Answer:
T = 1.07 s
a = 1.034 m/s2
Explanation:
Metric unit conversion
m = 145 g = 0.145 kg
x = 3 cm = 0.03 m
Suppose this is a simple harmonic motion, then its period T can be calculated using the following equation
[tex]T = 2\pi\sqrt{\frac{m}{k}}[/tex]
[tex]T = 2\pi\sqrt{\frac{0.145}{5}} = 1.07 s[/tex]
The maximum acceleration would occurs when spring is at maximum stretching length, aka 0.03m
The spring force at that point would be
[tex]F_s = kx = 5*0.03 = 0.15 N[/tex]
According to Newton's 2nd law, the acceleration at this point would be
[tex]a = F_s/m = 0.15 / 0.145 = 1.034 m/s^2[/tex]
Two objects moving in opposite directions with the same speed v undergo a totally inelastic collision, and half the initial kinetic energy is lost. Find the ratio of their masses.
Answer:
[tex]\frac{m}{M} =(\frac{2}{\sqrt{2} -1+1}) ^{-1}\\\frac{m}{M} =0.171572875[/tex]
Explanation:
We name M the larger mas, and m the smaller mass, so base on this we write the following conservation of momentum equation for the collision:
[tex]P_i=P_f\\M\,v_i -m\,v_i=(M+m)\,v_f\\(M-m)\,v_i = (M+m)\,v_f\\\\\frac{v_i}{v_f} = \frac{(M+m)}{(M-m)}[/tex]
We can write this in terms of what we are looking for (the quotient of masses [tex]\frac{m}{M}[/tex]:
[tex]\frac{v_i}{v_f} = \frac{(1+\frac{m}{M} )}{(1-\frac{m}{M} )}[/tex]
We use now the information about Kinetic Energy of the system being reduced in half after the collision:
[tex]K_i=2\,K_f\\\frac{1}{2} (M+m)\,v_i^2= 2*\frac{1}{2} (M+m)v_f^2\\v_i^2=2\,v_f^2\\\frac{v_i^2}{v_f^2} =2[/tex]
We can combine this last equation with the previous one to obtain:
[tex]\frac{v_i}{v_f} = \frac{(1+\frac{m}{M} )}{(1-\frac{m}{M} )}\\(\frac{v_i}{v_f})^2 = \frac{(1+\frac{m}{M} )^2}{(1-\frac{m}{M} )^2}\\2=\frac{(1+\frac{m}{M} )^2}{(1-\frac{m}{M} )^2}[/tex]
where solving for the quotient m/M gives:
[tex]\frac{m}{M} =(\frac{2}{\sqrt{2} -1+1}) ^{-1}\\\frac{m}{M} =0.171572875[/tex]
My Notes You push a box of mass 24 kg with your car up to an icy hill slope of irregular shape to a height 5.7 m. The box has a speed 12.1 m/s when it starts up the hill, the same time that you brake. It then rises up to the top (with no friction) with a horizontal velocity before immediately falling off a sheer cliff to the ground (with no drag). (a) What is the speed of the box at the top of the hill?
Answer:
The speed of the box at the top of the hill will be 5.693m/s.
Explanation:
The kinetic energy of the box at the bottom of the hill is
[tex]K.E = \dfrac{1}{2}mv^2[/tex]
putting in [tex]m =24kg[/tex] and [tex]v = 12.1m/s[/tex] we get
[tex]K.E = \dfrac{1}{2}(24kg)(12.1)^2\\\\K.E = 1756.92J[/tex]
Now, the potential energy this box gains as it rises [tex]h =5.7m[/tex] up the hill is
[tex]P.E = mgh[/tex]
[tex]P.E = (24kg)(10ms/s^2)(5.7m)\\\\P.E = 1368[/tex]
Therefore, the energy left [tex]E_{left}[/tex] in the box at the top if the hill will be
[tex]E_{left} =K.E - P.E = 1756.92J-1368J\\[/tex]
[tex]\boxed{E_{left} = 388.92J}[/tex]
This left-over energy must appear as the kinetic energy of the box at the top of the hill (where else could it go? ); therefore,
[tex]\dfrac{1}{2}mv_t^2= 388.92J[/tex]
putting in numbers and solving for [tex]v_t[/tex] we get:
[tex]\boxed{v_t = 5.693m/s.}[/tex]
Thus, the speed of the box at the top of the hill is 5.693m/s.
The diagram shows two balls before they collide.
2 balls with grey arrows pointing to them from the outside. The left ball has below it m subscript 1 = 0.6 kilograms v subscript 1 = 0.5 meters per second. The right ball has below it m subscript 2 = 0.5 kilograms v subscript 2 = negative 0.2 meters per second.
What is the momentum of the system after the collision?
Answer:
B.
Explanation:
I'm taking the test right now.
The final momentum of the system is 0.4kgm/s
According to the law conservation of momentum, the momentum of the system before the collision and after the collision remains consevred.
if [tex]m_{1}[/tex] is the mass of one ball and its initial and final velocities be [tex]v_{1}[/tex] and [tex]v_{1f}[/tex] respectively, and
if [tex]m_{2}[/tex] is the mass of one ball and its initial and final velocities be [tex]v_{2}[/tex] and [tex]v_{2f}[/tex] respectively
then, [tex]m_{1}v_{1} + m_{2}v_{2}=m_{1}v_{1f} + m_{2}v_{2f}[/tex]
Momentum after collision = [tex]m_{1}v_{1} + m_{2}v_{2}[/tex][tex]=0.6*0.5+0.5*0.2[/tex]
Momentum after collision = 0.4 kgm/s
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A planet is discovered orbiting the star 51 Peg with a period of four days (0.01 years). 51 Peg has the same mass as the Sun. Mercury's orbital period is 0.24 years, and Venus's is 0.62 years.
The average orbital radius of this planet is:
a) less than Mercury's.
b) between Mercury's and Venus's.
c) greater than Venus's.
Answer:
a) less than Mercury's.
Explanation:
For orbital time period of a planet , the expression is
T² = 4π² R³ / GM
T is time period , R is radius of orbit , G is universal gravitational constant , M is mass of the star or sun
T² ∝ R³
As radius of orbit increases , time period increases . The given planet is making around a star whose mass is equal that of sun so M is same as sun .
The given planet has time period equal to .01 years which is less than that of Mercury and Venus , hence its R will be less than orbit of both of them or less than mercury's .
A physics student hurries through their lab, releasing the bob of a simple pendulum from a height, and allowing it to swing. He measures the period of the pendulum and calculates that g = 8 m/s2. What might the student have done wrong? Select all that apply. The length of the pendulum string was too long, so the equation for the period of a pendulum was no longer valid in this case. The string wasn't taut when he released the bob, causing the bob to move erratically. The student only timed one cycle, introducing a significant timing error. The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case. The mass of the bob was too large, so the equation for the period of a pendulum was no longer valid in this case. Instead of releasing the bob from rest, the student threw the bob downward.
Answer:
The string wasn't taut when he released the bob, causing the bob to move erratically.
The student only timed one cycle, introducing a significant timing error.
The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case.
Explanation:
The period of a simple pendulum is given by the formula
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]
where
T is the period of oscillation
L is the length of the pendulum
g is the acceleration due to gravity
Therefore, it is possible to measure the value of [tex]g[/tex] in an experiment, by taking a pendulum, measuring its length L, and measuring its period of oscillation T. Re-arranging the equation above, we get the value of g as:
[tex]g=(\frac{2\pi}{T})^2 L[/tex]
Here the value of g measured in the experiment is [tex]8 m/s^2[/tex] instead of [tex]9.8 m/s^2[/tex]. Let's now analyze the different options:
The length of the pendulum string was too long, so the equation for the period of a pendulum was no longer valid in this case. --> FALSE. There is no constraint on the length of the pendulum.
The string wasn't taut when he released the bob, causing the bob to move erratically. --> TRUE. This is possible, as if the string is not taut, the pendulum would not start immediately its oscillation, so the period would be larger causing a smaller value measured for g.
The student only timed one cycle, introducing a significant timing error. --> TRUE. This is also impossible: in fact, we can get a more accurate measurement of the period if we measure several oscillations (let's say 10), and then we divide the total time by 10.
The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case. --> TRUE. The formula written above for the period of the pendulum is valid only for small angles.
The mass of the bob was too large, so the equation for the period of a pendulum was no longer valid in this case. --> FALSE. The equation that gives the period of the pendulum does not depend on the mass.
Instead of releasing the bob from rest, the student threw the bob downward. --> FALSE. In fact, this force would have been applied only at the very first moment, but then later the only force acting on the pendulum is the force of gravity, so the formula of the period would still be valid.
Answer:
The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case.
The string wasn't taut when he released the bob, causing the bob to move erratically.
The student only timed one cycle, introducing a significant timing error.
Instead of releasing the bob from rest, the student threw the bob downward.
Explanation:
The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case. your equation only works for angles > 20 degrees
The string wasn't taut when he released the bob, causing the bob to move erratically. this could lead to errors in either direction
The student only timed one cycle, introducing a significant timing error. unforseen events like wind could heavily influence one cycle.
Instead of releasing the bob from rest, the student threw the bob downward. this would artificially decrease your period (T) and therefor throw the other variables in your equation off. since L is a constant g would be forced to change.
he uniform 110-kg beam is freely hinged about its upper end A and is initially at rest in the vertical position withθ= 0. Determine the initial angular accelerationα of the beam and the magnitudeFAof the force supported by the pin at A due to the application of the force P = 350 N on the attached cab
Answer:
The initial angular acceleration of the beam is [tex]{\bf{1}}{\bf{.3056}}\,\frac{{{\bf{rad}}}}{{{{\bf{s}}^{\bf{2}}}}}[/tex]
The magnitude of the force at A is 832.56N
Explanation:
Here, m is the mass of the beam and l is the length of the beam.
[tex]I =\frac{1}{3} ml[/tex]
[tex]I=\frac{1}{3} \times110\times4^2\\I=586.67kgm^2[/tex]
Take the moment about point A by applying moment equilibrium equation.
[tex]\sum M_A =I \alpha[/tex]
[tex]P \sin45^0 \times 3 =I \alpha[/tex]
Here, P is the force applied to the attached cable and [tex]\alpha[/tex] is the angular acceleration.
Substitute 350 for P and 586.67kg.m² for I
[tex]350 \sin 45^0 \times3=568.67 \alpha[/tex]
[tex]\alpha =1.3056rad/s^2[/tex]
The initial angular acceleration of the beam is [tex]{\bf{1}}{\bf{.3056}}\,\frac{{{\bf{rad}}}}{{{{\bf{s}}^{\bf{2}}}}}[/tex]
Find the acceleration along x direction
[tex]a_x = r \alpha[/tex]
Here, r is the distance from center of mass of the beam to the pin joint A.
Substitute 2 m for r and 1.3056rad/s² for [tex]\alpha[/tex]
[tex]a_x = 2\times 1.3056 = 2.6112m/s^2[/tex]
Find the acceleration along the y direction.
[tex]a_y = r \omega ^2[/tex]
Here, ω is angular velocity.
Since beam is initially at rest,ω=0
Substitute 0 for ω
[tex]a_y = 0[/tex]
Apply force equilibrium equation along the horizontal direction.
[tex]\sum F_x =ma_x\\A_H+P \sin45^0=ma_x[/tex]
[tex]A_H + 350 \sin45^0=110\times2.6112\\\\A_H=39.75N[/tex]
Apply force equilibrium equation along the vertical direction.
[tex]\sum F_y =ma_y\\A_V-P \cos45^0-mg=ma_y[/tex]
[tex]A_v +350 \cos45^0-110\times9.81=0\\A_V = 831.61\\[/tex]
Calculate the resultant force,
[tex]F_A=\sqrt{A_H^2+A_V^2} \\\\F_A=\sqrt{39.75^2+691.61^2} \\\\= 832.56N[/tex]
The magnitude of the force at A is 832.56N
Answer:
a) Initial angular acceleration of the beam = 1.27 rad/s²
b) [tex]F_{A} = 851.11 N[/tex]
Explanation:
[tex]tan \theta = \frac{opposite}{Hypothenuse} \\tan \theta = \frac{3}{3} = 1\\\theta = tan^{-1} 1 = 45^{0}[/tex]
Force applied to the attached cable, P = 350 N
Mass of the beam, m = 110-kg
Mass moment of the inertia of the beam about point A = [tex]I_{A}[/tex]
Using the parallel axis theorem
[tex]I_{A} = I_{G} + m(\frac{l}{2} )^{2} \\I_{G} = \frac{ml^{2} }{12} \\I_{A} = \frac{ml^{2} }{12} + \frac{ml^{2} }{4} \\I_{A} = \frac{ml^{2} }{3}[/tex]
Moment = Force * Perpendicular distance
[tex]\sum m_{A} = I_{A} \alpha\\[/tex]
From the free body diagram drawn
[tex]\sum m_{A} = 3 Psin \theta\\ 3 Psin \theta = \frac{ml^{2} \alpha }{3}[/tex]
P = 350 N, l = 3+ 1 = 4 m, θ = 45°
Substitute these values into the equation above
[tex]3 * 350 * sin 45 = \frac{110 * 4^{2}* \alpha }{3} \\\alpha = 1.27 rad/s^{2}[/tex]
Linear acceleration along the x direction is given by the formula
[tex]a_{x} = r \alpha[/tex]
r = 2 m
[tex]a_{x} = 2 * 1.27\\a_{x} = 2.54 m/s^{2}[/tex]
the linear acceleration along the y-direction is given by the formula
[tex]a_{y} = r w^{2}[/tex]
Since the beam is initially at rest, w = 0
[tex]a_{y} = 0 m/s^{2}[/tex]
General equation of motion along x - direction
[tex]F_{x} + Psin \theta = ma_{x}[/tex]
[tex]F_{x} + 350sin45 = 110 * 2.54\\F_{x} = 31.913 N[/tex]
General equation of motion along y - direction
[tex]F_{x} + Pcos \theta - mg = ma_{y}[/tex]
[tex]F_{y} + 350cos45 - 110*9.8 = m* 0\\F_{y} = 830.513 N[/tex]
Magnitude [tex]F_{A}[/tex] of the force supported by the pin at A
[tex]F_{A} = \sqrt{F_{x} ^{2} + F_{y} ^{2} } \\F_{A} = \sqrt{31.913 ^{2} + 850.513 ^{2} } \\F_{A} = 851.11 N[/tex]
A bowling ball (which we can regard as a uniform sphere) has a mass of 3.63 kg and a diameter of 0.216 m. A baseball has a mass of 0.145 kg. If you connect these two balls with a lightweight rod, what must be the distance between the center of the bowling ball and the center of the baseball so that the system of the two balls and the rod will balance at the point where the rod touches the surface of the bowling ball?
Answer:
Distance will be 2.81 m
Explanation:
Detailed explanation and calculation is shown in the image below.
In the formula 4h²o how many total hydrogen atoms are there
Which liquid is the most viscous?
Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.57. Express your answer using two significant figures.
To move a 32 kg crate with a coefficient of static friction of 0.57 with the floor, the force necessary to start the crate moving is 180 N after rounding to two significant figures.
Explanation:To find the force necessary to start the crate moving, we need to use the concept of static friction. The force of static friction is given by the equation fs=μsN, where fs is the static friction, μs is the coefficient of static friction and N is the normal force. The normal force is equal to the product of the mass and gravitational acceleration (N = mg), where m is the mass and g is the acceleration due to gravity. Given that the mass of the crate is 32 kg and the coefficient of static friction is 0.57, we first calculate N = (32kg)(9.8m/s²) = 313.6 N.
Then, by substituting these values into the equation, we have fs = (0.57)(313.6 N) =178.85 N. Therefore, the force necessary to start the crate moving is 178.85 N, but we express the answer using two significant figures, so it is 180 N.
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5. 3.4 kg of water is heated from 78 C to boiling and it is all turned to steam at
100C. How much heat did it take?
Answer:
7.7 MJ
Explanation:
Let water specific heat be c = 0.004186 J/kgC and specific latent heat of vaporization be L = 2264705 J/kg
There would be 2 kinds of heat to achieve this:
- Heat to change water temperature from 78C to boiling point:
[tex]H_1 = mc\Delta t = 3.4 * 0.004186 * (100 - 78) = 0.313 J[/tex]
- Heat to turn liquid water into steam:
[tex]H_2 = mL = 3.4 * 2264705 \approx 7.7 \times 10^6J[/tex] or 7.7 MJ
So the total heat it would take is [tex]H_1 + H_2 = 7.7 \times 10^6 + 0.313 \approx 7.7 MJ[/tex]
A falcon is soaring over a prairie, flying at a height of 49.0 m with a speed of 13.5 m/s. The falcon spots a mouse running along the ground and dives to catch its dinner. Ignoring air resistance, and assuming the falcon is only subject to the gravitational force as it dives, how fast will the falcon be moving the instant it is 5.00 m above the ground
Answer:
32.3 m/s
Explanation:
We can solve this problem by applying one of Newton's equations of motion:
[tex]v^2 = u^2 + 2gs[/tex]
where v = final velocity of falcon
u = initial velocity of falcon
g = acceleration due to gravity ([tex]9.8 m/s^2[/tex])
s = distance moved by falcon
From the question, we have that:
u = 13.5 m/s
s = initial height - new height = 49.0 - 5.0 = 44.0 m
Hence, to find the velocity when it has traveled 44 m towards the ground (5 m above the ground):
[tex]v^2 = 13.5^2 + (2 * 9.8 * 44)\\\\\\v^2 = 182.25 + 862.4\\\\\\v^2 = 1044.65\\\\\\v = \sqrt{1044.65} \\\\\\v = 32.3 m/s[/tex]
The velocity of the falcon at the instant when it is 5.0 m above the ground is 32.3 m/s
In the United States, the average person eats about 3000 food Calories per day. What is the average power of this energy intake? 14.5 watts 14.5 watts 145 watts 145 watts 1450 watts 1450 watts 14 , 500 watts
Answer:
Explanation:
3000 Calorie = 3000 x 1000 calorie
1 calorie = 4.2 Joule
3000 x 1000 calorie = 4.2 x 3000 x 1000 J
= 12.6 x 10⁶ J
It is taken in one day
time period of one day = 24 x 60 x 60 second
= 86400 s
energy given by this intake in 86400s is 12.6 x 10⁶ J
power = energy / time
= 12.6 x 10⁶ / 86400 J/s
= 145.8 J/s or W
= 145 W.
Final answer:
The average power of energy intake for a person consuming 1,300 to 3,000 kcals per day translates to approximately 63 to 145 watts. This calculation is done by converting calories to joules and then dividing by the total number of seconds in a day.
Explanation:
In order to calculate the average power of the energy intake from food, we can convert the amount of energy consumed per day into watts. For a range of 1,300 to 3,000 kcal per day, we need to convert these values to joules, since 1 kcal = 4.184 kJ. Therefore, 1,300 kcal converts to 1,300 x 4.184 kJ = 5,439.2 kJ and 3,000 kcal to 3,000 x 4.184 kJ = 12,552 kJ.
Next, we divide each amount by the number of seconds in one day to find the average power in watts. There are 86,400 seconds in a day. Thus, the power for 1,300 kcal/day is 5,439.2 kJ / 86,400 s = approximately 63 watts, and for 3,000 kcal/day is 12,552 kJ / 86,400 s = approximately 145 watts.
The power range for 1,300 to 3,000 kcal/day is thus approximately 63 watts to 145 watts.
A horse shoe magnet is placed on a mass balance such that a uniform magnetic field of magnitude B runs between it from North to South. A coil of resistance R is connected to a battery which supplies a potential difference of V across the coil and is suspended such that a section of the coil of length L meters lies between it with current running from East to West. The mass balance measures a mass of M. • What is the measured change in mass due to the effect of Fmag? • What is the total measured mass of the magnets? Keep in mind the effect of Newton's third law.
Answer:
(a) Measured change in mass (Δm) = BVL/Rg
(b) Total measured mass M' = M - BVL/Rg
Explanation:
Current (I) across is coil is given by the formula;
I = V/R ------------------------1
The magnetic force is given by the formula;
Fb = B*I*L -------------------2
Putting equation 1 into equation 2, we have;
Fb = B*V*L/R -------------------3
Change in mass (Δm) is given as:
Δm = Fb/g -----------------------4
Putting equation 3 into equation 4, we have;
Δm = BVL/Rg
Therefore, change in mass (Δm) = BVL/Rg
2. Since B runs from North to South and current running from East to West, then the magnetic force is directed upward.
Therefore,
Total measure mass M' = M - BVL/Rg
A simple pendulum consisting of a small object of mass m attached to a string of length l has a period T.
A pendulum with what combination of object mass m and string length l will also have period T?
Any pendulum that will have the same period with mass, m, and length, L, must have the ratio of (L/g) and the ratio of its mass to force constant (m/k) must also be equal to this ratio.
For a simple pendulum, the period is given as
[tex]\bold {T = 2\pi \sqrt{\dfrac L{g}}}[/tex]
This is also given as
[tex]\bold {T = 2\pi \sqrt{\dfrac m{k}}}[/tex]
where
T = period of oscillation
m = mass of the pendulum
L = length
g = acceleration due to gravity
k = force constant
Equate these equations,
[tex]\bold {T = 2\pi \sqrt{\dfrac L{g}}} = \bold {T = 2\pi \sqrt{\dfrac m{k}}}\\\\\bold {\bold { \dfrac L{g} = \dfrac m{k}}} }\\\\\bold {\bold { \dfrac m{L} = \dfrac k{g}}} }[/tex]
So, any pendulum that will have the same period with mass, m, and length, L, must have the ratio of (L/g) and the ratio of its mass to force constant (m/k) must also be equal to this ratio.
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Two long, straight wires are separated by a distance of 9.15 cm . One wire carries a current of 2.79 A , the other carries a current of 4.36 A .Is the force per meter exerted on the 4.36-A wire greater than, less than, or the same as the force per meter exerted on the 2.79-A wire
Answer:
The force is the same
Explanation:
The force per meter exerted between two wires carrying a current is given by the formula
[tex]\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}[/tex]
where
[tex]\mu_0[/tex] is the vacuum permeability
[tex]I_1[/tex] is the current in the 1st wire
[tex]I_2[/tex] is the current in the 2nd wire
r is the separation between the wires
In this problem
[tex]I_1=2.79 A\\I_2=4.36 A\\r = 9.15 cm = 0.0915 m[/tex]
Substituting, we find the force per unit length on the two wires:
[tex]\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(2.79)(4.36)}{2\pi (0.0915)}=2.66\cdot 10^{-5}N[/tex]
However, the formula is the same for the two wires: this means that the force per meter exerted on the two wires is the same.
The same conclusion comes out from Newton's third law of motion, which states that when an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A (action-reaction). If we apply the law to this situation, we see that the force exerted by wire 1 on wire 2 is the same as the force exerted by wire 2 on wire 1 (however the direction is opposite).
How could you increase the width (in centimeters) of the central maximum of the diffraction pattern on the screen? There may be more than one correct method, so select all those that would work. In each case, assume you only do the one named thing and make no other changes.
Answer: 1) use of light with longer wavelenght will increase the width of the central maxima
2) increasing the number of slit will increase the width of the central maxima
3) narrowing the slit will increase the central maxima.
In order to determine the mass moment of inertia of a flywheel of radius 600 mm, a 12-kg block is attached to a wire that is wrapped around the flywheel. The block is released and is observed to fall 3 m in 4.6 s. To eliminate bearing friction from the computation, a second block of mass 24 kg is used and is observed to fall 3 m in 3.1 s. Assuming that the moment of the couple due to friction remains constant, determine the mass moment of inertia of the flywheel.
Answer:
Explanation:
Given that,
When Mass of block is 12kg
M = 12kg
Block falls 3m in 4.6 seconds
When the mass of block is 24kg
M = 24kg
Block falls 3m in 3.1 seconds
The radius of the wheel is 600mm
R = 600mm = 0.6m
We want to find the moment of inertia of the flywheel
Taking moment about point G.
Then,
Clockwise moment = Anticlockwise moment
ΣM_G = Σ(M_G)_eff
M•g•R - Mf = I•α + M•a•R
Relationship between angular acceleration and linear acceleration
a = αR
α = a / R
M•g•R - Mf = I•a / R + M•a•R
Case 1, when y = 3 t = 4.6s
M = 12kg
Using equation of motion
y = ut + ½at², where u = 0m/s
3 = ½a × 4.6²
3 × 2 = 4.6²a
a = 6 / 4.6²
a = 0.284 m/s²
M•g•R - Mf = I•a / R + M•a•R
12 × 9.81 × 0.6 - Mf = I × 0.284/0.6 + 12 × 0.284 × 0.6
70.632 - Mf = 0.4726•I + 2.0448
Re arrange
0.4726•I + Mf = 70.632-2.0448
0.4726•I + Mf = 68.5832 equation 1
Second case
Case 2, when y = 3 t = 3.1s
M= 24kg
Using equation of motion
y = ut + ½at², where u = 0m/s
3 = ½a × 3.1²
3 × 2 = 3.1²a
a = 6 / 3.1²
a = 0.6243 m/s²
M•g•R - Mf = I•a / R + M•a•R
24 × 9.81 × 0.6 - Mf = I × 0.6243/0.6 + 24 × 0.6243 × 0.6
141.264 - Mf = 1.0406•I + 8.99
Re arrange
1.0406•I + Mf = 141.264 - 8.99
1.0406•I + Mf = 132.274 equation 2
Solving equation 1 and 2 simultaneously
Subtract equation 1 from 2,
Then, we have
1.0406•I - 0.4726•I = 132.274 - 68.5832
0.568•I = 63.6908
I = 63.6908 / 0.568
I = 112.13 kgm²
In Problem 1, the rake angle was changed to a neutral rake angle. If the friction angle remains the same, determine: (a) the shear plane angle (b) the chip thickness (c) the shear strain for the operation.
Answer: seen in the explanations
Explanation:
Since no values for this problem is given, I will just give the formulas to solve the problem in mechanics of orthogonal plane cutting.
a) shear plane angle will be calculated using:
tan ϕ = r Cos α / 1- r sin α
Where :
r= chip thickness ratio
α = rake angle
ϕ = shear angle.
B) chip thickness
To get the chip thickness, we must calculate the chip thickness ratio using the formula:
Rt = Lc/ L
Where:
Lc = length of chip formed
L= uncut chip length.
Use the answer to solve for chip thickness with the formula;
Tc = t/ Rt
Where :
t = depth of cut
Rt = chip thickness ratio
C) shear strain for the operation will be solved using;
ε = cot β• + tan β•
Where:
ε = shear strain
β• = orthogonal shear angle.
20 POINTS! TRUE OR FALSE:
If a stationary box does not move when you push against it is because of kinetic friction.
A) TRUE
B) FALSE
Answer:
False.
Explanation:
Kinetic energy means it must move
Answer: I believe that it is A) True
Disclaimer: I'm not quite sure. We learned about this recently though so possibly I'm right. Good luck though!
Explanation:
A disk 8.04 cm in radius rotates at a constant rate of 1 220 rev/min about its central axis. (a) Determine its angular speed. rad/s (b) Determine the tangential speed at a point 3.02 cm from its center. m/s (c) Determine the radial acceleration of a point on the rim. magnitude km/s2 direction (d) Determine the total distance a point on the rim moves in 2.02 s. m
Answer:
A) 128 rad/s
B) 3.87 m/s
C) 1317.17 m/s²
D) 20.79m
Explanation:
A) We are given the angular speed as 1220 rev/min. Now let's convert it to rad/s.
1220 (rev/min) x (2πrad/1rev) x (1min/60sec) = (1220 x 2π)/60 rad/s = 127.76 rad/s ≈ 128 rad/s
B) Formula for tangential speed is given as;
v = ωr
We know that ω = 128 rad/s
Also, r = 3.02cm = 0.0302m
Thus, v = 128 x 0.0302 = 3.87 m/s
C) Formula for radial acceleration is given as;
a_c = v²/r
From earlier, v = ωr
Thus, a_c = v²/r = (ωr)²/r = ω²r
On the rim, r = 8.04cm = 0.0804
a_c = 128² x 0.0804 = 1317.17 m/s²
D) We know that; distance/time = speed
Thus, distance = speed x time
D = vt
From earlier, v = ωr
Thus, D = ωrt
Plugging in the relevant values ;
D = 128 x 0.0804 x 2.02 = 20.79m
A bubble of air has a volume of 16 cm3 when at a depth of 6.5 m. As the bubble rises, it expands. What is the bubble’s volume just below the surface of the water? (Note: Assume the temperature of the air in the bubble doesn’t change, in which case pressure times volume is constant.)
Final answer:
The volume of the bubble just below the surface of the water is 16 cm3.
Explanation:
To solve this problem, we can use Boyle's law, which states that the pressure times the volume of a gas is constant if the temperature remains constant. In this case, the volume of the air bubble at the bottom of the water is 16 cm3 and the depth is 6.5 m. As the bubble rises, the pressure decreases, causing the volume to increase. We can set up a proportion to solve for the volume just below the surface of the water.
Using the formula P1V1 = P2V2, where P1 is the pressure at the bottom of the water, V1 is the initial volume, P2 is the pressure just below the surface, and V2 is the volume just below the surface, we can plug in the values and solve for V2.
Since the temperature of the air in the bubble doesn't change, we can assume that the pressure times the volume is constant. Therefore, we have:
P1V1 = P2V2
Using the given values, we have:
(1 atm)(16 cm3) = (?, just below the surface) V2
To find V2, we can rearrange the equation:
V2 = (P1V1) / P2
Since the pressure just below the surface is atmospheric (1 atm), we can substitute that value in:
V2 = (1 atm)(16 cm3) / (1 atm)
V2 = 16 cm3
Therefore, the volume of the bubble just below the surface of the water is 16 cm3.
Astronaut Jennifer's lifeline to her spaceship comes loose and she finds herself stranded, "floating" 100 m from the mothership. She suddenly throws her 2.00-kg wrench at 20 m/s in a direction away from the ship. If she and her spacesuit have a combined mass of 200 kg, how long does it take her to coast back to her spaceship
Answer:
The time it will take the astronaut to coast back to her ship is 500 seconds
Explanation:
Mass of wrench, m₁ = 2 kg
Mass of astronaut with spacesuit, m₂ = 200 kg
Velocity of wrench, v₁ = 20 m/s
Velocity of astronaut with spacesuit v₂
Distance between the astronaut and the ship is 100 m
As linear momentum is conserved
[tex]m_1v_1+m_2v_2=0\\\\\Rightarrow v_2=\frac{m_1v_1}{m_2}\\\\\Rightarrow v_2=\frac{2\times 20}{200}\\\\\Rightarrow v_2=0.2\ m/s[/tex]
[tex]Time = \frac{ Distance}{Speed}[/tex]
[tex]Time=\frac{100}{0.2}=500\ s[/tex]
Hence, The time it will take the astronaut to coast back to her ship is 500 seconds
By conserving momentum, we find that Astronaut Jennifer will move at a velocity of 0.2 m/s after throwing the wrench. It will take her 500 seconds to travel the 100 meters back to her spaceship.
Explanation:To determine the time it takes for Astronaut Jennifer to return to her spaceship, we will use the conservation of momentum principle. Since there are no external forces acting on the system in space, the momentum before and after Jennifer throws the wrench is conserved.
Let's calculate the velocity of Jennifer after she throws the wrench using the following equation:
Momentum before = Momentum after
(mass of Jennifer) × (initial velocity of Jennifer) + (mass of wrench) × (initial velocity of wrench) = (mass of Jennifer) × (final velocity of Jennifer) + (mass of wrench) × (final velocity of wrench)
Since Jennifer and the wrench start at rest, their initial velocities are 0, so:
0 + 0 = 200 kg × (final velocity of Jennifer) + 2.00 kg × (-20 m/s)
Therefore, the final velocity of Jennifer v_j is:
200 kg × v_j = -2.00 kg × (-20 m/s)
v_j = (2.00 kg × 20 m/s) / 200 kg
v_j = 0.2 m/s
Now, we need to find out how long it takes her to cover 100 meters at this velocity:
Time = Distance / Velocity
Time = 100 m / 0.2 m/s
Time = 500 s
It would take Jennifer 500 seconds to coast back to her spaceship.
why is the motion of simple pendulum not strictly a simple harmonic motion
Final answer:
The motion of a simple pendulum is not strictly a simple harmonic motion. While it exhibits some characteristics of simple harmonic motion, such as periodic motion and a restoring force, the restoring force is not linearly proportional to the displacement.
Explanation:
The motion of a simple pendulum is not strictly a simple harmonic motion because it does not follow Hooke's law. Simple harmonic motion is oscillatory motion for a system that can be described only by Hooke's law. However, a simple pendulum is governed by the law of conservation of energy and the restoring force is not directly proportional to the displacement.
While a simple pendulum exhibits some characteristics of simple harmonic motion, such as periodic motion and a restoring force, it deviates from true simple harmonic motion because the restoring force is not linearly proportional to the displacement. Instead, the restoring force for a simple pendulum is proportional to the sine of the displacement angle.
For example, when a pendulum is displaced to one side, it experiences a restoring force in the opposite direction. But the magnitude of the restoring force does not increase linearly with the displacement. Instead, it follows the relationship F = -mg sin(theta), where F is the restoring force, m is the mass of the pendulum bob, g is the acceleration due to gravity, and theta is the displacement angle.
Which one of the following substances can be separated into several different elements?
A: Air
B: Iron
C: hydrogen
D: nickel
The substance that can be separated into several different elements is air. The correct option is A.
Air, which is a mixture mainly composed of nitrogen and oxygen, among other gases, can thus be separated into its component elements. Substances like Iron (Fe), Hydrogen (H), and Nickel (Ni) are elements and cannot be broken down into simpler substances by ordinary chemical means.
It is proposed that future space stations create an artificial gravity by rotating. Suppose a space station is constructed as a 1000-m-diameter cylinder that rotates about its axis. The inside surface is the deck of the space station. What rotation period will provide "normal" gravity?
To solve this problem, apply the concepts related to the centripetal acceleration as the equivalent of gravity, and the kinematic equations of linear motion that will relate the speed, the distance traveled and the period of the body to meet the needs given in the problem. Centripetal acceleration is defined as,
[tex]a_c = \frac{v^2}{r}[/tex]
Here,
v = Tangential Velocity
r = Radius
If we rearrange the equation to get the velocity we have,
[tex]v = \sqrt{a_c r}[/tex]
But at this case the centripetal acceleration must be equal to the gravitational at the Earth, then
[tex]v = \sqrt{gr}[/tex]
[tex]v = \sqrt{(9.8)(\frac{1000}{2})}[/tex]
[tex]v = 70m/s[/tex]
The perimeter of the cylinder would be given by,
[tex]\phi = 2\pi r[/tex]
[tex]\phi = 2\pi (500m)[/tex]
[tex]\phi = 3141.6m[/tex]
Therefore now related by kinematic equations of linear motion the speed with the distance traveled and the time we will have to
[tex]v = \frac{d}{t} \rightarrow \text{ But here } d = \phi[/tex]
[tex]v = \frac{\phi}{t} \rightarrow t = \frac{\phi}{t}[/tex]
[tex]t = \frac{3141.6}{70m/s}[/tex]
[tex]t = 44.9s[/tex]
Therefore the period will be 44.9s
Answer:
[tex]T \approx 44.88\,s[/tex]
Explanation:
"Normal" gravity is equal to 9.807 meters per squared second and cylinder must rotate at constant speed in order to simplify the equation of acceleration, which is in the radial direction. The centrifugal acceleration experimented by people allow them to be on the inside surface.
[tex]g = \omega^{2}\cdot R[/tex]
The angular speed required to provide "normal" gravity is:
[tex]\omega = \sqrt{\frac{g}{R} }[/tex]
[tex]\omega = \sqrt{\frac{9.807\,\frac{m}{s^{2}} }{500\,m} }[/tex]
[tex]\omega \approx 0.14\,\frac{rad}{s}[/tex]
The rotation period is:
[tex]T = \frac{2\pi}{\omega}[/tex]
[tex]T = \frac{2\pi}{0.14\,\frac{rad}{s} }[/tex]
[tex]T \approx 44.88\,s[/tex]
Hola comastas muy bein
Answer:
v:
xddddddddddddddddd
Answer: hola
Explanation:спасибо за очки
The unit of current, the ampere, is defined in terms of the force between currents. Two 1.0-meter-long sections of very long wires a distance 5.0 mm apart each carry a current of 1.0 AA.What is the force between them? (If the force between two actual wires has this value, the current is defined to be exactly 1 A.)
The force between two parallel conductors carrying identical currents of one ampere over a length of one meter and separated by a distance of 5 mm is 4 x 10^-5 Newtons.
Explanation:The force between two parallel conductors carrying a current is used to define the unit of current, the ampere. According to the definition, one ampere of current through each of two parallel conductors of infinite length, separated by one meter in empty space free of other magnetic fields, causes a force of exactly 2x10^-7 N/m on each conductor. But the wires in your question are shorter and closer together.
In this case, a simplified formula can be used which is F = (2 * k * I1 * I2 * L) / r, where k = 10^-7 N/A², I1 and I2 are the currents in the wires, L the length of the wires, and r is the distance between the wires. With I1 = I2 = 1 A, L = 1 m, and r = 5 mm = 0.005 m, the formula gives: F = (2 * 10^-7 N/A² * 1 A * 1 A * 1 m) / 0.005 m = 4 * 10^-5 N.
Therefore, the force between these two wires is 4 * 10^-5 Newtons.
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The force between the two sections of wire carrying a current of 1.0 A each and separated by a distance of 5.0 mm is 1x10*-9 N per meter
Explanation:The force between two 1.0-meter-long sections of very long wires carrying a current of 1.0 A each can be calculated using the definition of the ampere. According to the definition, one ampere of current through each of two parallel conductors separated by one meter in empty space causes a force of exactly 2x10-7 N/m on each conductor. In this case, the wires are separated by a distance of 5.0 mm, which is equivalent to 0.005 m. Therefore, the force between the wires is 2x10-7 N/m * 0.005 m = 1x10-9 N per meter of separation.
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You lift a 25-kg child 0.80 m, slowly carry him 10 m to the playroom, and finally set him back down 0.80 m onto the playroom floor. What work do you do on the child for each part of the trip and for the whole trip
To solve this problem we will apply the work theorem which is expressed as the force applied to displace a body. Considering that body strength is equivalent to weight, we will make the following considerations
[tex]\text{Mass of the child} = m = 25kg[/tex]
[tex]\text{Acceleration due to gravity} = g = 9.81m/s^2[/tex]
[tex]\text{Height lifted} = h = 0.80m (Upward)[/tex]
Work done to upward the object
[tex]W = mgh[/tex]
[tex]W = (25)(9.81)(0.8)[/tex]
[tex]W = 196.2J[/tex]
Horizontal Force applied while carrying 10m,
[tex]F = 0N[/tex]
[tex]W = 0J[/tex]
Height descended in setting the child down
[tex]h' = -0.8m (Downwoard)[/tex]
[tex]W = mgh'[/tex]
[tex]W = (25)(9.81)(-0.80)[/tex]
[tex]W = -196.2J[/tex]
For full time, assuming that the total value of work is always expressed in terms of its symbol, it would be zero, since at first it performs the same work that is later complemented in a negative way.
To calculate the work done on the child during different parts of the trip and for the whole trip, we need to consider the weight of the child and the vertical distances involved. The work done in lifting the child up and setting the child back down is equal to the weight of the child multiplied by the vertical distance. The work done in carrying the child horizontally is zero.
Explanation:In this problem, we need to calculate the work done on the child during different parts of the trip and for the whole trip. Work, W, is defined as the product of force and displacement, W = Fd. Since the child is being lifted vertically, the work done in lifting the child up is equal to the weight of the child multiplied by the vertical distance lifted, Wup = mgh. The work done in carrying the child horizontally is zero since the displacement is perpendicular to the force. Finally, the work done in setting the child back down is also equal to the weight of the child multiplied by the vertical distance lowered, Wdown = mgh. Therefore, the total work done on the child for the whole trip is the sum of the work done in lifting the child up and setting the child back down, Wtotal = Wup + Wdown.
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The roads are icy, and you observe a head-on collision on Summit, at the corner with Rhodes: a 1ton car swerves out of his lane and slides through a stop sign at 34mph straight into a 3ton SUV traveling at 13mph in the other direction. The car and the SUV crumple from the collision, and stick together. What is the final velocity, in MPH (you don't need to enter MPH in your answer) , of the SUV/car entanglement (the positive direction is the direction the car was initially going)?
Answer:
v = -1.3 mph
Explanation:
Assuming no external forces acting during the collision, total momentum must be conserved, so the following condition must be met:[tex]p_{init} = p_{final} (1)[/tex]
The initial momentum is the sum of the momenta of both vehicles, taking into account their relative velocities as they are going in opposite directions. If we take as positive the direction the car was initially going, we can write the following expression:[tex]p_{init} = m_{car} * v_{car} - m_{SUV} * v_{SUV}[/tex]
Replacing by the values of the masses of both vehicles and their speeds, we have:[tex]p_{init} = 1t*34 mph - 3t*13 mph = - 5t*mph (2)[/tex]
This must be equal to the final momentum of the car/SUV entanglement, as follows:[tex]p_{final} =( m_{car} + m_{SUV} )* v_{final} (3)[/tex]
Replacing in (3) for the masses, and equating (1) and (3), we can solve for vfinal, as follows:[tex]v_{final} = \frac{p_{init}}{(m_{car} + m_{SUV} } = \frac{-5t*mph}{4t} = -1.3 mph[/tex]
This means that the car/SUV entanglement will move in the opposite direction that the car was initially going.