Answer:
the magnitude of the tension on the cable necessary to keep the system in equilibrium is 1060.9 N
Explanation:
Given that;
Length L = 10 m
mass of beam m_b = 150 kg; weight W_beam = 150×9.8
mass of sign m = 75 kg
distance of sign hung from the beam from the wall d = 2.50 m
angle ∅ = 60°
g = 9.8 m/s²
Now,
Torque acting at one end of the beam will be;
[tex]T_{net}[/tex] = Tsin∅ × L - mg(d)-W × (L/2)
for equilibrium, [tex]T_{net}[/tex] = 0
therefore, 0 = Tsin∅ × L - mg(d)-W × (L/2)
so we substitute
Tsin(60°) × 10 - 75×9.8(2.50) - 150 × 9.8× (10/2) = 0
Tsin(60°) × 10 - 1837.5 - 7350 = 0
Tsin(60°) × 10 - 9187.5 = 0
Tsin(60°) × 10 = 9187.5
divide both side by 10
Tsin(60°) = 918.75
T × 0.8660 = 918.75
T = 918.75 / 0.8660
T = 1060.9 N
Therefore, the magnitude of the tension on the cable necessary to keep the system in equilibrium is 1060.9 N
The magnitude of the tension on the cable that keep the system in equilibrium is 1060.9 N.
Torque acting at one end of the beam,
= Tsin∅ × L - mg(d)-W × (L/2)
When equilibrium = 0
Tsin∅ × L - mg(d)-W × (L/2) = 0
Where,
L - Length = 10 m
m - mass of sign bord= 75 kg
g- gravitational accelaration = 9.8 m/s²
W - weight of beam = 150×9.8 = 1470 kg
Put the values in the formula,
Tsin(60°) × 10 - 75×9.8(2.50) - 150 × 9.8× (10/2) = 0
Tsin(60°) × 10 - 1837.5 - 7350 = 0
Tsin(60°) × 10 - 9187.5 = 0
Tsin(60°) × 10 = 9187.5
Tsin(60°) = 918.75
T × 0.8660 = 918.75
T = 918.75 / 0.8660
T = 1060.9 N
Therefore, the magnitude of the tension on the cable that keep the system in equilibrium is 1060.9 N.
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The diameter of the Earth is about 13,000km. The diameter of the moon is about 3,500km. If the student makes her model of the Earth 26mm in diameter, what should be the diameter of her model of the moon in mm?
What is measurement ?
A car completes a journey in 2 hours. Its average speed over this journey was 45 mph. Calculate the distance travelled by the car in miles.
how many carbon (C) atoms are in a molecule of CH2O?
Answer:1 carbon atom
Explanation:AP3X
Answer:
1
Explanation:
The left fielder throws the baseball home from 60 m away. The ball's horizontal velocity is 30 m/s, and its vertical velocity is sufficient to keep it in the air until it reaches the catcher at home plate. At the same time, a base runner is running toward home plate. His velocity is 5 m/s, and he is 5 m from home plate. If the base runner's velocity stays constant and air resistance isn't a factor, which will arrive at home plate first, the base runner or the baseball?
Answer:
The base runner
Explanation:
To know the correct answer to the question, we shall determine the time taken for the baseball and the base runner to get to the home plate. This is illustrated below:
For the baseball:
Horizontal velocity (u) = 30 m/s
Horizontal distance (s) = 60 m
Time (t) =?
s = ut
60 = 30 × t
Divide both side by 30
t = 60 / 30
t = 2 s
Thus, it will take the baseball 2 s to get to the home plate.
For the base runner:
Horizontal velocity (u) = 5 m/s
Horizontal distance (s) = 5 m
Time (t) =?
s = ut
5 = 5 × t
Divide both side by 5
t = 5 / 5
t = 1 s
Thus, it will take the base runner 1 s to get to the home plate.
SUMMARY:
Time taken for the baseball to get to the home plate = 2 s
Time taken for the base runner to get to the home plate = 1 s
From the calculations made above, we can conclude that the base runner will arrive at the home plate first because it took him 1 s to get to the home plate whereas the baseball took 2 s to get there.
50 Points, It is about Inquiry, Experiment, Controlled Variable, Independent Variable, Hypothesis, and Dependent Variable.
Answer:
1) dependent
2) Experiment
3) Controlled variable
4) hypothesis
5) Inquiry
Explanation:
The dependent variables are those variables that change when the experimental or independent variable is being manipulated.
An experiment is a carefully designed study that tends to establish a cause and effect relationship between two or more variables.
A control variable is any variable which is held constant or unchanged throughout the course of the investigation(experiment).
Hypothesis is a precise statement which can be tested and predicts the outcome of the study.
The independent variable is the variable that is being manipulated.
Identify the labeled parts in the figure.
i. Which of the labeled parts is a chemical symbol?
ii. Which of the labeled parts is a coefficient?
iii. Which of the labeled parts is the number of atoms of the element?
iv. Which of the labeled parts indicates that only one atom of the element is present in the substance?
Answer:
I. a, c, f and h
II. e
III. b, d, g and i
IV. i
Explanation:
I. Chemical symbols are simple abbreviations used to represent various elements or compound. They consist entire of alphabet.
For the diagram given above, the labelled parts which represent chemical symbol are: a, c, f and h
II. Coefficients are numbers written before the chemical symbol of elements or compound.
For the diagram given above, the labelled part which represent Coefficient is: e
III. Number of atoms of element present in a compound is simply obtained by taking note of the numbers written as subscript in the chemical formula of the compound.
For the diagram given above, the labelled part which represent the number of atoms of the element are: b, d, g and i
IV. When no number is written as subscript in the formula of the element in the compound, it means the element has just 1 atom in the compound.
For the diagram given above, the labelled part which indicates that only 1 atom of the element is present is: i
Answer:
i. a, c, f, h
ii. e
iii. b, d, g, i
iv. i
Can someone help me grasp the concept of solving this please
Answer:
Q= m × c × change in tempQ= 1000 cal/kg × C Q= (5kg) × (1000 ) × (100-13) Q= 435000Hence 4.35 × 10^5 Ask any questions!I also do quizzes and exams if your interested!What force balances gravity as the balloon sticks to the sticks to the wall?
Answer:
thats answer
Explanation:
please give me a brainlyliest
A person on shore sees light
coming from a fish underwater. In
air, the ray travels at 21.2 deg. At
what angle did the ray travel
underwater?
Resources
Help
(Water n = 1.33, Air n = 1.00)
t
ti
(Unit = deg)
Answer:
15.7
Explanation:
Since N1 sin(angle1)= N2 sin(angle2), you can set up the equation to find the missing angle using inverse sin( 1.00 sin(21.2)/ 1.33). You would then get that the other angle is equal to 15.7 or 15.8 if you round up. I hope this helped! :)
Answer:
15.7
Explanation:
Acellus
regular reflection is the reflection of light on a surface
on smooth surface like mirror.
A certain sprinter has a top speed of 11.3 m/s. If the sprinter starts from rest and accelerates at a constant rate, he is able to reach his top speed in a distance of 12.8 m. He is then able to maintain his top speed for the remainder of a 100 m race. (a) What is his time for the 100 m race? (b) In order to improve his time, the sprinter tries to decrease the distance required for him to reach
h his top speed. What must this distance be if he is to achieve a time of 9.75 s for the race?
Answer:
Explanation:
initial velocity u = 0
final velocity v = 11.3 m /s
distance covered s = 12.8 m
v² = u² + 2 a s
11.3² = 0 + 2 x a x 12.8
a = 4.99 m /s²
again ,
v = u + a t
11.3 = 0 + 4.99 t
t = 2.26 s .
Rest of the sprint will be covered with uniform velocity .
Distance covered = 100 - 12.8 = 87.2 m
speed = 11.3 m /s
time taken = 87.2 / 11.3 = 7.7 s
Total time of 100 m sprint = 7.7 + 2.26 = 9.96 m .
b )
Let the time taken to reach the top speed be t .
acceleration a = 11.3 / t
distance covered s = 1/2 a t²
= .5 x (11.3 / t) x t²
= 5.65 t
Rest of the distance = 100 - 5.65 t
time taken to cover rest of the distance = (100 - 5.65 t ) / 11.3
Total time = (100 - 5.65 t / 11.3 ) + t = 9.75
100 - 5.65 t + 11.3 t = 11.3 x 9.75
100 + 5.65 t = 110.175
5.65 t = 10.175
t = 1.8
acceleration a = 11.3 / t
= 11.3 / 1.8
= 6.278 m /s²
distance covered in 1.8 s
s = 1/2 a t²
= .5 x 6.278 x 1.8²
= 10.17 m .
is found in wood or straw?*
2.
Sucrose
2.Xylose
3.Maltose
4.Galactose
There are two categories of ultraviolet light. Ultraviolet A (UVA) has a wavelength ranging from 320 nmnm to 400 nmnm. It is not so harmful to the skin and is necessary for the production of vitamin D. UVB, with a wavelength between 280 nmnm and 320 nmnm, is much more dangerous, because it causes skin cancer.
Required:
a. Find the frequency ranges of UVA and UVB.
b. What are the ranges of the wave numbers for UVA and UVB?
Answer:
a) Remember the relationship:
v = f*λ
where:
v = velocity of the wave, remember that for an electromagnetic wave, this always is equal to the speed of light, then:
v = c = 3*10^8 m/s
f is the frequency of the wave
λ is the wavelength.
So if we want to compute the frequency, we have the equation:
f = c/λ
then:
We know that UVA has the range from 320 nm to 400nm, converting these to meters we hav:
1nm = 1m*10^(-9)
then:
320mm = (320*10^(-9)) m = 0.00000032 m
400nm = (400*10^(-9)) m = 0.00000040 m
Replacing these in our equation we can find that the range of frequencies is:
f = (3*10^8 m/s)/( 0.00000032 m) = 9.375*10^(14) Hz
f = (3*10^8 m/s)/( 0.00000040 m) = 7.5*10^(14) Hz
Then the range of frequencies is:
( 7.5*10^(14) Hz to 9.375*10^(14) Hz)
Similar for the case of UVB:
the wavelengths are:
280 nm to 320 nm
Rewriting these in meters we get:
280nm = 280*10^(-9) m = 0.00000028m
320nm = 320*10^(-9) m = 0.00000032 m
Then the frequencies are:
f = (3*10^8 m/s)/( 0.00000032 m) = 9.375*10^(14) Hz
f = (3*10^8 m/s)/( 0.00000028 m) = 1.07*10^(15) Hz
The range of frequencies for UVB is:
(9.375*10^(14) Hz to 1.07*10^(15) Hz)
B) We want to know the range of wavenumbers for both types of waves.
The wave number is calculated as:
1/λ
Which represents the number of waves in a given unit of length.
Then the range for UVA will be:
1/320nm = 0.0031 nm^-1
1/400nm = 0.0025 nm^-1
Then the range is:
( 0.0025 nm^-1, 0.0031 nm^-1)
And for the case of UVB we will have:
1/320 nm = 0.0031 nm^-1
1/280 nm = 0.0036 nm^-1
Then the range is:
(0.0031 nm^-1, 0.0036 nm^-1)
I 2016 there were fewer pedestrians killed in traffic crashes than during the previous year
A weightlifter lifts a weight of 500 N from the ground over her head, a distance of 1.8 m. How much work has been done to move the weight?
find the moment of inertia of a point mass 0.005g at aperpendicular distance of 3m from its axis of rotation.
Answer:
the moment of inertia is 4.5 × 10⁻⁵ kg.m²
Explanation:
Given that;
point mass m = 0.005 g = ( 0.005 / 1000 ) = 5 × 10⁻⁶ kg
perpendicular distance r = 3m
We know that a point mass doesn't have a moment of inertia around its own axis but, but using the parallel axis theorem, a moment of inertia around a distant axis of rotation can be determined using;
[tex]I_{}[/tex] = mr²
so we substitute
[tex]I_{}[/tex] = (5 × 10⁻⁶ kg) × (3 m)²
[tex]I_{}[/tex] = (5 × 10⁻⁶ kg) × 9 m²
[tex]I_{}[/tex] = 4.5 × 10⁻⁵ kg.m²
Therefore; the moment of inertia is 4.5 × 10⁻⁵ kg.m²
The moment of inertia of given point mass is 4.5 × 10⁻⁵ kgm² at a perpendicular distance of 3 m.
The moment of inertia of given point mass can be determined by,
[tex]I = mr^2[/tex]
Where,
[tex]I[/tex]- moment of inertia
[tex]m[/tex]- mass = 0.005 g = ( 0.005 / 1000 ) = 5 × 10⁻⁶ kg
[tex]r[/tex] - perpendicular distance = 3 m
Put the values in the formula,
[tex]I = (5 \times 10^{-6}{\rm \ kg}) \times (3 {\rm \ m})^2\\\\I = 5 \times 10^{-6}{\rm \ kg} \times 9 {\rm \ m}\\\\I = 4.5 \times 10^{-5} kgm^2[/tex]
Therefore; the moment of inertia of given point mass is 4.5 × 10⁻⁵ kgm².
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Binding energy is the energy needed to
a. keep the electrons spinning around the nucleus of an atom.
b. keep the atoms in a covalent compound together.
c. keep the atoms in example of an element together.
d. hold the nucleus of an atom together.
Answer:
Answer C
Explanation:
What is the velocity of this graph between
points C and D?
Position x (m)
A
0
1 2 3
Time t (s)
A. -0.50 m/s
B. -1.0 m/s
C. -2.0 m/s
D. -4.0 m/s
The velocity of the given graph will be 2m/s.
Speed is a scalar quantity that measures how fast an object is moving. It is the rate of change of distance traveled by an object over a specific period of time. Speed does not consider the direction of motion; it only reflects the magnitude of the velocity.
Mathematically, speed (v) is calculated as:
Speed (v) = Distance traveled (d) / Time taken (t)
where:
Speed (v) is measured in units like meters per second (m/s), kilometers per hour (km/h), miles per hour (mph), etc.
Distance traveled (d) is the total path length covered by the object in a given timeframe.
Time taken (t) represents the duration it took the object to cover that distance.
According to graph,
speed = 6-2/ 3- 1
= 4/ 2
= 2 m/s
Therefor, the velocity of the given graph will be 2m/s.
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S
m
On a frictionless toy race track, a 0.035 kg toy car moving right at 0.30 collides with another 0.040 kg toy
m
car moving left at 0.20 After the collision, the 0.035 kg car moves left at 0.20
S
m
S
What is the final speed of the 0.040 kg toy car?
Answer: 0.24 m/s
Explanation:
While sliding a couch across a floor, Hannah and Andrea exerts forces FH and FA on the couch. Hannah's force is due north with a magnitude of FH = 59 N and Andrea's force is θ = 22° east of north with a magnitude of FA = 155 N. In this problem, use a coordinate system with y directed north and x directed east.
Part (a) Find the net force in the y-direction in Newtons.
Part (b) Find the net force in the x-direction in Newtons.
Part (c) Calculate the angle in degrees north of east of the net force exerted on the couch by Hannah and Andrea, F HAF→HA.
Part (d) Hannah and Andrea's housemates, David and Stephanie disagree with the move and want to prevent its relocation. Their combined force F DSF→DS must be equal and opposite to that of FHAF→HA. What is the magnitude in Newtons of the force F DSF→DS which will prevent the relocation?
Answer:
a) 202.7 N
b) 58.1 N
c) 74.1º N of E.
d) 210.9 N
Explanation:
a)
The net force exerted in the y-direction, will be the sum of FH (which is directed northwards) and the y-component of FA.Since the magnitude of FA is 155 N and the angle of FA with the y-axis, is 22º (E of N), we can find the N-S component of FA, just applying the the definition of cosine, to the triangle defined by FA, the y- axis and a segment parallel to the x- axis between FA and the y-axis, as follows:[tex]F_{Ay} = F_{A} * cos \theta = 155 N* cos 22 = 143.7 N (1)[/tex]
⇒ Fy = FH + FAy = 59 N + 143.7 N = 202.7 N (2)
b)
We can proceed exactly in the same way for the x-axis. Since FH is directed due North, it has no component along the x-axis.So, Fx is directly the component of FA along the x-axis, which can be found applying the definition of sine to the same triangle than in a) as follows:[tex]F_{x} = F_{A} * sin \theta = 155 N* sin 22 = 58.1 N (3)[/tex]
c)
Taking the same triangle than in a) and b), we can apply the definition of tangent, in order to find the angle between F and the x-axis, as follows:[tex]tg \theta = \frac{F_{y}}{F_{x}} = \frac{202.7N}{58.1N} = 3.5 (4)[/tex]
⇒ θ = tg⁻¹ (3.5) = 74.1º N of E. (5)
d)
In order to be equal and opposite to the combined force FH+FA, it must have the same magnitude.This magnitude can be found applying the Pythagorean Theorem to the same triangle that we used in a), b) and c):[tex]F_{DS} = \sqrt{(F_{x} ^{2} +F_{y} ^{2})} = \sqrt{(58.1N)^{2}) + (202.7N)^{2} } = 210.9 N (6)[/tex]
A boy throws a rock with an initial horizontal velocity of 17.0 m/s and an initial vertical velocity of 21.0 m/s. How high above the boy's hand is the rock after 2.8 s?
Answer:
53.2
Explanation:
You can use the kinematic equation: displacement of x = (initial velocity + final velocity)*t/2
Subsititing: 17+21 = 38 * 2.8/2 = 53.2
Note: Displacement = distance between the 2 points
As a space shuttle moves through the dilute ionized gas of Earth's ionosphere, the shuttle's potential is typically changed by -1.5 V during one revolution. Assuming the shuttle is a conducting sphere of radius 11 m, estimate the amount of charge it collects.
Which plate is the South American plate?
Answer:
The south American plate
what's an equation that represents the relationship between speed distance and time
Answer:
The formula for speed is speed = distance ÷ time. To work out what the units are for speed, you need to know the units for distance and time. In this example, distance is in metres (m) and time is in seconds (s), so the units will be in metres per second (m/s).
Someone please help asap What general trend does electronegativity follow on the periodic table?
A. It alternates between high and low values.
B. It decreases from left to right.
C. It increases from left to right.
D. It varies randomly.
Electronegativity increases from left to right in the periodic table.
What is Electronegativity?Electronegativity is a chemical characteristic that defines an atom's or functional group's capacity to attract electrons to itself. An atom's electronegativity is influenced by its atomic number as well as the distance between its valence electrons and the charged nuclei.What is Periodic table?A table of chemical elements arranged in atomic number order, commonly in rows, with elements having comparable atomic structures (and hence chemical characteristics) appearing in vertical columns.
Electronegativity in periodic tableElectronegativity increases as you move from left to right across a period on the periodic table, and drops as you move down a group.
As a result, the most electronegative elements appear on the periodic table's top right, while the least electronegative elements appear on the bottom left.
Hence, the correct option is C.
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You toss a ball into the air and note the time interval between the ball leaving your hand and reaching its highest position. While you are doing this, a construction worker being lifted on a hydraulic platform at constant speed also notes the time interval needed for the ball to reach its highest position. Is the time interval reported by the worker longer, shorter, or the same as the interval you report?
Answer:
It is longer
Explanation:
According to the theory of special relativity, moving clocks run slower. So, the construction worker moving at a constant speed observers a time much longer than the time I observe since I am stationary. If t is the time observed by me and v is the speed of the construction worker, then, the time observed by the construction worker, t' is given by
t' = t/√[1 - (v/c)²] where c = speed of light
So, the construction worker reports a longer time interval than me since his time runs slower.
Suppose that the acceleration of a model rocket is proportional to the difference between 160 ft/sec and the rocket's velocity. If it is initially at rest and its initial acceleration is 280 ft/sec22, how long will it take to accelerate to 128 ft/s
Answer:
Explanation:
Given ,
dv / dt = k ( 160 - v )
dv / ( 160 - v ) = kdt
ln ( 160 - v ) = kt + c , where c is a constant
when t = 0 , v = 0
Putting the values , we have
c = ln 160
ln ( 160 - v ) = kt + ln 160
ln ( 160 - v / 160 ) = kt
(160 - v ) / 160 = [tex]e^{kt}[/tex]
1 - v / 160 = [tex]e^{kt }[/tex]
v / 160 = 1 - [tex]e^{kt }[/tex]
v = 160 ( 1 - [tex]e^{kt }[/tex] )
differentiating ,
dv / dt = - 160k [tex]e^{kt }[/tex]
acceleration a = - 160k [tex]e^{kt }[/tex]
given when t = 0 , a = 280
280 = - 160 k
k = - 175
a = - 160 x - 175 [tex]e^{kt}[/tex]
a = 28000 [tex]e^{kt}[/tex]
when a = 128 t = ?
128 = 28000 [tex]e^{kt}[/tex]
[tex]e^{kt }[/tex] = .00457
Acceleration of a rocket model.
As acceleration is the change in the magnitude and direction of the moving body and refers to the increase in the velocity which varies over time. The acceleration of the model rocket is proportional to the r difference between the 160 feet/sec.
Thus the answer is 0.0457
The initial velocity of the rockset at the rest and the initial acceleration is about 280 fett/ sec then the length of acceleration for the rokst to reach 128 feet / sec. Will be calculated by the dv / dt = k ( 160 - v ).ln ( 160 - v ) = kt + c , where c is a constant, when t = 0 , v = 0 then putting the values , we have c = ln 160ln ( 160 - v ) = kt + ln 160, ln ( 160 - v / 160 ) = kt(160 - v ) / 160 = 1 - v / 160 = v / 160 = 1 - v = 160 ( 1 - ) acceleration a = - 160k given when t = 0 , a = 280 The 280 = - 160 k, k = - 175 a = - 160 x - 175 Hence a = 2800Thus 128 = 28000 = .00457.Learn more about acceleration.
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Please give a step by step explanation: 30 points up for grabs.
Answer:
Honestly, I don't know ¯\_(ツ)_/¯
Explanation:
A light spring of force constant 3.85 N/m is compressed by 8.00 cm and held between a 0.250 kg block on the left and a 0.500 kg block on the right. Both blocks are at rest on a horizontal surface. The blocks are released simultaneously so that the spring tends to push them apart. What is the maximum velocity each block attains if the coefficient of kinetic friction between each block and the surface is:_______.
A) 0
B) 0.100
C) 0.463
Assume that the coefficient of static friction is larger than that for kinetic friction.
Answer:
A) v1 = -0.256 m/s
v2 = 0.128 m/s
B) v1 = -0.0642 m/s
v2 = 0 m/s
C) v1 = v2 = 0 m/s
Explanation:
We are given;
Spring constant; k = 3.85 N/m
Distance compressed; x = 8 cm = 0.08 m
Mass of left block; m1 = 0.25 kg
Mass of right block; m2 = 0.5 kg
A) From conservation of energy;
½kx² = ½m1•v1² + ½m2•(v2)²
Also from conservation of linear momentum, we know that;
m1v1 = m2v2
Now, v2 = m1•v1/m2
Plugging this for v2 in the first equation gives;
½kx² = ½m1•v1² + ½m2•(m1•v1/m2)²
Making v1 the subject, we have;
v1 = √(kx²(m2))/(m1•m2 + (m1)²))
v1 = √[(3.85 × 0.08² × 0.5)/((0.25 × 0.5) + 0.5²))
v1 = 0.256 m/s
This is the left block which is in the negative x direction and so v1 = -0.256 m/s
We saw that v2 = m1•v1/m2
Thus; v2 = (0.25 × 0.256)/0.5
v2 = 0.128 m/s
B) Force exerted by spring is;
F_s = kx = 3.85 × 0.08
F_s = 0.308 N
Normal forces will be calculated for both blocks as;
N1 = m1•g = 0.25 × 9.8 = 2.45 N
N2 = m2•g = 0.5 × 9.8 = 4.9 N
Let's calculate force of static friction for both blocks;
F_s1 = μN1 and F_s2 = μN2
We are given coefficient of friction as
μ = 0.1.
Thus;
F_s1 = 0.1 × 2.45 = 0.245 N
F_s2 = 0.1 × 4.9 = 0.49 N
F_s1 is lesser than F_s. Thus let's calculate the new compression;
x_1 = 0.245/3.85
x_1 = 0.06364 m
Thus, change in compression is;
Δx = x - x_1
Δx = 0.08 - 0.06364
Δx = 0.01636 m
From conservation of energy, since our coefficient of friction is not zero and we have frictional force, then we use the equation;
½k(x)² - F_s1•Δx = ½m1•v1² + ½k(x_1)²
Making v1 the subject, we have;
v_1 = √(k(x² - x_1²) - 2F_s1•Δx)/m1
v_1 = √(3.85(0.08² - 0.06364²) - 2(0.245 × 0.01636)/0.25
v1 = 0.0642 m/s
Since in negative x direction, then
v1 = -0.0642 m/s
F_s2 is greater than F_s. Thus, it means the right side block will not move and velocity is zero. v2 = 0 m/s
C) Coefficient of friction is now 0.463.
Thus;
F_s1 = 0.463 × 2.45 = 1.13435 N
F_s2 = 0.463 × 4.9 = 2.2687 N
They are both greater than F_s and thus no motion in both cases.
So v1 = v2 = 0 m/s