file............
Friction is a (1)______
that (2)______
motion. It acts in a direction (3)______
to the motion of
the object. There is (4)______
exerted on an object that
moves on a (5)_____
than on a (6)_____
Friction opposes the motion of an object as it moves across a
surface. An object slows down (7)______
on a rough
surface than on a smooth surface.
Answer:
friction is motion heat that collect hot vibration
g:p:e = gravitational field strength x height
gravitational field strength = 9.8N/kg
Answer:
kdfihiyfxnlL chapter kar kaam
Explanation:
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: Ánh nắng mặt trời có cường độ đồng đều với bước sóng nằm trong vùng khả khiến 430nm-690nm đến đập vuông góc với một bản mỏng nước có bề dày 320nm, chiết suất 1,33 lơ lửng trong không khí. Tìm bước sóng thích hợp để ánh sáng phản từ bản mỏng là sáng nhất đối với người quan sát
In June 1985, a laser beam was sent out from the Air Force Optical Station on Maui and reflected back from the shuttle Discovery as it passed by 354 km overhead. The diameter of the central maximum of the beam at the shuttle position was 500 nm. What is the effective diameter of the circular laser aperture at the Maui ground station
This question is incomplete, the complete question is;
In June 1985, a laser beam was sent out from the Air Force Optical Station on Maui and reflected back from the shuttle Discovery as it passed by 354 km overhead. The diameter of the central maximum of the beam at the shuttle position was said to be 9.1 m, and the beam wavelength was 500 nm.
What is the effective diameter of the circular laser aperture at the Maui ground station
Answer:
the effective diameter of the circular laser aperture at the Maui ground station is 4.747 cm
Explanation:
Given the data in the question;
Separation between observer and point L = 354 km = 354000 m
Linear separation D = 9.1 m
wavelength λ = 500 nm = 500 × 10⁻⁹ m
Now, for small angles;
θ = D / L
θ = 9.1 m / 354000 m
θ = 2.57 × 10⁻⁵ rad
For a circular aperture;
sinθ = ( 1.22 × λ ) / d
for small angles;
θ = ( 1.22 × λ ) / d
so
θ = 2 × θ
θ = 2 × [( 1.22 × λ ) / d]
we substitute
2.57 × 10⁻⁵ = 2 × [( 1.22 × 500 × 10⁻⁹ ) / d]
2.57 × 10⁻⁵ = 0.00000122 / d
d = 0.00000122 / 2.57 × 10⁻⁵
d = 0.04747 m
d = ( 0.04747 × 100 )m
d = 4.747 cm
Therefore, the effective diameter of the circular laser aperture at the Maui ground station is 4.747 cm
To calculate the final enthalpy of the overall chemical equation, which step must occur?
Answer:
Explanation:
Reverse the second equation, and change the sign of the enthalpy
When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the condenser. Suppose you have a discharging condenser and the instantaneous rate of change of the voltage is -0.01 of the voltage (in volts per second). How many seconds does it take for the voltage to decrease by 90 %?
Answer:
460.52 s
Explanation:
Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that
dV/dt ∝ V
dV/dt = kV
separating the variables, we have
dV/V = kdt
integrating both sides, we have
∫dV/V = ∫kdt
㏑(V/V₀) = kt
V/V₀ = [tex]e^{kt}[/tex]
Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V
Since dV/dt = kV
-0.01V = kV
k = -0.01
So, V/V₀ = [tex]e^{-0.01t}[/tex]
V = V₀[tex]e^{-0.01t}[/tex]
Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀
So, V = 0.1V₀
Thus
V = V₀[tex]e^{-0.01t}[/tex]
0.1V₀ = V₀[tex]e^{-0.01t}[/tex]
0.1V₀/V₀ = [tex]e^{-0.01t}[/tex]
0.1 = [tex]e^{-0.01t}[/tex]
to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have
㏑(0.01) = -0.01t
So, t = ㏑(0.01)/-0.01
t = -4.6052/-0.01
t = 460.52 s
Consider the following possibilities and select the correct choice.
1. Tx Ty > Tz
2. Tx Ty < Tz
3. Tx Ty = Tz
Answer:
Tx not but mybe
Explanation:
for that reason its just trying to help
The peak value of the electric field component of an electromagnetic wave is E. At a particular instant, the intensity of the wave is of 0.020 W/m2. If the electric field were increased to 5E, what would be the intensity of the wave?
Answer:
[tex]I_2=0.50 w/m^2[/tex]
Explanation:
From the question we are told that:
initial Intensity [tex]I_1=0.020 w/m^2[/tex]
Final Electric field [tex]E_2=5E[/tex]
Generally the equation for Relation ship between intensity and Electric field is mathematically given by
[tex]\frac{I_1}{I_2}= \frac{E_1^2}{E_2^2}[/tex]
Therefore
[tex]I_2=\frac{I_1}{ \frac{E_1^2}{E_2^2}}[/tex]
[tex]I_2=\frac{0.020}{ \frac{E^2}{5E^2}}[/tex]
[tex]I_2=0.50 w/m^2[/tex]
Effects of global warming is
A-decrease in temperature
B-melting of polar ice caps
C-breathing problems
Answer:
B- the melting of polar ice caps
Explanation:
As the world's temperature increases, polar ice caps will no longer be able to remain solid.
URGENT
The component of solid-state physics that works with and/or heats metals and alloys to give them certain desired
shapes or properties is..
Answer:
the is metallurgy .....
what advantage does hovercraft have over a boar or a road vehicle?
Answer:
The advantages of Hovercraft:
They can travel over almost any non-porous surface.
They can operate to and from any unprepared beach or slipway.
They take fast, direct routes compared to a conventional marine vessel.
Kinetic energy is the energy an object has due to its
die hele
Answer:
There are a couple of interesting things about kinetic energy that we can see from the equation.
Kinetic energy depends on the velocity of the object squared. This means that when the velocity of an object doubles, its kinetic energy quadruples. A car traveling at 60 mph has four times the kinetic energy of an identical car traveling at 30 mph, and hence the potential for four times more death and destruction in the event of a crash.
Kinetic energy must always be either zero or a positive value. While velocity can have a positive or negative value, velocity squared is always positive.
Kinetic energy is not a vector. So a tennis ball thrown to the right with a velocity of 5 m/s, has the exact same kinetic energy as a tennis ball thrown down with a velocity of 5 m
A certain light truck can go around a flat curve having a radius of 150m with a max speed if 26.5. What is the max speed it can go around a curve having a radius of 76.0m
Answer:
The maximum speed is 18.86 m/s.
Explanation:
initial radius, r = 150 m
maximum speed, v = 26.5 m/s
new radius, r' = 76 m
Let the new maximum speed is v'.
The formula of the maximum speed is
[tex]tan\theta = \frac{v^2}{rg}[/tex]
So,
[tex]\frac{v'^2}{v^2}=\frac{r'}{r}\\\\\frac{v'^2}{26.5\times 26.5}=\frac{76}{150}\\\\v=18.86 m/s[/tex]
sanaysay tungkol sa pangangalaga ng mga endangered animals
The steps to determine the sum are shown. (6.74x104)+(8.95 x 104) Step 1. Rearrange the expression: (6.74+8.95) 104 Step 2. Add the coefficients: (15.69) 104 Step 3. Write in scientific notation: 1.569x 10 What is the value of k in Step 3? =
Answer:
We want to solve the sum:
6.74*10⁴ + 8.95*10⁴
first, we take the common factor 10⁴ out, so we get:
(6.74 + 8.95)*10⁴
Now we solve the sum:
(15.66)*10⁴
Now we want to rewrite it in exponential form, wo we can rewrite it as:
(15.66)*10⁴ = (1.566*10)*10⁴ = (1.566)*10*10⁴ = (1.566)*10⁴⁺¹ = 1.566*10⁵
k = 5.
According to Coulomb's law, rank the interactions between charged particles from highest potential energy to lowest potential energy.
a. 1+ charge and 1- charge seperated by 200pm
b. 1+ charge and 1+ charge seperated by 100pm
c. 1+ charge and 1- charge seperated by 100pm
d. 2+ charge and 1- charge seperated by 100pm
According to Coulomb's law, rank the interactions between charged particles from highest potential energy to lowest potential energy.
Highest potential energy to lowest potential energy.
b. 1+ charge and 1+ charge seperated by 100 pm
a. 1+ charge and 1- charge seperated by 200 pm
c. 1+ charge and 1- charge seperated by 100 pm
d. 2+ charge and 1- charge seperated by 100 pm
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How does gravity affect your ability to live on a planet?
A 1.2-kg mass suspended from a spring of spring constant 22 N.m-1 executes simple harmonic motion of amplitude 5 cm. Assuming that the mass is at the equilibrium posiiton at t = 0, what is its displacement at t = 1.0 s?
Answer:
[tex]d =3.7*10^{-3} m[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=1.2kg[/tex]
Spring constant [tex]\mu=22Nm^{-1}[/tex]
Amplitude [tex]A=5cm=0.05m[/tex]
Generally the equation for displacement d is mathematically given by
[tex]d = Asin(\omega t)[/tex]
Where
[tex]\omega=angular\ velocity[/tex]
[tex]\omega=\sqrt{k/m}[/tex]
[tex]\omega=\sqrt{22/1.2}[/tex]
[tex]\omega=4.2817rads^{-1}[/tex]
Therefore
[tex]d = 0.05*sin(4.2817*1)[/tex]
[tex]d =3.7*10^{-3} m[/tex]
Hunter is studying a new form of plastic chip that can be added onto a credit card, linking it to a bank account
separate from the main card. This chip works by passing an electric current through a sensing coil. What application
of solid-state physics is Hunter most likely working with?
O electromagnetism
O metallurgy
O quantum mechanics
O crystallography
Answer:
electromagnetism....