Answer:
1,642.8 coulombs and 1.39*10²¹ electrons pass through any cross section the width of the wire.
Explanation:
Electric current is the circulation of electric charges in an electric circuit, while electric current intensity (I) is the amount of electricity or electric charge (Q) that circulates through a circuit in unit time (t).
Then, the intensity of electric current is expressed as:
[tex]I=\frac{Q}{t}[/tex]
Where:
I is the intensity expressed in Amps (A) Q is the electric charge expressed in Coulombs (C) t is the time expressed in seconds (s)In this case:
I= 7.4 AQ= ?t= 3.7 min= 222 s (being 1 min= 60 s)Replacing:
[tex]7.4 A=\frac{Q}{222 s}[/tex]
Solving:
Q= 7.4 A* 222 s
Q= 1,642.8 C
A Coulomb represents about 6.24*10¹⁸ electrons, so you can apply the following rule of three: if 1 C represents 6.24*10¹⁸ electrons, 222 C how many electrons does it represent?
[tex]electrons=\frac{222 C*6.24*10^{18} }{1C}[/tex]
electrons= 1.39*10²¹
1,642.8 coulombs and 1.39*10²¹ electrons pass through any cross section the width of the wire.
Which answer is it
0.32 m/s
3.2 m/s
9.8 m/s
40 m/s
Answer:
3.2 m/s
Explanation:
you take the amount of blocks height wise, and the number of blocks with wise, and divide. height over with, so in this case 16/5. 16 divided by 5 is 3.2
A segment of wire of total length 3 m carries a 10 A current and is formed into a semicircle. Determine the magnitude of the magnetic field (in micro-tesla) at the center of the circle along which the wire is placed.
Answer:
[tex]B=3.29\ \mu T[/tex]
Explanation:
Given that,
Length of a wire is 3 m
Current in the wore, I = 10 A
The wire formed a semicircle.
We need to find the magnitude of the magnetic field (in micro-tesla) at the center of the circle along which the wire is placed.
The magnetic field at the center of semicircle is given by :
[tex]B=\dfrac{\mu_o I}{4R}[/tex]
R is radius of semicircle,
[tex]R=\dfrac{3}{\pi}=0.954\ m[/tex]
So,
[tex]B=\dfrac{4\pi \times 10^{-7}\times 10I}{4(0.954)}\\\\B=3.29\times 10^{-6}\ T\\\\B=3.29\ \mu T[/tex]
So, the magnitude of the magnetic field is [tex]3.29\ \mu T[/tex].
Two identical objects of the same size and shape are dropped. One is dropped from a table, while the other is dropped from the top of a tall building. Which will hit the ground with more impact?
Answer:
These two objects hit the ground at the same time. But their impact velocity will be different
because they are dropped from different heights.
The one that is dropped from the building will have more impact when it hits the ground because gravity causes this onject to fall toward the ground at a faster and faster velocity the longer the object falls. In fact, its velocity increases by 9.8 m/s2.
Explanation:
Initial velocity of the object u=0
Final velocity of the object when it reaches the ground v=A m/s
Using v^2−u^2=2gS where g=10m/s^2
A^2 −0^2=2(10)H
A^2=20*H H- heigh from the table
Object falling from the building will have the same initial velocity but different heigh.
T>H
so
B^2=20*T
20*T>20*H
so B^2>A^2
it equal to B>A so object falling from the building will have more impact than falling from the table.
What is the average velocity of the cart? Is the velocity increasing decreasing or constant as the cart moves along the air track?
Answer:
It is increasing .60 m/s
Explanation:
Which is the right one?
(Show work)
a mass of .4 kg is raised by a vertical distance of .450 m in the earth's gravitational field. what is the change in its gravitational potential energy
Answer:
E = 1.76 J
Explanation:
Given that,
Mass of an object, m = 0.4 kg
It moves by a vertical distance of 0.45 m in the Earth's gravitational field.
We need to find the change in its gravitational potential energy. It can be given by the formula as follow :
[tex]E=mgh\\\\E=0.4\times 9.8\times 0.45\\\\E=1.76\ J[/tex]
So, the change in its gravitational potential energy is 1.76 J.
a man in a gym is holding an 8.0 kg weight at arm's length, a distance of 0.55 m from his shoulder joint What is the torque about his shoulder joint due to the weight if his arm is horizontal
Answer:
[tex]the \: torque= 44 \: Nm.[/tex]
Explanation:
[tex] torque \: \boxed{t} \: is \: the \: product \: of \:one \: of \: the \: force \: - raduis \: \\ and \: sine \: of \: the \: angle \: of \: inclination : \\ torque \: \boxed{t} = fr \sin( \alpha ) \\ f = w = mg = 80. \times 10 = 80 \\ r = 0.55 \\ \alpha = 90 \\ \sin( \alpha ) = 1 \\ therefore \: the \: torque \: \boxed{t} \: is : \\ 80 \times 0.55 \times 1 = 44 \: Nm.[/tex]
♨Rage♨
A man is holding a weight at arm's length, a distance apart from his shoulder joint, it will experience torque. The torque about his shoulder joint due to the weight if his arm is horizontal is 43.12 N.m .
What is torque?Torque is the rotation caused about a point when a force is applied at a distance from the point.
The mass of man is 8.0 kg and the distance from the shoulder joint is 0.55 m. The torque applied will be the product of force and distance.
τ =F x r
and Force = weight of the body = mass x acceleration due to gravity
Substitute the values to get the torque.
τ = 8 kg x 9.8 m/s² x 0.55 m
τ = 43.12 N.m
Hence, the torque about his shoulder joint due to the weight if his arm is horizontal is 43.12 N.m
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A golf ball rolls up a hill toward a miniature-golf hole. Assume the direction toward the hole is positive. If the golf ball starts with a speed of 2.0 m/s and slows at a constant rate of 0.50 m/s2. What is the golf ball’s velocity if the constant acceleration continues for 6.0 s? *
Answer:
0.2533333334
Explanation:
It is simple, just do the math
What would be the luminosity of the Sun if its surface temperature were 4000 K and its radius were 2.0 AU ?
Answer:
0.04558
Explanation:
Luminosity is defined as the total amount of energy a star can produce in a second. Luminosity is given by the formula
σ * A * T⁴
where;
σ = Stefan-Boltzmann constant
= 5.670367 * 10⁻⁸
A= Surface Area
=4πr²
T= Temperature in Kelvin
In the question,
Temperature= 4000K
Radius= 2.0 A.U
Substituting the variables into the equation,
0.000000056704 * 4(3.14)(2.0²) * 4000⁴
= 0.000000056704 * 50.24 * 16000
= 0.04558
A current of 4sin4t A flows through a 5-F capacitor. Find the voltage v(t) across the capacitor given that v(0)
Answer:
[tex]v(t) = \frac{-cos4t}{5} + \frac{6}{5} Volts[/tex]
Explanation:
Given
current I = 4sin4t
capacitance C = 5F
v(0) = 1
The voltage across the capacitor is expressed as:
[tex]v(t) = \frac{1}{C} \int\limits^t_{t_0} {i(t)} \, dt + v(t_0)[/tex]
Given v(0) = 1, means at t = 0, v(t0) = 1
Substitute
[tex]v(t) = \frac{1}{5} \int\limits^t_{0} {4sin4t} \, dt + 1\\v(t) = \frac{1}{5} [\frac{-4cos4t}{4} ] + 1\\\\v(t) = \frac{1}{5} (-cos4t)+1 + \frac{1}{5}\\v(t) = \frac{-cos4t}{5} + \frac{6}{5}[/tex]
A 3.05 kg block is sliding across level ground and feels a 5.27 N friction force. What is the coefficient of friction?
Answer:
0.176
Explanation:
(5.27)/(3.05*9.8)
It is indeed 0.176 for the answer.
The speed of light in vacuum is exactly 299,792,458 m/s. A beam of light has a wavelength of 621 nm in vacuum. This light propagates in a liquid whose index of refraction at this wavelength is 1.18.What is the speed of this light in the liquid
Answer:
2.54*10^8 m/s
Explanation:
Given that
Speed of light in space, c = 299792458 m/s
Wavelength of the beam of light, λ = 621 nm or 621*10^-9 m
Index of refraction of the liquid, n = 1.18
Speed of light in the liquid, v = ?
The index of refraction of an optical material, n is given by the following formula.
n = c/v, and if we substitute directly, we have
1.18 = 299792458 / v, if we make v subject of formula, we have
v = 299792458 / 1.18
v = 2.54*10^8 m/s
The gravity on the moon is 1/6 as strong as that of earth.An object on earth weighs 1 Newton another object on the moon also weighs 1 Newton.Which of the two objects has the greater mass
Answer:
The object on the moon has greater mass
Explanation:
Given;
weight of the object on Earth, W₁ = 1 N
wight of the object on Moon, W₂ = 1 N
gravity on Earth, g₁ = 9.8 m/s²
gravity on the Moon, g₂ = g₁/6 = 1.633 m/s²
Apply Newton's second law of motion;
Weight of the object on Earth is given by;
W₁ = m₁g₁
m₁ = W₁/g₁
m₁ = 1 / 9.8
m₁ = 0.102 kg
Weight of the object on Moon is given by;
W₂ = m₂g₂
m₂ = W₂ / g₂
m₂ = 1 / 1.633
m₂ = 0.612 kg
Therefore, the object on the moon has greater mass.
Additional Practice on Speed
1. A group of campers and one group leader left a campsite in a canoe.
They traveled at an average rate of 10 km/h. Two hours later, the other
group leader left the campsite in a motorboat. He traveled at an average
rate of 22 km/h. How long had each boat traveled when the motorboat
caught up with the canoe?
a.
canoe: 2.2 h; motorboat: 1 h
b. canoe: 2.2 h; motorboat: 10 h
c. O canoe: 3 2 h; motorboat: 12 h
3
3
d. O canoe: 4.4 h; motorboat: 2 h
Answer:
canoe = 3 h 40 min. motorboat = 1 h 40 min
Explanation:
As both boats travel at a constant speed, we can apply the definition of average velocity to both, so we can get:[tex]v_{b} = \frac{\Delta x_{b} }{\Delta t_{b} } (1)[/tex]
[tex]v_{mb} = \frac{\Delta x_{mb} }{\Delta t_{mb} } (2)[/tex]
Now, when the motorboat caught up with the canoe, distance travelled by both boats were the same, so, solving (1) and (2) for Δx, we get:[tex]v_{b} *\Delta t_{b} = v_{mb} *\Delta t_{mb} (3)[/tex]
if we take the initial time for the canoe to be t₀ =0, the initial time for the motorboat will be tmb₀ = 2 Hs.So, replacing in (3) we have:[tex]v_{b} *t_{b} = v_{mb} *(t_{b} - 2 hs.) = 10 km/h* t_{b} = 22 km/h* (t_{b} - 2 hs.)[/tex]
Solving for tb:[tex]t_{b} = \frac{-44 }{-12} hs = 3h 40 min.[/tex]
tmb = tb - 2 hs. = 1h 40 min.
In an old-fashioned television tube, an electron () starting from rest experiences a force of 4.0 × 10–15 N over a distance of 50 cm. Ignoring the relativistic effects, the final speed of the electron is:
Answer:
The final speed of the electron is 6.626 x 10⁷ m/s
Explanation:
force applied to the electron, f = 4 x 10⁻¹⁵ N
distance traveled by the electron, d = 50 cm = 0.5 m
The work done on the electron;
W = F x d
W = (4 x 10⁻¹⁵ N) x (0.5 m)
W = 2 x 10⁻¹⁵ J
The kinetic energy of the electron is given by;
[tex]K.E = \frac{1}{2}m_ev^2[/tex]
Apply work - energy theorem;
K.E = W
[tex]\frac{1}{2}m_ev^2 = 2 *10^{-15}\\\\ v^2 = \frac{2( 2 *10^{-15})}{m_e}\\\\v= \sqrt{\frac{2( 2 *10^{-15})}{m_e}}\\\\ v = \sqrt{\frac{2( 2 *10^{-15})}{9.11*10^{-31}}}\\\\v = 6.626*10^7 \ m/s[/tex]
Therefore, the final speed of the electron is 6.626 x 10⁷ m/s
You are given two vectors A⃗ =−3.00i^+7.00j^ and B⃗ =5.00i^+2.00j^. Let counterclockwise angles be positive.
1- What angle θA, where 0∘≤θA<360∘, does A⃗ make with the +x-axis?
2-What angle θB, where 0∘≤θB<360∘, does B⃗ make with the +x-axis?
Express your answer in degrees.
3-Vector C⃗ is the sum of A⃗ and B⃗ , so C⃗ =A⃗ +B⃗ . What angle θC, where 0∘≤θC<360∘, does C⃗ make with the +x-axis?
Express your answer in degrees
Answer:
Ax = -3 By= 7 tan phi = -7 / 3 phi = -66.8
Since theta is in the second quadrant theta = 180 - 66.8 = 113.2 deg
Bx = 5 By = 2 tan theta = .4 theta = 21.8 deg
Cx = Ax + Bx = 2 Cy = Ay + By = 9
tan theta = 9 / 2 = 4.5 theta = 77.5 deg
1) The solution is [tex]\theta_{1} \approx 113.141^{\circ}[/tex].
2) The solution is [tex]\theta_{1} \approx 21.874^{\circ}[/tex].
3) The solution is [tex]\theta_{1} \approx 77.467^{\circ}[/tex].
1) In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:
[tex]\cos \theta = \frac{\vec A\,\bullet \, \vec u}{\|\vec A\|\cdot \|\vec u\|}[/tex] (1)
Where [tex]\|\vec A\|[/tex] and [tex]\|\vec u\|[/tex] are the norms of [tex]\vec A[/tex] and [tex]\vec u[/tex].
If we know that [tex]\vec A = (-3, 7)[/tex] and [tex]\vec u = (1, 0)[/tex], then the angle that [tex]\vec A[/tex] make with the x-axis is:
[tex]\|\vec A\| = \sqrt{(-3)^{2}+7^{2}}[/tex]
[tex]\|\vec A\| = \sqrt{58}[/tex]
[tex]\|\vec u\| = 1[/tex]
[tex]\cos \theta = \frac{(-3)\cdot (1)}{\sqrt{58} \cdot 1}[/tex]
[tex]\cos \theta \approx -0.393[/tex]
The solution is [tex]\theta_{1} \approx 113.141^{\circ}[/tex].
2) In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:
[tex]\cos \theta = \frac{\vec B\,\bullet \, \vec u}{\|\vec B\|\cdot \|\vec u\|}[/tex] (1)
Where [tex]\|\vec B\|[/tex] and [tex]\|\vec u\|[/tex] are the norms of [tex]\vec B[/tex] and [tex]\vec u[/tex].
If we know that [tex]\vec B = (5, 2)[/tex] and [tex]\vec u = (1, 0)[/tex], then the angle that [tex]\vec A[/tex] make with the x-axis is:
[tex]\|\vec B\| = \sqrt{5^{2}+2^{2}}[/tex]
[tex]\|\vec B\| = \sqrt{29}[/tex]
[tex]\|\vec u\| = 1[/tex]
[tex]\cos \theta = \frac{(5)\cdot (1)}{\sqrt{29} \cdot 1}[/tex]
[tex]\cos \theta \approx 0.928[/tex]
The solution is [tex]\theta_{1} \approx 21.874^{\circ}[/tex].
3) In this case, we need to find the value of [tex]\vec C[/tex] by vectorial sum before calculating the angles with respect to x-axis:
[tex]\vec C = \vec A + \vec B = (-3, 7) + (5, 2)[/tex]
[tex]\vec C = (2, 9)[/tex]
In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:
[tex]\cos \theta = \frac{\vec C\,\bullet \, \vec u}{\|\vec C\|\cdot \|\vec u\|}[/tex] (1)
Where [tex]\|\vec C\|[/tex] and [tex]\|\vec u\|[/tex] are the norms of [tex]\vec C[/tex] and [tex]\vec u[/tex].
If we know that [tex]\vec B = (2, 9)[/tex] and [tex]\vec u = (1, 0)[/tex], then the angle that [tex]\vec A[/tex] make with the x-axis is:
[tex]\|\vec C\| = \sqrt{2^{2}+9^{2}}[/tex]
[tex]\|\vec C\| = \sqrt{85}[/tex]
[tex]\|\vec u\| = 1[/tex]
[tex]\cos \theta = \frac{(2)\cdot (1)}{\sqrt{85} \cdot 1}[/tex]
[tex]\cos \theta \approx 0.217[/tex]
The solution is [tex]\theta_{1} \approx 77.467^{\circ}[/tex].
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A dart is loaded into a compressed spring. Starting from rest the spring is released and shoots the dart up into the air. From the instant the dart is released (before it starts to move) until it reaches its maximum height, what is the Net work done on the dart
Answer:
Wnet = 0
Explanation:
The work-energy theorem says that the total net work done on an object, is equal to the change in the kinetic energy of that object.In this case, the total change in kinetic energy of the dart is zero, due to the initial status is at rest (loaded into a compressed spring) and the final is also at rest (when it reaches to its maximum height).The total work done on the dart can be written as follows:[tex]W_{net} = \frac{1}{2} k* \Delta x^{2} - m*g*h_{max} (1)[/tex]
Now, in absence of friction, all the elastic potential energy becomes gravitational potential energy, so the following is true:[tex]\frac{1}{2} k* \Delta x^{2} = m*g*h_{max} (2)[/tex]
From (1) and (2) we conclude that Wnet = 0.I HAVE A QUESTION ABOVE CAN ANYONE PLEASE HELP ME I WOULD REALLY APPRECIATE THE HELP I HAVE BEEN ON THIS QUESTION FOREVER i will give you the brainlist !!!!!
Answer:
answer is C
Explanation:
Answer: Monetary policy used during times of recession
Explanation:
Help please it’s urgent
A fisherman notices that one wave passes the bow of his anchored boat
every 3 seconds. He measured the wavelength to be 8.5 meters. How fast
are the waves traveling?
1.)2.83m/s
2.)0.283m/s
3.)28.3m/s
An 800 N table is slid across the kitchen floor by pushing with a force of 100 N. If the table moves at a constant speed, what is the friction force with the floor?
A. 100 N
B. 8 N
C. 800 N
D. over 800 N
Answer:
A. 100 N
Explanation:
Given
Table = 800N
Pushing Force = 100N
Required
Determine the friction force with the floor
From the question, we understand that the table moves at a constant speed.
In this kind of situation, the friction force equals to the pushing force.
Hence:
Friction Force = Pushing Force
Friction Force = 100N
Find μsμsmu_s, the coefficient of static friction between the rod and the rails. Express the coefficient of static friction in terms of variables given in the introduction.
Answer:
The question is incomplete. However the complete question is :
A Rail Gun uses electromagnetic forces to accelerate a projectile to very high velocities. The basic mechanism of acceleration is relatively simple and can be illustrated in the following example. A metal rod of mass m and electrical resistance R rests on parallel horizontal rails (that have negligible electric resistance), which are a distance L apart. The rails are also connected to a voltage source V, so a current loop is formed.
The vertical magnetic field, initially zero, is slowly increased. When the field strength reaches the value [tex]$B_0$[/tex], the rod, which was initially at rest, begins to move. Assume that the rod has a slightly flattened bottom so that it slides instead of rolling. Use g for the magnitude of the acceleration due to gravity.
Find μs, the coefficient of static friction between the rod and the rails.
Express the coefficient of static friction in terms of variables given in the introduction.
So the answer is : [tex]$\mu_s=\frac{\Delta V LB_0}{Rmg}$[/tex]
Explanation:
It is given : B = [tex]$B_0$[/tex]
Therefore, using force on the current carrying wire is:
[tex]$\mu_s mg = B_0IL$[/tex]
[tex]$\mu_s= \frac{ILB_0}{mg}$[/tex]
Therefore,
[tex]$\mu_s=\frac{\Delta V LB_0}{Rmg}$[/tex]
The coefficient of static friction in terms of variables will be [tex]\rm \mu_s =\frac{ \triangle VLB_0}{Rmg} \\\\[/tex].
What is the cofficient of static friction?The ratio of the greatest static friction force (F) between the surfaces in contact before movement begins to the normal (N) force is the coefficient of static friction.
The given data in the problem is;
[tex]\rm B=B_0[/tex]
The force on a current-carrying wire is balanced by the friction force.
The force on a current-carrying wire is given by;
[tex]\rm \mu_s mg= B_0IL \\\\ \rm \mu_s=\frac{ILB_0}{mg}\\\\ \rm \mu_s= \frac{\triangle VLB_0}{Rmg}[/tex]
Hence the coefficient of static friction in terms of variables will be [tex]\rm \mu_s =\frac{ \triangle VLB_0}{Rmg} \\\\[/tex].
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A half-eaten apple turns brown which of these is evidence a chemical change took place A.th3e apple is softer b.the apple changed color C.half the apple was eaten D.the apple is warmer \
Answer:
b the apple changed color
if a cross country runner covers 347 meters in 134 seconds what is her speed
A truck is traveling at an initial velocity
of +39m/s and it starts slowing down smoothly for 3.3s. It covers a distance of 45m while slowing down. What is the acceleration?
Answer:
Approximately [tex]-15.37m/s^2[/tex]
Explanation:
The acceleration can be found using the formula:
[tex]x_{f} = x_{i} + v_{i} (t)+\frac{1}{2} (a)(t^{2} )[/tex]
The work is as shown:
[tex]45m = 0 + 39 m/s (3.3s)+\frac{1}{2} a(3.3s)^2[/tex]
[tex]45m=128.7m + 5.445s^2 a[/tex]
[tex]-83.7m=5.445s^2a[/tex]
[tex]a = -15.37190083m/s^2[/tex]
a is about -15.37[tex]m/s^2[/tex]
Besides being asleep and being awake list the different states of consciousness
Answer:
bias.
Consciousness.
Hypnosis.
Priming.
Sleep.
Trance.
Explanation:
what vehicle are runaway ramps designed for
Answer:truck
Explanation:
truck
What is the mass of an object if a force of 17N causes it to accelerate at 1.5 m/sec/sec
A. 11.38
B. 11.3 kg
C. 12.3 kg
D. 14.3 kg
Answer:
B.11.3 kg
Explanation :
F=ma here force is given as 17N and acceleration as 1.5 m/sec/sec using the formula mass can be calculated
∴ F=ma
17N =m × 1.5 m[tex]s^{-2}[/tex]
[tex]m=\frac{F}{a}[/tex]
[tex]m= \frac{17 kg m/sec/sec}{1.5 m/sec/sec}[/tex]
∴ m=11.3 kg
All living creatures have some form of sleep that they engage in at regular periods, making sleep a(n) __________ function. A. environmental B. biological C. species-specific D. psychological
The answer is B. biological
Answer:
ans: a biological function
Answer:
B
Explanation:
EDGE2020
What is the minimum compression of the spring necessary for the block to make it to the top of the hill on the other side?
Answer:
sciencephysicsphysics questions and answersA 12-kg Block Is Pressed Against A Spring (spring Constant 870 N/m ), Compressing It Some Distance. ...
Question: A 12-kg Block Is Pressed Against A Spring (spring Constant 870 N/m ), Compressing It Some Distance. The Block Is Released From Rest And Slides Across A Track As Shown In (Figure 1). While Most Of The Track Is Frictionless, There Is A 55-cm Section Of Track That Has A Coefficient Of Friction With The Block Of 0.4. A Bit Further On, The Track Ascends ...
A 12-kg block is pressed against a spring (spring constant 870 N/m ), compressing it some distance. The block is released from rest and slides across a track as shown in (Figure 1). While most of the track is frictionless, there is a 55-cm section of track that has a coefficient of friction with the block of 0.4. A bit further on, the track ascends into a hill that is 40-cm tall.
What is the minimum compression of the spring necessary for the block to slide past the section with friction?
What is the minimum compression of the spring necessary for the block to make it to the top of the hill on the other side? 40cm 12kg Ww.w.wwwww 55cm
Explanation:
A block with mass 0.50 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.20 m . When released, the block moves on a horizontal tabletop for 1.00 m before coming to rest. The spring constant k is 100 N/m. What is the coefficient of kinetic friction between the block and the tabletop
Answer:
μ = 0.41
Explanation:
[tex]\frac{1}{2} kx^2 = \mu mgd[/tex]
[tex]\mu = (\frac{1}{2} kx^2)/mgd[/tex]
[tex]\mu = (\frac{1}{2} (100)(0.2)^2)/(0.5)(9.81)(1)[/tex]
[tex]\mu = 0.41[/tex]
The coefficient of kinetic friction between the block and the tabletop is 0.41.
The given parameters;
mass of the block, m = 0.5 kgextension of the spring, x = 0.2 mspring constant, K = 100 N/mThe coefficient of kinetic friction between the block and the tabletop by applying the principle of conservation of energy as shown below.
Fd = Uₓ
μmgd = ¹/₂kx²
where;
μ is the coefficient of kinetic frictionThe coefficient of kinetic friction between the block and the tabletop;
[tex]\mu_k = \frac{kx^2}{mgd} \\\\\mu_k = \frac{100\times 0.2^2}{2\times 0.5\times9.8 \times 1 } \\\\\mu_k = 0.41[/tex]
Thus, the coefficient of kinetic friction between the block and the tabletop is 0.41.
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