A horizontal force of P=100 N is just sufficient to hold the crate from sliding down the plane, and a horizontal force of P=350 N is required to just push the crate up the plane. Determine the coefficient of static friction between the plane and the crate, and find the mass of the crate.
Answers
"down/up the plane" suggests an inclined plane, but no angle is given so I'll call it θ for the time being.
The free body diagram for the crate in either scenario is the same, except for the direction in which static friction is exerted on the crate. With the P = 100 N force holding up the crate, static friction points up the incline and keeps the crate from sliding downward. When P = 350 N, the crate is pushed upward, so static friction points down. (see attached FBDs)
Using Newton's second law, we set up the following equations.
• p = 100 N
∑ F (parallel) = f + p cos(θ) - mg sin(θ) = 0
∑ F (perpendicular) = n - p sin(θ) - mg cos(θ) = 0
• P = 350 N
∑ F (parallel) = P cos(θ) - F - mg sin(θ) = 0
∑ F (perpendicular) = N - P sin(θ) - mg cos(θ) = 0
(where n and N are the magnitudes of the normal force in the respective scenarios; ditto for f and F which denote static friction, so that f = µn and F = µN, with µ = coefficient of static friction)
Solve for n and N :
n = p sin(θ) + mg cos(θ)
N = P sin(θ) - mg cos(θ)
Substitute these into the corresponding equations containing µ, and solve for µ :
µ = (mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ))
µ = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))
Next, you would set these equal and solve for m :
(mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ)) = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))
...
Once you find m, you back-substitute and solve for µ, but as you might expect the result will be pretty complicated. If you take a simple angle like θ = 30°, you would end up with
m ≈ 36.5 kg
µ ≈ 0.256
The coefficient of static friction between the plane and the crate is μ = 0.256 and the mass of the crate is m=36.4 kg.
From the given,
The force that opposes the crate by sliding is P = 100N
In X-axis, the sum of forces is zero.
ΣF = 0
Pcosθ - mgsinθ-Ff = 0
Ff = Pcosθ - mgsinθ
In Y-axis
Psinθ - mgcosθ - N = 0
N = Psinθ-mgcosθ
Frictional force, Ff = μN, μ is the coefficient of friction
Ff = μN
Pcos30- mgsin30 + μ( Psin30+mgcos30) = 0
μ = mgsin30-Pcos30/Psin30+mgcos30 ------1
The block is sliding with the horizontal force, F = 350N
X-axis
P₂cosθ - mgsinθ-Ff = 0
Y-axis
P₂sinθ - mgcosθ - N = 0
N = P₂sinθ-mgcosθ
μ = P₂cos30-mgsin30/P₂sin30-mgcos30 -----2
Equate equations 1 and 2
mgsin30-Pcos30/Psin30+mgcos30 =P₂cos30-mgsin30/P₂sin30-mgcos30
4.905m-86.6/50+8.49 = 303.1-4.905m/175+8.49
41.7m² + 123m - 1.516×10⁴ = 0
-41.7m² +2330m -1.516×10⁴(4.905-86.6)(175+8.49) =(303.1-4.905)(50+8.49)
83.4m² - 2207m -3.03×10⁴ = 0
m= 36.4 kg
Hence, the mass of the crate is 36.4 Kg.
Substitute the value of m in equation 1,
μ = 4.905(36.4) - 86.6 / 50 + 8.49
μ = 0.256
Thus, the coefficient of static friction is 0.256.
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Related Questions
Describe an imaginary process that satisfies the second law but violates the first law of thermodynamics.
Answers
Answer:
Explanation:
First last of thermodynamics, just discusses the changes that a system is undergoing and the processes involved in it. It explains conservation of energy for a system undergoing changes or processes.
Second law of thermodynamics helps in defining the process and also the direction of the processes. It tells about the possibility of a process or the restriction of a process. It states that the entropy of a system always increases.
For this to occur the energy contained by a body has to diminish without converting to work or internal energy. So imagine a machine which works with less than efficiency, this means there are losses but they don’t show up anywhere. But the energy is obtained from a higher energy source to lower.
The easy way to do this is with an imaginary device that extracts zero-point energy to heat a quantity of gas. Energy is being created, so the first law is violated, and the entropy of the system is increasing as the gas heats up.
First law is violated since the energy conversion don't apply but the direction of work is applied so second law is satisfied.
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when an ideal capacitor is connected across an ac voltage supply of variable frequency, the current flowing
a) is in phase with voltage at all frequencies
b) leads the voltage with a phase independent of frequency
c) leads the voltage with a phase which depends on frequency
d) lags the voltage with a phase independent of frequency
what would be the correct option?
Answers
Answer:
(b)
Explanation:
The voltage always lags the current by 90°, regardless of the frequency.
A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotates horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 26.5 m/s2 with a beam of length 5.89 m , what rotation frequency is required
Answers
Answer:
The angular acceleration is 4.5 rad/s^2.
Explanation:
Acceleration, a = 26.5 m/s2
length, L = 5.89 m
The angular acceleration is
[tex]\alpha =\frac{a}{L}\\\\\alpha = \frac{26.5}{5.89}=4.5 rad/s^2[/tex]
A wire 54.6 cm long carries a 0.480 A current in the positive direction of an x axis through a magnetic field with an x component of zero, a y component of 0.000420 T, and a z component of 0.0130 T. Find the (a) x, (b) y, and (c) z components of the magnetic force on the wire.
Answers
Answer:
wire 66.0 cm long carries a 0.750 A current in the positive direction of an x axis through a magnetic field $$\vec { B } = ( 3.00 m T ) \hat { j } ...
Top answer · 1 vote
Two pistons are connected to a fluid-filled reservoir. The first piston has an area of 3.002 cm2, and the second has an area of 315 cm2. If the first cylinder is pressed inward with a force of 50.0 N, what is the force that the fluid in the reservoir exerts on the second cylinder?
Answers
Answer:
The force on the second piston is 5246.5 N .
Explanation:
Area of first piston, a = 3.002 cm^2
Area of second piston, A = 315 cm^2
Force on first piston, f = 50 N
let the force of the second piston is F.
According to the Pascal's law
[tex]\frac{f}{a} = \frac{F}{A}\\\\\frac{50}{3.002}=\frac{F}{315}\\\\F = 5246.5 N[/tex]
A student has to work the following problem: A block is being pulled along at constant speed on a horizontal surface a distance d by a rope supplying a force F at an angle of elevation q. The surface has a frictional force acting during this motion. How much work was done by friction during this motion? The student calculates the value to be –Fd sinq. How does this value compare to the correct value?
a. It is the correct value.
b. It is too high.
c. It is too low.
d. The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.
Answers
Answer:
D
The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.
Explanation:
Since block moves with constant speed
So, frictional force
f = FCosq
Work done by friction
W = - fd
W = - fd Cos q
The answer may be greater or less than - fdSinq. It depends on the value of q which is less than, or equal to 45°.
Action and reaction are equal in magnitude and opposite in direction.Then Why do not balance each other
Answers
Answer:
Action and reaction are equal in magnitude and opposite in direction but they do not balance each other because they act on different objects so they don't cancel each other out.
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why is the water drawn from the bottom of the dam rather than the top?
Answers
Answer:
because minerals can be gotten from the bottom
Explanation:
it's self explanatory
A point charge of -3.0 x 10-C is placed at the origin of coordinates. Find the clectric field at the point 13. X= 5.0 m on the x-axis.
Answers
Answer:
-1.0778×10⁻¹⁰ N/C
Explanation:
Applying,
E = kq/r²................ equation 1
Where E = elctric field, q = charge, r = distance, k = coulomb's law
From the question,
Given: q = -3.0×10 C, r = 5.0 m
Constant: k = 8.98×10⁹ Nm²/C²
Substitute these values in equation 1
E = (-3.0×10)(8.98×10⁹)/5²
E = -1.0778×10⁻¹⁰ N/C
Hence the electric field on the x-axis is -1.0778×10⁻¹⁰ N/C
A car accelerates at 2 meters/s/s. Assuming the car starts from rest how far will it travel in 10 seconds
Answers
Answer:
Distance = velocity x time, so 10 m/s X 10 s = 100 m
Explanation:
If you accelerate at 2 m/s^2 for 10 seconds, at the end of the 10 seconds you are moving at a rate of 20 m/s.
V(f) = V(i) + a*t
Final velocity = initial velocity + acceleration x time
Your average velocity will be half of your final, because you accelerated at a constant rate. So your average velocity is 10 m/s.
Distance = velocity x time, so 10 m/s X 10 s = 100 m
Answer:
100 m
Explanation:
Given,
Initial velocity ( u ) = 0 m/s
Acceleration ( a ) = 2 m/s^2
Time ( t ) = 10 sec s
To find : Displacement ( s ) = ?
By 2nd equation of motion,
s = ut + at^2 / 2
= ( 0 ) ( 10 ) + ( 2 ) ( 10 )^2 / 2
= 0 + ( 2 ) ( 100 ) / 2
= 200 / 2
s = 100 m
A football quarterback runs 15.0 m straight down the playing field in 3.00 s. He is then hit and pushed 3.00 m straight backward in 1.71 s. He breaks the tackle and runs straight forward another 24.0 m in 5.20 s. Calculate his average velocity (in m/s) for the entire motion. (Assume the quarterback's initial direction is positive. Indicate the direction with the sign of your answer.)
Answers
Answer:
Average Velocity = 3.63 m/s
Explanation:
First, we will calculate the total displacement of the quarterback, taking forward direction as positive:
Total Displacement = 15 m - 3 m + 24 m = 36 m
Now, we will calculate the total time taken for this displacement:
Total Time = 3 s + 1.71 s + 5.2 s = 9.91 s
Therefore, the average velocity will be:
[tex]Average\ Velocity = \frac{Total\ Displacement}{Total\ Time}\\\\Average\ Velocity = \frac{36\ m}{9.91\ s}[/tex]
Average Velocity = 3.63 m/s
Which is the definition of refraction?
1)the blocking of light waves vibrating in a particular plane
2) the bending of a light wave as it passes at an angle from one medium to another
3) a false or distorted image causing the gradual distortion of light through hot air
the redirection of light by tiny particles as it passes through a medium
Answers
Answer:
2) The bending of a light wave as it passes at an angle from one medium to another .
Hope it is helpful to you ☺️
The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.
Answers
Complete question:
A transverse wave on a rope is given by [tex]y \ (x, \ t) = (0.75 \ cm) \ cos \ \pi[(0.400 \ cm^{-1}) x + (250 \ s^{-1})t][/tex]. The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.
Answer:
The tension on the rope is 1.95 N
Explanation:
The general equation of a progressive wave is given as;
[tex]y \ (x,t) = A \ cos(kx \ + \omega t)[/tex]
Compare the given equation with the general equation of wave, the following parameters will be deduced.
A = 0.75 cm
k = 0.400π cm⁻¹
ω = 250π s⁻¹
The frequency of the wave is calculated as;
ω = 2πf
2πf = 250π
2f = 250
f = 250/2
f = 125 Hz
The wavelength of the wave is calculated as;
[tex]\lambda = \frac{2\pi}{k} \\\\\lambda = \frac{2\pi }{0.4 \pi} = 5 \ cm = 0.05 \ m[/tex]
The velocity of the wave is calculated as;
v = fλ
v = 125 x 0.05
v = 6.25 m/s
The tension on the rope is calculated as;
[tex]v = \sqrt{\frac{T}{\mu}} \\\\where;\\\\T \ is \ the \ tension \ of \ the \ rope\\\\\mu \ is \ the \ mass \ per \ unit \ length = 0.05 \ kg/m\\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (6.25)^2\times (0.05)\\\\T = 1.95 \ N[/tex]
Therefore, the tension on the rope is 1.95 N
An object moving along a horizontal track collides with and compresses a light spring (which obeys Hooke's Law) located at the end of the track. The spring constant is 52.1 N/m, the mass of the object 0.250 kg and the speed of the object is 1.70 m/s immediately before the collision.
(a) Determine the spring's maximum compression if the track is frictionless.
?? m
(b) If the track is not frictionless and has a coefficient of kinetic friction of 0.120, determine the spring's maximum compression.
??m
Answers
(a) As it gets compressed by a distance x, the spring does
W = - 1/2 (52.1 N/m) x ²
of work on the object (negative because the restoring force exerted by the spring points in the opposite direction to the object's displacement). By the work-energy theorem, this work is equal to the change in the object's kinetic energy. At maximum compression x, the object's kinetic energy is zero, so
W = ∆K
- 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²
==> x ≈ 0.118 m
(b) Taking friction into account, the only difference is that more work is done on the object.
By Newton's second law, the net vertical force on the object is
∑ F = n - mg = 0
where n is the magnitude of the normal force of the track pushing up on the object. Solving for n gives
n = mg = 2.45 N
and from this we get the magnitude of kinetic friction,
f = µn = 0.120 (2.45 N) = 0.294 N
Now as the spring gets compressed, the frictional force points in the same direction as the restoring force, so it also does negative work on the object:
W (friction) = - (0.294 N) x
W (spring) = - 1/2 (52.1 N/m) x ²
==> W (total) = W (friction) + W (spring)
Solve for x :
- (0.294 N) x - 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²
==> x ≈ 0.112 m
For the 0.250 kg object moving along a horizontal track and collides with and compresses a light spring, with a spring constant of 52.1 N/m, we have:
a) The spring's maximum compression when the track is frictionless is 0.118 m.
b) The spring's maximum compression when the track is not frictionless, with a coefficient of kinetic friction of 0.120 is 0.112 m.
a) We can calculate the spring's compression when the object collides with it by energy conservation because the track is frictionless:
[tex] E_{i} = E_{f} [/tex]
[tex] \frac{1}{2}m_{o}v_{o}^{2} = \frac{1}{2}kx^{2} [/tex] (1)
Where:
[tex]m_{o}[/tex]: is the mass of the object = 0.250 kg
[tex]v_{o}[/tex]: is the velocity of the object = 1.70 m/s
k: is the spring constant = 52.1 N/m
x: is the distance of compression
After solving equation (1) for x, we have:
[tex] x = \sqrt{\frac{m_{o}v_{o}^{2}}{k}} = \sqrt{\frac{0.250 kg*(1.70 m/s)^{2}}{52.1 N/m}} = 0.118 m [/tex]
Hence, the spring's maximum compression is 0.118 m.
b) When the track is not frictionless, we can calculate the spring's compression by work definition:
[tex] W = \Delta E = E_{f} - E_{i} [/tex]
[tex] W = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} [/tex] (2)
Work is also equal to:
[tex] W = F*d = F*x [/tex] (3)
Where:
F: is the force
d: is the displacement = x (distance of spring's compression)
The force acting on the object is given by the friction force:
[tex] F = -\mu N = -\mu m_{o}g [/tex] (4)
Where:
N: is the normal force = m₀g
μ: is the coefficient of kinetic friction = 0.120
g: is the acceleration due to gravity = 9.81 m/s²
The minus sign is because the friction force is in the opposite direction of motion.
After entering equations (3) and (4) into (2), we have:
[tex]-\mu m_{o}gx = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2}[/tex]
[tex]\frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} + \mu m_{o}gx = 0[/tex]
[tex] \frac{1}{2}52.1 N/m*x^{2} - \frac{1}{2}0.250 kg*(1.70)^{2} + 0.120*0.250 kg*9.81 m/s^{2}*x = 0 [/tex]
Solving the above quadratic equation for x
[tex] x = 0.112 m [/tex]
Therefore, the spring's compression is 0.112 m when the track is not frictionless.
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What is the minimum angular spread (in rad) of a 534 nm wavelength manganese vapor laser beam that is originally 1.19 mm in diameter
Answers
Answer:
Minimum angular spread (in rad) = 547.45 x 10⁻⁶ rad
Explanation:
GIven;
Wavelength of manganese vapor laser beam = 534 nm = 534 x 10⁻⁹ m
Diameter = 1.19 mm = 1.19 x 10⁻³ m
Find:
Minimum angular spread (in rad)
Computation:
Minimum angular spread (in rad) = 1.22[Wavelength / Diameter]
Minimum angular spread (in rad) = 1.222[(534 x 10⁻⁹) / (1.19 x 10⁻³)]
Minimum angular spread (in rad) = 2[448.73 x 10⁻⁶]
Minimum angular spread (in rad) = 547.45 x 10⁻⁶ rad
A nerve impulse travels along a myelinated neuron at 90.1 m/s.
What is this speed in mi/h?
Answers
Answer:
201.5537 mph
Explanation:
Given the following data;
Speed = 90.1 m/s
Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.
Mathematically, speed is given by the formula;
Speed = distance/time
To convert this value into miles per hour;
Conversion;
1 meter = 0.000621 mile
90.1 meters = 90.1 * 0.000621 = 0.05595 miles
1 metre per second = 2.237 miles per hour
90.1 meters per seconds = 90.1 * 2.237 = 201.5537 miles per hour
90.1 m/s = 201.5537 mph
A 1200 kg car traveling east at 4.5 m/s crashes into the side of a 2100 kg truck that is not moving. During the collision, the vehicles get stuck together. What is their velocity after the collision? A. 2.9 m/s east B. 1.6 m/s east m C. 2.6 m/s east D. 1.8 m/s east
Answers
Answer:
Explanation:
This is a simple Law of Momentum Conservation problem of the inelastic type. The equation for this is
[tex][m_1v_1+m_2v_2]_b=[(m_1+m_2)v]_a[/tex] Filling in:
[tex][1200(4.5)+2100(0)]=[(1200+2100)v][/tex] which simplifies to
5400 + 0 = 3300v
so v = 1.6 m/s to the east, choice B
(b) Name the devices used to measure the volume of liquid.
Answers
Answer:
Liquid volume is usually measured using either a graduated cylinder or a buret. As the name implies, a graduated cylinder is a cylindrical glass or plastic tube sealed at one end, with a calibrated scale etched (or marked) on the outside wall.
3. Calculate the force it would take to accelerate a 50 ka bike at a rate of 3 m/s2 (6 points)
Answers
Answer:
150 N
Explanation:
Given that,
Acceleration (a) = 3 m/s²Mass of the bike (m) = 50 kgWe are asked to calculate force required.
[tex]\longrightarrow[/tex] F = ma
[tex]\longrightarrow[/tex] F = (50 × 3) N
[tex]\longrightarrow[/tex] F = 150 N
A long, straight, vertical wire carries a current upward. Due east of this wire, in what direction does the magnetic field point
Answers
The magnetic field of the wire will be directed towards west. Using right thumb rule one can get the direction of field lines.
a microwave operates at a frequency of 2400 MHZ. the height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. assume that microwave energy is generated uniformly on the uipper surface. What is the power output of the oven
Answers
Complete question is;
A microwave oven operates at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. Assume that microwave energy is generated uniformly on the upper surface of the cavity and propagates directly
downward toward the base. The base is lined with a material that completely absorbs microwave energy. The total microwave energy content of the cavity is 0.50 mJ.
Answer:
Power ≈ 600,000 W
Explanation:
We are given;
Frequency; f = 2400 Hz
height of the oven cavity; h = 25 cm = 0.25 m
base area; A = 30 cm by 30 cm = 0.3m × 0.3m = 0.09 m²
total microwave energy content of the cavity; E = 0.50 mJ = 0.5 × 10^(-3) J
We want to find the power output and we know that formula for power is;
P = workdone/time taken
Formula for time here is;
t = h/c
Where c is speed of light = 3 × 10^(8) m/s
Thus;
t = 0.25/(3 × 10^(8))
t = 8.333 × 10^(-10) s
Thus;
Power = (0.5 × 10^(-3))/(8.333 × 10^(-10))
Power ≈ 600,000 W
the spring was compressed three times farther and then the block is released, the work done on the block by the spring as it accelerates the block is
Answers
Answer:
The work done on the block by the spring as it accelerates the block is 4kx².
Explanation:
Let initial distance is x.
It was compressed three times farther and then the block is released, new distance is 3x.
The work done in compressing the spring is given by :
[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)[/tex]
[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\W=\dfrac{1}{2}k((3x)^2-x^2)\\\\W=\dfrac{1}{2}k((9x^2-x^2)\\\\W=\dfrac{1}{2}k\times 8x^2\\\\W=4kx^2[/tex]
So, the work done on the block by the spring as it accelerates the block is 4kx².
Help me plssssssss cause I’m struggling
Answers
Answer:
I am pretty sure it is C
Explanation:
It can be found all over the universe
Two pendulums have the same dimensions (length {L}) and total mass (m). Pendulum A is a very small ball swinging at the end of a uniform massless bar. In pendulum B, half the mass is in the ball and half is in the uniform bar.
1. Find the period of pendulum A for small oscillations.
2. Find the period of pendulum B for small oscillations.
Answers
Answer:
1) [tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex], 2) [tex]T_{B} \approx 1.137\cdot T_{A}[/tex], where [tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex].
Explanation:
1) Pendulum A is a simple pendulum, whose period ([tex]T_{A}[/tex]) is determined by the following formula:
[tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex] (1)
Where:
[tex]l[/tex] - Length of the massless bar.
[tex]g[/tex] - Gravitational acceleration.
2) Pendulum B is a physical pendulum, whose period ([tex]T_{B}[/tex]) is determined by the following formula:
[tex]T_{B} = 2\pi \cdot \sqrt{\frac{I_{O}}{m\cdot g\cdot l} }[/tex] (2)
Where:
[tex]m[/tex] - Total mass of the pendulum.
[tex]g[/tex] - Gravitational acceleration.
[tex]l[/tex] - Length of the uniform bar.
[tex]I_{O}[/tex] - Moment of inertia of the pendulum with respect to its suspension axis.
The moment of inertia can be found by applying the formulae of the moment of inertia for a particle and the uniform bar and Steiner's Theorem:
[tex]I_{O} = \frac{1}{2} \cdot m\cdot l^{2}+\frac{1}{24}\cdot m\cdot l^{2} + \frac{3}{4}\cdot m\cdot l^{2}[/tex]
[tex]I_{O} = \frac{31}{24}\cdot m\cdot l^{2}[/tex] (3)
By applying (3) in (2) we get the following expression:
[tex]T_{B} = 2\pi \cdot \sqrt{\frac{\frac{31}{24}\cdot m \cdot l^{2} }{m\cdot g \cdot l} }[/tex]
[tex]T_{B} = 2\pi \cdot \sqrt{\frac{31\cdot l}{24\cdot g} }[/tex]
[tex]T_{B} = \sqrt{\frac{31}{24} } \cdot \left(2\pi \cdot \sqrt{\frac{l}{g} }\right)[/tex]
[tex]T_{B} \approx 1.137\cdot T_{A}[/tex]
1. The period of pendulum A for small oscillations is
[tex]T_A=2\pi\sqrt{\dfrac{L}{g}}[/tex]
2. The period of pendulum B for small oscillations.
[tex]T_B=1.137.T_A[/tex]
What is simple harmonic motion?Simple harmonic motion is the periodic motion or back and forth motion of any object with respect to its equilibrium or mean position. The restoring force is always acting on the object which try to bring it to the equilibrium.
1) Pendulum A is a simple pendulum, whose period () is determined by the following formula:
[tex]T_A=2\pi\sqrt{\dfrac{L}{g}}[/tex]
Where:
l - Length of the massless bar.
g - Gravitational acceleration.
2) Pendulum B is a physical pendulum, whose period () is determined by the following formula:
[tex]T_A=2\pi\sqrt{\dfrac{I_o}{mgl}}[/tex] .............................2
Where:
m - Total mass of the pendulum.
g - Gravitational acceleration.
l - Length of the uniform bar.
Io- Moment of inertia of the pendulum with respect to its suspension axis.
The moment of inertia can be found by applying the formulae of the moment of inertia for a particle and the uniform bar and Steiner's Theorem:
[tex]I_o=\dfrac{1}{2}ml^2+\dfrac{1}{24}ml^2+\dfrac{3}{4}ml^2[/tex]
[tex]I_o=\dfrac{31}{24}ml^2[/tex]..................................3
By applying (3) in (2) we get the following expression:
[tex]T_B=2\pi\sqrt{\dfrac{\frac{31}{24}ml^2}{mgl}[/tex]
[tex]T_B=2\pi\sqrt{\dfrac{31l}{24g}}[/tex]
[tex]T_B=\sqrt{\dfrac{31}{24}}. (2\pi\sqrt{\dfrac{l}{g}})[/tex]
[tex]TB=1.137.T_A[/tex]
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Three 30 g metal balls, one of aluminum, copper and lead, are placed in a large beaker of hot water for a few minutes. [The specific heats of aluminum, copper, and lead are 903, 385, and 130 J / (kg ° C), respectively].
to. Which of the balls, if any, will reach the highest temperature? Explain.
b. Which of the balls, if any, will have the most heat energy? Explain.
Answers
Answer:
The answer is below
Explanation:
Specific heat capacity is an intensive property of a material. The specific heat of a material is the amount of energy required to raise the temperature of one unit mass m of material by one unit of temperature.
a) Temperature is inversely proportional to specific heat capacity. If the same amount of heat is applied to all three balls, the ball that will reach the highest temperature is the ball with the least specific heat capacity.
Hence lead will have the highest temperature since it has the least specific heat capacity.
b) The quantity of heat is directly proportional to the specific heat capacity. Hence if all balls experience the same temperature change, the ball that have the most energy will be that with the highest specific heat capacity.
Hence aluminum will have the most heat since it has the highest specific heat capacity.
If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. What will its speed be, in meters per second, when it reaches the positive plate in this case
Answers
Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.
Thermometers and Temperature Scales
While traveling outside the United States, you feel sick. A companion gets you a thermometer, which says your temperature is 40.9. What scale is that on? What is your Fahrenheit temperature? Should you seek medical help?
Answers
Answer:
105.62°F
Explanation:
When the body temperature having fever is measured to be 40.9 on a scale then it must be a Celsius scale thermometer because 37°C is the normal temperature of a healthy human. In case of fever the given temperature is measured on a standard Celsius scale.
The relation between Fahrenheit and Celsius scale is:
[tex]\frac{C}{5}=\frac{F-32}{9}[/tex]
[tex]F=\frac{9C}{5} +32[/tex]
[tex]F=105.62^{o}F[/tex]
It is a high fever and an immediate medical help must be taken.
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. How much of the sample is left after 1.98 x 10^4 seconds?
Answers
[tex]A=2.01×10^{16}\:\text{nuclei}[/tex]
Explanation:
Given:
[tex]\lambda = 4.96×10^3 s[/tex]
[tex]A_0 = 3.21x10^{17}[/tex] nuclei
t = 1.98×10^4 s
[tex]A=A_02^{-\frac{t}{\lambda}}[/tex]
[tex]A=(3.21×10^{17}\:\text{nuclei}) \left(2^{-\frac{1.98×10^4}{4.96×10^3}} \right)[/tex]
[tex]\:\:\:\:\:\:\:=2.01×10^{16}\:\text{nuclei}[/tex]
(a) If half of the weight of a flatbed truck is supported by its two drive wheels, what is the maximum acceleration it can achieve on wet concrete where the coefficient of kinetic friction is 0.5 and the coefficient of static friction is 0.7.
(b) Will a metal cabinet lying on the wooden bed of the truck slip if it accelerates at this rate where the coefficient of kinetic friction is 0.3 and the coefficient of static friction is 0.55?
(c) If the truck has four-wheel drive, and the cabinet is wooden, what is it's maximum acceleration (in m/s2)?
Answers
Answer:
a) a = 27.44 m / s², b) a = 5.39 m / s², c) a = 156.8 m / s², cabinet maximum acceleration does not change
Explanation:
a) In this exercise the wheels of the truck rotate to provide acceleration, but the contact point between the ground and the 2 wheels remains fixed, therefore the coefficient of friction for this point is static.
Let's apply Newton's second law
we set a regency hiss where the x axis is in the direction of movement of the truck
Y axis y
N- W = 0
N = W = m g
X axis
2fr = m a
the expression for the friction force is
fr = μ N
fr = μ m g
we substitute
2 μ m g = m /2 a
a = 4 μ g
a = 4 0.7 9.8
a = 27.44 m / s²
b) let's look for the maximum acceleration that can be applied to the cabinet
fr = m a
μ N = ma
μ m g = m a
a = μ g
a = 0.55 9.8
a = 5.39 m / s²
as the acceleration of the platform is greater than this acceleration the cabinet must slip
c) the friction force is in the four wheels as well
With when the truck had two-wheel Thracian the weight of distributed evenly between the wheels, in this case with 4-wheel Thracian the weight must be distributed among all
applying Newton's second law
4 fr = (m/4) a
16 mg = (m) a
a = 16 g
a = 16 9.8
a = 156.8 m / s²
cabinet maximum acceleration does not change
A 31 kg block is initially at rest on a horizontal surface. A horizontal force of 83 N is required to set the block in motion. After it is in motion, a horizontal force of 55 N i required to keep it moving with constant speed. From this information, find the coefficients of static and kinetic friction
Answers
Answer:
The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.
Explanation:
By Newton's Laws of Motion and definition of maximum friction force, we derive the following two formulas for the static and kinetic coefficients of friction:
[tex]\mu_{s} = \frac{f_{s}}{m\cdot g}[/tex] (1)
[tex]\mu_{k} = \frac{f_{k}}{m\cdot g}[/tex] (2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, no unit.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.
[tex]f_{s}[/tex] - Static friction force, in newtons.
[tex]f_{k}[/tex] - Kinetic friction force, in newtons.
[tex]m[/tex] - Mass, in kilograms.
[tex]g[/tex] - Gravitational constant, in meters per square second.
If we know that [tex]f_{s} = 83\,N[/tex], [tex]f_{k} = 55\,N[/tex], [tex]m = 31\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the coefficients of friction are, respectively:
[tex]\mu_{s} = \frac{83\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{s} = 0.273[/tex]
[tex]\mu_{k} = \frac{55\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{k} = 0.181[/tex]
The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.