Answer:
0.563 kg-m²/s
Explanation:
Bill Nye's angular momentum L = Iω where I = rotational inertia = 1/2MR² (since we are considering him to be cylinder with his axis of rotation from is top to bottom- that is vertical through his height) where M = his mass = 84.0 kg and R = his radius = 16.0 cm = 0.16 m and ω = angular speed = 5.00 rpm = 5.00 × 2π/60 = π/6 rad/s = 0.524 rad/s
L = Iω
L = 1/2MR²ω
Substituting the values of the variables into the equation, we have
L = 1/2MR²ω
L = 1/2 × 84.0 kg × (0.16 m)² × 0.524 rad/s
L = 1/2 × 84.0 kg × 0.0256 m² × 0.524 rad/s
L = 1/2 × 1.12681 kg-m²rad/s
L = 0.5634 kg-m²rad/s
L ≅ 0.563 kg-m²/s
A cold block of metal feels colder than a block of wood at the same temperature. Why? A hot block of metal feels hotter than a block of wood at the same temperature. Again, why? Is there any temperature at which the two blocks feel equally hot or cold? What temperature is this?
: In general, metals feel colder or hotter to the touch than other materials at the same temperature because they're good thermal conductors. This means they easily transfer heat to colder objects or absorb heat from warmer objects
It is because the metal conducted heat faster than it feels colder than the wood, which conducted heat slower. Even tho they are similar temperatures. The metal will feel colder than the wood because of the thermal conductivity of the metal, compared to the wood.
define projectile in your own .
Answer:
a body which was thrown in space ,moves under the influence of gravity only is defined as projectile.
Answer:
projectile is defined as a body thrown in space , moves under the influence of gravity .hope it is you
Two long, straight wires are separated by a distance of 32.2 cm. One wire carries a current of 2.75 A, the other carries a current of 4.33 A. (a) Find the force per meter exerted on the 2.75-A wire. (b) Is the force per meter exerted on the 4.33-A wire greater than, less than, or the same as the force per meter exerted on the 2.75-A wire
Answer:
a)[tex]\frac{F_1}{L}=1.95*10^-^5N[/tex]
b)[tex]\frac{F_2}{L}=1.95*10^-^5N[/tex]
Explanation:
From the question we are told that:
Distance between wires [tex]d=32.2[/tex]
Wire 1 current [tex]I_1=2.75[/tex]
Wire 2 current [tex]I_2=4.33[/tex]
a)
Generally the equation for Force on [tex]l_1[/tex] due to [tex]I_2[/tex] is mathematically given by
[tex]F_1=I_1B_2L[/tex]
Where
B_2=Magnetic field current by [tex]I_2[/tex]
[tex]B_2=\frac{\mu *i_2}{2\pi d}[/tex]
Therefore
[tex]F_1=I_1B_2L[/tex]
[tex]F_1=I_1(\frac{\mu *i_2*l_1}{2\pi d})L[/tex]
[tex]\frac{F_1}{L} =\frac{4*\pi*10^{-7}*2.75*4.33*100 }{2*\pi*12.2 }[/tex]
[tex]\frac{F_1}{L}=1.95*10^-^5N[/tex]
b)
Generally the equation for Force on [tex]I_2[/tex] due to [tex]I_1[/tex] is mathematically given by
[tex]F_2=I_2B_1L[/tex]
Where
B_1=Magnetic field current by [tex]I_2[/tex]
[tex]B_1=\frac{\mu *I_1}{2\pi d}[/tex]
Therefore
[tex]\frac{F_2}{L} =I_2(\frac{\mu *I_1*I_2}{2\pi d})[/tex]
[tex]\frac{F_2}{L}=1.95*10^-^5N[/tex]
How much energy must the brakes absorb to bring a 1200kg car from 30m/s to 15 m/s?
A cart of mass m is moving with negligible friction along a track with known speed v1 to the right. It collides with and sticks to a cart of mass 4m moving with known speed v2 to the right. Which of the two principles, conservation of momentum and conservation of mechanical energy, must be applied to determine the final speed of the carts, and why
Answer:
conservation of linear momentum
We were told that two objects became stuck together hence we have to use the principle of conservation of momentum to obtain the final velocities of the carts.
What is conservation of momentum ?The principle of conservation of momentum lets us know that the momentum before collision is equal to the momentum after collision. As such we can write; m1u1 + m2u2 = m1v1 + m2v2.
We can use this thus principle to obtain the final speeds of the carts since the two objects that collided became stuck together.
Learn more about conservation of momentum: https://brainly.com/question/11256472
If acceleration is zero what statement about velocity is true *
A)Velocity is zero
B)Velocity is constant
C)Velocity cannot be determined
D) Velocity is changing
Answer:
Option"B" is correct.
Explanation:
when a body move with constant velocity then acceleration is zero.
If the acceleration of an object is 0, it means the velocity of an object is constant and not changing with respect to time,
So, the correct option will be :
=》B)Velocity is constant
A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 s apart. Part A How far away did the impact occur? (Use vair=343m/s , vconcrete=3000m/s )
Answer:
The impact occured at a distance of 2478.585 meters from the person.
Explanation:
(After some research on web, we conclude that problem is not incomplete) The element "Part A" may lead to the false idea that question is incomplete. Correct form is presented below:
A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 seconds apart. How far away did the impact occur? (Sound speed in the air: 343 meters per second, sound speed in concrete: 3000 meters per second)
Sound is a manifestation of mechanical waves, which needs a medium to propagate themselves. Depending on the material, sound will take more or less time to travel a given distance. From statement, we know this time difference between air and concrete ([tex]\Delta t[/tex]), in seconds:
[tex]\Delta t = t_{A}-t_{C}[/tex] (1)
Where:
[tex]t_{C}[/tex] - Time spent by the sound in concrete, in seconds.
[tex]t_{A}[/tex] - Time spent by the sound in the air, in seconds.
By suposing that sound travels the same distance and at constant speed in both materials, we have the following expression:
[tex]\Delta t = \frac{x}{v_{A}}-\frac{x}{v_{C}}[/tex]
[tex]\Delta t = x\cdot \left(\frac{1}{v_{A}}-\frac{1}{v_{C}} \right)[/tex]
[tex]x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }[/tex] (2)
Where:
[tex]v_{C}[/tex] - Speed of the sound in concrete, in meters per second.
[tex]v_{A}[/tex] - Speed of the sound in the air, in meters per second.
[tex]x[/tex] - Distance traveled by the sound, in meters.
If we know that [tex]\Delta t = 6.4\,s[/tex], [tex]v_{C} = 3000\,\frac{m}{s}[/tex] and [tex]v_{A} = 343\,\frac{m}{s}[/tex], then the distance travelled by the sound is:
[tex]x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }[/tex]
[tex]x = 2478.585\,m[/tex]
The impact occured at a distance of 2478.585 meters from the person.
a rocket sled decelerates from 400m/s to 0 in 5 seconds. Calculate the acceleratio
Answer:
80m/s2
Explanation:
a = v-u/ t
=>a = 0- 400/ 5
= -400/5
= -80m/s2
v- sign is negligible
A coil of 10 turns is in the shape of an ellipse having a major axis of 13.0 cm and a minor axis of 6.00 cm. The coil rotates at 73rpm in a region in which the Earth's magnetic field is 55.0 microT. What is the maximum voltage induced in the coil if the axis of rotation of the coil is along its major axis and this axis of rotation is aligned perpendicular to the Earth's magnetic field
Answer:
the maximum voltage induced in the coil is 2.574 × 10⁻⁵ V
Explanation:
Given the data in the question;
Number of turns N = 10
major axis Ma = 13 cm = 0.13 m
a = 0.13/2 = 0.065 m
Minor axis Mi = 6 cm = 0.06 m
b = 0.06/2 = 0.03 m
we know that; 1 RPM = 0.10472 rad/s
rate of rotation R = 73rpm = 7.64 rad/s
Magnetic field = 55 uT
we know that, Area of ellipse = π × a × b
we substitute
A = π × 0.065 m × 0.03 m
A = 0.006126 m²
so
Maximum Voltage = N × Area × Magnetic field × rate of reaction
we substitute
Maximum Voltage = 10 × 0.006126 × ( 55 × 10⁻⁶ ) × 7.64
Maximum Voltage = 2.574 × 10⁻⁵ V
Therefore, the maximum voltage induced in the coil is 2.574 × 10⁻⁵ V
What is oscillating to form a light wave?
O Electric and Magnetic Fields
O Matter
O Light is only a particle, not a wave
O The Luminiferous Aether
If you weigh 690 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 15.0 km ? Take the mass of the sun to be ms = 1.99×1030 kg , the gravitational constant to be G = 6.67×10−11 N⋅m2/kg2 , and the free-fall acceleration at the earth's surface to be g = 9.8 m/s2 . Express your weight wstar in newtons.
Answer:
W' = 1.66 x 10¹⁴ N
Explanation:
First, we will calculate the mass:
[tex]W = mg[/tex]
where,
W = weight on earth = 690 N
m = mass = ?
g = acceleration due to gravity on earth = 9.8 m/s²
Therefore,
[tex]m = \frac{W}{g} = \frac{690\ N}{9.8\ m/s^2}\\\\m = 70.4\ kg[/tex]
Now, we will calculate the value of g on the neutron star:
[tex]g' = \frac{GM}{R^2}[/tex]
where,
g' = acceleration due to gravity on the surface of the neutron star = ?
G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of the Neutron Star = 1.99 x 10³⁰ kg
R = Radius of the Neutron Star = 15 km/2 = 7.5 km = 7500 m
Therefore,
[tex]g' = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(1.99\ x\ 10^{30}\ kg)}{(7500\ m)^2}\\\\g' = 2.36\ x\ 10^{12}\ m/s^2[/tex]
Therefore, the weight on the surface of the neutron star will be:
[tex]W' = mg'\\W' = (70.4\ kg)(2.36\ x\ 10^{12}\ m/s^2)[/tex]
W' = 1.66 x 10¹⁴ N
A lead fishing weight of a mass of 0.20 kg is tied to a fishing line that is 0.50 m long. The weight is then whirled in a vertical circle. The finishing line will break if its tension exceeds 100.0 N.
Required:
a. If the weight is whirled at higher and higher speeds, at what point in the vertical circle will the string break (top, bottom, or random position)?
b. At what speed will the string break?
Answer:
The solution of the given question is summarized in the below section.
Explanation:
The given values are:
Tension,
T = 100 N
mass,
m = 0.2 kg
length,
l = 0.5 m
Now,
(a)
Somewhere at bottom, string or thread breaks since string voltage seems to be the strongest around this stage.
then,
⇒ [tex]T-mg=\frac{mv^2}{l}[/tex]
or,
⇒ [tex]T=mg+\frac{mv^2}{l}[/tex]
(b)
As we know,
⇒ [tex]\frac{mv^2}{l}=T-mg[/tex]
or,
⇒ [tex]v^2=\frac{(T-mg)l}{m}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{(100-0.2\times 10)0.5}{0.2}[/tex]
⇒ [tex]=\frac{49}{0.2}[/tex]
⇒ [tex]v =\sqrt{245}[/tex]
⇒ [tex]=15.65 \ m/s[/tex]
a boat carving people more than its capacity is at risk of sinking, why?
Answer:
Explained below.
Explanation:
For a boat or any object to float on water, it's density must be less than that of water.
Now, when the maximum capacity of people to be carried by the boat is exceeded, it's possible that the maximum mass of people will also be exceeded depending on the mass of the people in the boat.
Now, we know that; density = mass/volume.
Thus, the higher the mass of the people, the higher the density and the higher the density, the more likely it is to be above that of water and the more likely it is to sink.
I am b o r e d, I am very very b o r e d!
I'm b o r e d with Lazarbeam Quarantine edition
episode 2352 because apparently the quarantining never ends :(
Which of the following best describes what occurs in a fission reaction?
A.
Two low mass nuclei are joined to form one nucleus.
B.
Electrons are shared between the nuclei.
C.
A single nucleus divides into two or more nuclei and gives off energy.
D.
A chemical reaction occurs between the nuclei.
Answer:
C.A single nucleus divides into two or more nuclei and gives off energy best describes what occurs in a fission reaction.
Answer:
C.
A single nucleus divides into two or more nuclei and gives off energy.
hope it is helpful to you
A 700-gram grinding wheel 22.0 cm in diameter is in the shape of a uniform solid disk. (We can ignore the small hole at the center.) When it is in use, it turns at a constant 215 rpm about an axle perpendicular to its face through its center. When the power switch is turned off, you observe that the wheel stops in 50.0 s with constant angular acceleration due to friction at the axle.
What torque does friction exert while this wheel is slowing down?
Solution :
Given :
Mass of grinding wheel, m = 700 g
= 0.7 kg
Diameter of the grinding wheel, d = 22 cm
= 0.22 m
Radius of the grinding wheel, r = 0.11 m
Initial angular velocity of grinding wheel, [tex]$\omega_0$[/tex] = 215 rpm
[tex]$=215 \ rpm \times \frac{2 \pi \ rad}{1 \ rev}\times \frac{1 \ min}{60 \ s}$[/tex]
where, [tex]$\pi = \frac{22}{7}$[/tex]
Time taken to stop, t = 50 s
Final angular velocity is [tex]$\omega$[/tex] = 0
Angular acceleration of the grinding wheel is given by :
[tex]$\alpha = \frac{\omega-\omega_0}{t}$[/tex]
[tex]$=\frac{0-215 \ rpm \times \frac{2 \pi \ rad}{1 \ rev}\times \frac{1 \ min}{60 \ s}}{50 \ s}$[/tex]
[tex]$=-0.45 \ rad/s^2$[/tex]
Magnitude of the angular acceleration of grinding wheel [tex]$\alpha$[/tex] [tex]$=-0.45 \ rad/s^2$[/tex]
Moment of inertia of the grinding wheel (solid disk),
[tex]$I=\frac{1}{2}mR^2$[/tex]
[tex]$=\frac{1}{2} \times 0.7 \times 0.11^2$[/tex]
[tex]$=4.235 \times 10^{-3} \ kgm^2$[/tex]
Torque exerted by friction while the wheel is slowing down is
[tex]$\tau = I \alpha$[/tex]
[tex]$=4.235 \times 10^{-3} \times 0.45$[/tex]
[tex]$=1.90 \times 10^{-3} \ Nm$[/tex]
A negative charge of 4.0 x 10 C and a positive charge of 7.0 x 10 C are separated by 0.15 m. What is the force between the two charges?
12. By convention (agreement of the scientific community for consistency)
magnetic field lines...
A. always start on the north pole and terminate (end) on the South Pole
B. start at infinity and point toward each pole
C. start at each pole and go outward
D. always start on the south pole and terminate (end) on the north pole.
Answer:
. always start on the north pole and terminate (end) on the South Pole
Explanation:
has a man he has married many women but has never been married before who is he
Answer:
Explanation:
The answer is a priest or a moulana
What happens is a series circuit when you increase the number of bulbs?
The bulbs will produce lesser light than their capacity, In short they will be dimmer because the the energy will get divided in the number of bulbs.
Lucy moves down the hall at 3.5 m/s. When he sees Luke coming, he slows down. After 4.0 s, he is moving at 2.1. m/s. What is his acceleration?
Answer:
Acceleration, a = 0.35 m/s²
Explanation:
Given the following data;
Initial velocity, u = 3.5 m/s²
Final velocity, v = 2.1 m/s²
Time, t = 4 secs
To find the acceleration, we would use the first equation of motion;
V = u - at (the sign is negative because Lucy is slowing down).
Substituting into the formula, we have;
2.1 = 3.5 - a(4)
2.1 = 3.5 - 4a
4a = 3.5 - 2.1
4a = 1.4
a = 1.4/4
a = 0.35 m/s²
10 POINTS! SPACE QUESTION!!
A heating coil operates on 220 V if it draws 15.0 A. Find it's resistance
Answer:
R ≈ 15 ohms
Explanation:
Using ohm's law equation,
I = V/R, to solve for the resistance of the heating coil.
R = V/I
Known:
V = 220 v = 220 kgm^2s^-3A^-1
I = 15 A
Unknown:
R =?
Solution:
R = (220 kgm^2s^-3A^-1)/ 15.0 A
R = 14.6 kgm^2s^-3A^-2
R ≈ 15 kgm^2s^-3A^-2
R ≈ 15 ohms
the pygmy shrew has an average mass of 2.0 g if 49 of these shrew are placed on a spring scale with a spring constant of 24 N/m , what is the springs displacement
Answer:
Spring's displacement, x = -0.04 meters.
Explanation:
Let the spring's displacement be x.
Given the following data;
Mass of each shrew, m = 2.0 g to kilograms = 2/1000 = 0.002 kg
Number of shrews, n = 49
Spring constant, k = 24 N/m
We know that acceleration due to gravity, g is equal to 9.8 m/s².
To find the spring's displacement;
At equilibrium position:
Fnet = Felastic + Fg = 0
But, Felastic = -kx
Total mass, Mt = nm
Fg = -Mt = -nmg
-kx -nmg = 0
Rearranging, we have;
kx = -nmg
Making x the subject of formula, we have;
[tex] x = \frac {-nmg}{k} [/tex]
Substituting into the formula, we have;
[tex] x = \frac {-49*0.002*9.8}{24} [/tex]
[tex] x = \frac {-0.9604}{24} [/tex]
x = -0.04 m
Therefore, the spring's displacement is -0.04 meters.
ou are making a clock out of a solid disk with a radius of 0.2m and mass of 0.1 kg that will be attached at its center of mass to the end of a uniform thin rod with a mass of 0.3 kg. Calculate The length of the rod such that the period of the system is 1 second.The distance from the pivot to theCOM of the system must be expressed in terms of L. The parallel axis theorem must be used for both therod and the solid disk in terms of L as well. There will be a quadratic equation to be solved
Answer:
the correct result is L = 0.319 m
Explanation:
This system is a physical pendulum whose angular velocity is
w² = [tex]\frac{M \ g \ d}{I}[/tex]
where d is the distance from the center of mass to the point of rotation and I is the moment of inertia of the system
The Moment of Inertia is a scalar, therefore an additive quantity
I = I_bar + I_disk
the moment of inertia of each element with respect to the pivot point can be found with the parallel axes theorem
let's use M for the mass of the bar and m for the mass of the disk
Bar
I_bar = I_{cm} + Md²
the moment of inertia of the center of mass is
I_{cm} = [tex]\frac{1}{12}[/tex] M L²
the distance from the center of mass
d = L / 2
we substitute
I_bar = [tex]\frac{1}{12}[/tex] M L² + M ([tex]\frac{L^2}{4}[/tex])
Disk
I_disk = I_{cm} + m d²
moment of inertia of the center of mass
I_{cm} = ½ m R²
the distance d is
d = L
we substitute
I_disk = 1/2 m R² + m L²
the total moment of inertia is
I = [tex]\frac{1}{12}[/tex] M L² +[tex]\frac{1}{4}[/tex] M L² + [tex]\frac{1}{2}[/tex] m r² + m L²
I = [tex]\frac{1}{4}[/tex] M L² + m L² + ½ m r²
I = L² (m + [tex]\frac{1}{4}[/tex] M) + ½ m r²
The position of the center of mass of the system can be found with the expressions
d_{cm} = [tex]\frac{1}{M} \sum r_i m_i[/tex]
d_{cm} = [tex]\frac{1}{m+M} \ ( M \frac{L}{2} + m L)[/tex]
d_{cm} = [tex]L \frac{m + M/2}{m +M }[/tex]
now we can substitute in the expression for the angular velocity
w² = (m + M) g L [tex]\frac{m + \frac{M}{2} }{m+M}[/tex] [tex]\frac{1}{L^2 (m+ \frac{M}{4} ) + \frac{1}{2} m r^2 }[/tex]
w² = g (m + [tex]\frac{1}{2}[/tex] M) [tex]\frac{L}{ L^2 ( m +\frac{1}{4} M ) + \frac{1}{2} m r^2}[/tex]
angular velocity and period are related
w = 2π/T
sustitute
4π²/T² = g (m + [tex]\frac{1}{2}[/tex] M) [tex]\frac{L}{ L^2 ( m +\frac{1}{4} M ) + \frac{1}{2} m r^2}[/tex]
L² (m + [tex]\frac{1}{4}[/tex] M) + ½ m r² = [tex]\frac{T^2}{4 \pi ^2 } \ g ( m + \frac{1}{2} M ) \ \ L[/tex]
we substitute the values and solve the second grade equation
L² (0.1 + [tex]\frac{1}{4}[/tex] 0.3) - [[tex]\frac{1^2}{4\pi ^2}[/tex] 9.8 (0.1 + 0.3/2) ] L + ½ 0.1 0.2² = 0
L² 0.175 - 0.06206 L + 0.002 = 0
the equation remains after simplifying
L² - 0.3546 La + 0.01143 = 0
solve us
L = [tex]\frac{0.3546 \ \pm \sqrt{ 0.3546^2 - 4 \ 0.01143 }}{2}[/tex]
L = [tex]\frac{0.3546 \ \pm \ 0.28288 }{2}[/tex]
L₁ = 0.319 m
L₂ = 0.036m
the correct result must have a value greater than the radius of the disk. The correct result is L = 0.319 m
A sled on ice moves in the way described . Friction is so small that it can be ignored. A person wearing spiked shoes standing on the ice can apply a force to the sled and push it along the ice. Choose the one force (A through G) which would keep the sled moving as described. Which force would keep the sled moving toward the right and speeding up at a steady rate (constant acceleration)
The question is not complete, so i have attached an image of the complete question and the image of the sled on the ice.
Answer:
A) 3
B) 7
C) 5
D) 5
E) 5
Explanation:
A) Looking at the image attached, The force that will continue to move the sled toward the right and speeding up at a steady rate would have to be a force acting towards the right and has constant strength (magnitude).
Thus, option 3 is correct.
B) The force that will continue to move the sled toward the right and at Constant velocity will be when no force is applied because at constant velocity, acceleration is zero and thus force will also be zero.
Thus, option 7 is correct.
C) The force that will slow the sled when moving to the right at steady rate(constant acceleration) would be an opposite force which means a force towards the left with a constant strength (magnitude).
Thus, option 5 is correct.
D) The force exerted on the sled by a person when the sled is accelerating to the right would be the normal force exerted on the person by the ice).
Thus, option 5 is correct.
E) The reaction to the normal force exerted on the sled by the ice when the sled is accelerating to the left would be the normal force exerted by the sled on the ice
Thus, option 5 is correct
What does it mean if the reflected beam is above the incident beam? What does it mean if reflected beam is below the incident beam?
Answer:
aim at prisma and will have all colors
Explanation:
Answer:
If the vision position is above the actual image location then the light travel from the object in such a way that the angle of incidence is less than the angle of reflected ray which means that the reflected beam is above the incident beam.
Explanation:
Objects A and B, of mass M and 2M respectively, are each pushed a distance d straight up an inclined plane by a force F parallel to the plane. The coefficient of kinetic friction between each mass and the plane has the same value μ.k At the highest point is:______
a. KEA > KEB
b. KEA = KEB
c. KEA < KEB
d. The work done by F on A is greater than the work done by F on B.
e. The work done by F on A is less than the work done by F on B.
Answer:
The correct answer is option (A) that is KEA > KEB .
Explanation:
Let us calculate -
If the object is straighten up and inclined plane , the work done is
[tex]W=F_d- F_f_r_id-F_gh[/tex]
[tex]W=F_d-\mu_kmgdcos\theta-mgdsin\theta[/tex]
The change in kinetic energy is ,
[tex]\Delta K=\frac{1}{2}mv^2-\frac{1}{2}m\nu_0^2[/tex]
At the top of the inclined plane , the velocity is zero
So,
[tex]\Delta K=\frac{1}{2} m(0)^2-\frac{1}{2}m\nu_0^2[/tex]
[tex]\Delta KE=-\frac{1}{2}m\nu_0^2[/tex]
From the work energy theorem , we have [tex]W=-\Delta K[/tex] in case of friction , so
[tex]\frac{1}{2}m\nu_0^2=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]
[tex]KE=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]
For object A-
[tex]KE_A=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]
For object B
[tex]KE_B= Fd -2\mu_kMgdcos\theta-2Mgdsin\theta[/tex]
[tex]KE_B= Fd -2(\mu_kMgdcos\theta-Mgdsin\theta)[/tex]
Thus , larger mass is going to mean less total work and a lower kinetic energy .
From the above results , we get
[tex]KE_A >KE_B[/tex]
Therefore , option A is correct .
How many times a minute does a boat bob up and down on ocean waves that have a wavelength of 35.6 m and a propagation speed of 4.68 m/s? (the answer may not be a whole number)
Answer:
It will bob 7.887640449 times a minute
Explanation:
I hope this is correct!!
A star with the same mass and diameter as the sun rotates about a central axis with a period of about 24.0 days. Suppose that the sun runs out of nuclear fuel and collapses to form a white dwarf star with a diameter equal to that of the earth. Assume the star acts like a solid sphere and that there is no loss of mass in the process. You will need some data from the inside front cover of you text. (a) What would be the new rotation period (s) of the star? (b) What is the ratio of final to initial kinetic energies (Kf /Ki)?
Answer:
a) w = 2.52 10⁷ rad / s, b) K / K₀ = 1.19 10⁴
Explanation:
a) We can solve this exercise using the conservation of angular momentum.
Initial instant. Before collapse
L₀ = I₀ w₀
Final moment. After the collapse
L_f = I w
angular momentum is conserved
L₀ = L_f
I₀ w₀ = I w (1)
The moment of inertia of a sphere is
I = 2/5 m r²
we take from the table the mass and diameter of the star
m = 1,991 10³⁰ kg
r₀ = 6.96 10⁸ m
r = 6.37 10⁶ m
to find the angular velocity let's use
w = L / T
where the length of a circle is
L = 2π r
T = 24 days (24 h / 1 day) (3600 s / 1h) = 2.0710⁶ s
we substitute
w = 2π r / T
wo = 2π 6.96 10⁸ / 2.07 10⁶
wo = 2.1126 10³ rad / s
we substitute in equation 1
w = [tex]\frac{I_o}{I}[/tex]
w = 2/5 mr₀² / 2/5 m r² w₀
w = ([tex]\frac{r_o}{r}[/tex]) ² wo
w = (6.96 10⁸ / 6.37 10⁶) ² 2.1126 10³
w = 2.52 10⁷ rad / s
b) the kinetic energy ratio
K = ½ m w²
K₀ = ½ m w₀²
K = ½ m w²
K / K₀ = (w / wo) ²
K / K₀ = 2.52 10⁷ / 2.1126 10³
K / K₀ = 1.19 10⁴