Answer:
Power = 576 Watts
Explanation:
The electrical power of an electric circuit can be defined as a measure of the rate at which energy is either produced or absorbed in the circuit.
Mathematically, electrical power is given by the formula;
[tex] Electrical \; power = current * voltage [/tex]
This ultimately implies that, the quantity (current times voltage ) is electrical power and it is measured (S.I units) in Watt (W).
Given the following data;
Resistance = 100 ohms
Voltage = 240 V
To find the power rating of the heater;
Power = V²/R
Where;
V is the voltage.
R is the resistance.
Substituting into the formula, we have;
Power = 240²/100
Power = 57600/100
Power = 576 Watts
Your dog is running around the grass in your back yard. He undergoes successive displacements 3.20 m south, 8.16 m northeast, and 15.6 m west. What is the resultant displacement
Answer:
D1 = 3.50 m, south; D2 = 8.20 m, northeast; D3 = 15.0 m, west. Converting all these displacements from east where zero degrees is at east or + x-axis, the converted displacements are: D1 = 3.50 m 270°; D2 = 8.20 m 45° and D3 = 15.0 m 180°. We then tabulate these vectors including there x and y components. The x-components are solved by magnitudes * cos of direction angle while the y-components of the three vectors are solved by magnitudes * sin of direction angle.
The resultant is computed by summing the components algebraically. The direction in degrees is the arc tangent of the sum of all y divided by the sum of all x.
Explanation:
g How much buoyancy force, in N, a person with a mass of 70 kg experiences by just standing in air
Answer:
686.7N
Explanation:
Given data
Mass= 70kg
We know that the buoyant force experienced by the person is equal to the weight of the person
Hence the weight is
Weight = mass* Acceleration
Weight= 70*9.81
Weight= 686.7N
Therefore the weight is 686.7N
A double-slit experiment is performed with light of wavelength 550 nm. The bright interference fringes are spaced 2.3 mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 360 nm?
Answer:
[tex]d_2=1.5*10^-3m[/tex]
Explanation:
From the question we are told that:
Initial Wavelength [tex]\lambda_1=550nm=550*10^{-9}[/tex]
Space 1 [tex]d_1=2.3*10^{-3}[/tex]
Final wavelength [tex]\lambda_2=360*10^{-9}[/tex]
Generally the equation for Fringe space at [tex]\lambda _2[/tex] is mathematically given by
[tex]d_2=\frac{d_1}{\lambdaI_1}*\lambda_2[/tex]
[tex]d_2=\frac{2.3*10^{-3}}{550*10^{-9}}*360*10^{-9}[/tex]
[tex]d_2=1.5*10^-3m[/tex]
(c) The ball leaves the tennis player's racket at a speed of 50 m/s and travels a
distance of 20 m before bouncing.
(i) Calculate how long it takes the ball to travel this distance.
(1 mark)
Answer:
t=0.417s
Explanation:
After the ball hits the racket it is in freefall(assume air resistance as negligible)
so a=-g
use
x-x0=v0t+1/2at^2
Plug in givens
20=50t-4.9t^2
Solve quadratic equation using quadratic formula
t= 0.417 seconds, (the other answer is extraneous because it is too big because in 1 second, the ball travels 50 meters)
12) If, after viewing a specimen at low power, you switch to high-dry power and, after using fine focus, cannot find the specimen, what things could you do to help yourself (before calling me over to assist you?)
Answer:
See the answer below
Explanation:
After seeing an object on a slide at the low-power objective of the microscope and it disappears on changing to high power, the following can be done to resolve the problem
1. Drop a few drops of immersion oil on the slide and view again under high the power objective.
2. If the object is still not visible after the action above, return the microscope to the low-power objective and make sure the object is refocused and centered. Then carefully change back to the high power objective and use the fine adjustment to bring it into focus.
What is life like in a cave camp? Do you think you would like to experience this? Why or why not?
Answer:
There's no risk of animals or bad weather interfering with your campsite, either. You don't even really need a tent. A sleeping pad, sleeping bag and a mindful eye to pick up everything you brought in is all you really need to enjoy overnight caving. Do your research
Explanation:
If you buy an amateur-sized reflecting telescope, say around 10 inches (25cm) aperture, it'll have something in it that sends the gathered starlight out the side of the telescope tube. What do we call this thing
Answer: objective lens
Explanation:
Light enters a refra
Light enters a telescope through a lens at the upper end, which focuses the light near the bottom of the telescope. An eyepiece then magnifies the image so that it can be viewed by the eye, or a detector like a photographic plate can be placed at the focus. The upper end of a reflecting telescope is open, and the light passes through to the mirror located at the bottom of the telescope. The mirror then focuses the light at the top end, where it can be detected. Alternatively, a second mirror may reflect the light to a position outside the telescope structure, where an observer can have easier access to it.
A solid piece of clear transparent material has an index of refraction of 1.61. If you place it into a clear transparent solution and it seems to disappear, approximately what is the index of refraction of the solution
Answer:
1.61
Explanation:
According to Oxford dictionary, refractive index is, ''the ratio of the velocity of light in a vacuum to its velocity in a specified medium.''
If the clear transparent solid disappears when dipped into the liquid, it means that the index of refraction of the solid and liquid are equal.
Hence, when a transparent solid is immersed in a liquid having the same refractive index, there is no refraction at the boundary between the two media. As long as there is no refraction between the two media, the solid can not be seen because the solid and liquid will appear to the eye as one material.
4. Consider a 1 kg block is on a 45° slope of ice. It is connected to a 0.4 kg block by a cable
and pulley. Does the 1 kg block move or down the slope? What is the net force on it and
its acceleration? (8 pts)
If an icy surface means no friction, then Newton's second law tells us the net forces on either block are
• m = 1 kg:
∑ F (parallel) = mg sin(45°) - T = ma … … … [1]
∑ F (perpendicular) = n - mg cos(45°) = 0
Notice that we're taking down-the-slope to be positive direction parallel to the surface.
• m = 0.4 kg:
∑ F (vertical) = T - mg = ma … … … [2]
Adding equations [1] and [2] eliminates T, so that
((1 kg) g sin(45°) - T ) + (T - (0.4 kg) g) = (1 kg + 0.4 kg) a
(1 kg) g sin(45°) - (0.4 kg) g = (1.4 kg) a
==> a ≈ 2.15 m/s²
The fact that a is positive indicates that the 1-kg block is moving down the slope. We already found the acceleration is a ≈ 2.15 m/s², which means the net force on the block would be ∑ F = ma ≈ (1 kg) (2.15 m/s²) = 2.15 N directed down the slope.
A 5 kg object is moving in one dimension along the x-axis with a speed of 2 m/s. An external impulse acts on the force causing the speed of the object to increase to 5 m/s. The impulse lasted for 3 s. What is the average net force (in N) exerted on the object
Answer:
The correct answer is "15 Kg.m/s".
Explanation:
Given values are:
Mass,
m = 5 Kg
Initial velocity,
u = 2 m/s
Final velocity,
v = 5 m/s
Now,
The magnitude of change in linear momentum will be:
= [tex]m\times (v - u)[/tex]
By substituting the values, we get
= [tex]5\times (5 - 2)[/tex]
= [tex]5\times 3[/tex]
= [tex]15 \ Kg.m/s[/tex]
An elevator with its occupants weighs 2400 N and is supported by a vertical cable. What is the tension in the cable if the elevator is moving up with its speed decreasing at a rate of 1.7
Answer:
Hope you find it useful. please correct me if I am wrong
The tension in the cable if the elevator is moving upward with its speed decreasing at a rate of 1.7 m/s² is equal to 1983.67 N.
What is tension?Tension can be described as a force acting along the length of a medium such as a rope, mainly a force carried by a flexible medium.
Tension can be defined as an action-reaction pair of forces acting at each end of the elements. The tension force is in every section of the rope in both directions, apart from the endpoints. Each endpoint of the rope experience tension and force from the weight attached.
Given the force due to the weight of the elevator = mg = 2400N
m = 2400/9.8 Kg
The elevator deaccelerating while moving upward, a = -1.7 m/s²
According to Newton's 3rd law: T - mg = ma
T - 2400 = (2400/9.8) × (-1.7)
T = 2400 - 416.32
T = 1983.67 N
Learn more about tension, here:
brainly.com/question/28965515
#SPJ5
if one branch of a 120-v power lines is protected by a 20-A fuse, will the fuse carry an 8-Ώ load
Answer:
No I won't.
It will carry 6ohm load.
Explanation:
It obeys ohms law therefore V=IR
120=20R
R=120/20
R= 6 ohms
A heat engine exhausts 3 000 J of heat while performing 1 500 J of useful work. What is the efficiency of the engine
efficiency=work output/work input×100
since it exhausts(use up)3000j of heat that's the work input and the 1500j is the work input
efficiency=1500/3000×100
=50%
A car is moving with a velocity of45m/sis brought to rest in 5s.the distance travelled by car before it comes to rest is
Answer:
The car travels the distance of 225m before it comes to rest.
Explanation:
Given,
v = 45m/s
t = 5s
Therefore,
d = v × t
= 45 × 5
= 225m
Which of the following statements is correct about the magnitude of the static friction force between an object and a surface?
a. Static friction depends on the mass of the object.
b. Static friction depends on the shape of the object.
c. Static friction depends on what the object is made of but not what the surface is made of.
d. None of the above is correct.
Answer:
Static friction depends on the mass of the object.
Explanation:
Friction is the force between two surfaces in contact. The force of friction between two surfaces in contact depends on;
1) nature of the object and the surface(how rough or smooth the surfaces are)
2)surface area of the object and the surface
3) mass of the object
Since;
F=μmg
Where;
μ= coefficient of static friction
m= mass of the object
g= acceleration due to gravity
Hence, as the mass of the object increases, the magnitude of static friction force between an object and a surface increases and vice versa.
An electron moving in the y direction, at right angles to a magnetic field, experiences a magnetic force in the -x direction. The direction of the magnetic field is in the
Answer:
The direction of magnetic field is along + Z axis.
Explanation:
The direction of motion of electron is along y axis.
The magnetic force is along - X axis.
The force on the charged particle moving in the magnetic field is
[tex]\overrightarrow{F} = q (\overrightarrow{v}\times \overrightarrow{B})\\\\- F \widehat{i} = - q (v \widehat{j}\times \overrightarrow{B})\\[/tex]
So, the direction of the magnetic field is along + Z axis.
A chimpanzee sitting against his favorite tree gets up and walks 51 m due east and 39 m due south to reach a termite mound, where he eats lunch. (a) What is the shortest distance between the tree and the termite mound
Answer:
64.20m
Explanation:
As we can see from the image I have attached below, the route that the chipanzee makes forms a right triangle. In this case, the shortest distance is represented by x in the image, which is the hypotenuse. To find this value we use the Pythagorean theorem which is the following.
[tex]a^{2} +b^{2} = c^{2}[/tex]
where a and b are the length of the two sides and c is the length of the hypotenuse (x). Therefore, we can plug in the values of the image and solve for x
[tex]51^{2} +39^{2} =x^{2}[/tex]
2,601 + 1,521 = [tex]x^{2}[/tex]
4,122 = [tex]x^{2}[/tex] ... square root both sides
64.20 = x
Finally, we see that the shortest distance is 64.20m
The mass of the sun is 2*10^30 kg and its radius is
6.96*10^8 m. what is the weight of 1kg mass on the
surface of the sun.
Explanation:
Distance d=1.5×108 km=1.5×1011 m
Mass of the sun, m=2×1030 kg
Mass of the earth, M=6×1024 kg
Force of gravitation, F=G×d2m×M
F=6.7×10−11×(1.5×1011)22×1030×6×1024=3.57×1022 N
The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.180 × 10-3 m3. Into this cavity is placed 1.100 × 10-3 m3 of benzene [β = 1240 × 10-6 (C°)-1]. Both the copper and the benzene have the same temperature. By what amount ΔT should the temperature of the sphere and the benzene within it be increased, so that the liquid just begins to spill out?
Answer:
The answer is "[tex]60.74^{\circ}[/tex]".
Explanation:
Cavity and benzene should be extended in equal quantities.
[tex]\to 1.18 \times 10^{-3}\times (1+ \Delta T \times 0.000051) = 1.1\times 10^{-3} \times (1+ \Delta T \times 0.00124)\\\\\to (\frac{1.18}{1.1})\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 - 1- \Delta T \times 0.00124=0\\\\[/tex]
[tex]\to 0.072+ \Delta T \times 0.000054672 - \Delta T \times 0.00124=0\\\\ \to 0.072+ \Delta T ( 0.000054672 -0.00124)=0\\\\ \to \Delta T ( 0.000054672 -0.00124)= -0.072\\\\ \to \Delta T = -\frac{0.072}{( 0.000054672 -0.00124)}\\\\ \to \Delta T = -\frac{0.072}{-0.001185328 }\\[/tex]
[tex]\to \Delta T = \frac{0.072}{0.001185328 }\\\\ \to \Delta T = 60.74^{\circ}\\[/tex]
What happens to the acceleration if you triple the force that you apply to the painting with your hand? (Use the values from the example given in the previous part of the lecture.) Submit All Answers Answer: Not yet correct, tries 1/5 3. A driver slams on the car brakes, and the car skids to a halt. Which of the free body diagrams below best matches the braking force on the car. (Note: The car is moving in the forward direction to the right.] (A) (B) (C) (D) No more tries. Hint: (Explanation) The answer is A. The car is moving to the right and slowing down, so the acceleration points to the left. The only significant force acting on the car is the braking force, so this must be pointing left because the net force always shares the same direction as the object's acceleration. 4. Suppose that the car comes to a stop from a speed of 40 mi/hr in 24 seconds. What was the car's acceleration rate (assuming it is constant). Answer: Submit Al Answers Last Answer: 55 N Only a number required, Computer reads units of N, tries 0/5. 5. What is the magnitude (or strength) of the braking force acting on the car? [The car's mass is 1200 kg.) Answer: Submit Al Answers Last Answer: 55N Not yet correct, tries 0/5
Answer:
2) when acceleration triples force triples, 3) a diagram with dynamic friction force in the opposite direction of movement of the car
4) a = 2.44 ft / s², 5) fr = 894.3 N
Explanation:
In this exercise you are asked to answer some short questions
2) Newton's second law is
F = m a
when acceleration triples force triples
3) Unfortunately, the diagrams are not shown, but the correct one is one where the axis of movement has a friction force in the opposite direction of movement, as well as indicating that the car slips, the friction coefficient of dynamic.
The correct answer is: a diagram with dynamic friction force in the opposite direction of movement of the car
4) let's use the scientific expressions
v = v₀ - a t
as the car stops v = 0
a = v₀ / t
let's reduce the magnitudes
v₀ = 40 mile / h ([tex]\frac{5280 ft}{1 mile}[/tex]) ([tex]\frac{1 h}{3600 s}[/tex]) = 58.667 ft / s
a = 58.667 / 24
a = 2.44 ft / s²
5) let's use Newton's second law
fr = m a
We must be careful not to mix the units, we will reduce the acceleration to the system Yes
a = 2.44 ft / s² (1 m / 3.28 ft) = 0.745 m / s²
fr = 1200 0.745
fr = 894.3 N
A body of mass 2kg is released from from a point 100m above the ground level. calculate kinetic energy 80m from the point of released.
Answer:
1568J
Explanation:
Since the problem states 80 m from the point of drop, the height relative to the ground will be 100-80=20m.
Use conservation of Energy
ΔUg+ΔKE=0
ΔUg= mgΔh=2*9.8*(20-100)=-1568J
ΔKE-1568J=0
ΔKE=1568J
since KEi= 0 since the object is at rest 100m up, the kinetic energy 20meters above the ground is 1568J
How do the magnitudes of the currents through the full circuits compare for Parts I-III of this exercise, in which resistors are combined in series, in parallel, and in combination
Answer: hello tables and data related to your question is missing attached below are the missing data
answer:
a) I = I₁ = I₂ = I₃ = 0.484 mA
b) I₁ = 0.016 amps
I₂ = 0.0016 amps
I₃ = 7.27 * 10^-4 amps
c) I₁ = 1.43 * 10^-3 amp
I₂ = 0.65 * 10^-3 amps
Explanation:
A) magnitude of current for Part 1
Resistors are connected in series
Req = r1 + r2 + r3
= 3300 Ω ( value gotten from table 1 ) ,
V = 1.6 V ( value gotten from table )
hence I ( current ) = V / Req = 1.6 / 3300 = 0.484 mA
The magnitude of current is the same in the circuit
Vi = I * Ri
B) magnitude of current for part 2
Resistors are connected in parallel
V = 1.6 volts
Req = [ ( R1 * R2 / R1 + R2 ) * R3 / ( R1 * R2 / R1 + R2 ) + R3 ]
= [ ( 100 * 1000 / 100 + 1000) * 2200 / ( 100 * 1000 / 100 + 1000 ) + 2200]
= 87.30 Ω
For a parallel circuit the current flow through each resistor is different
hence the magnitude of the currents are
I₁ = V / R1 = 1.6 / 100 = 0.016 amps
I₂ = V / R2 = 1.6 / 1000 = 0.0016 amps
I₃ = V / R3 = 1.6 / 2200 = 7.27 * 10^-4 amps
C) magnitude of current for part 3
Resistors are connected in combination
V = 1.6 volts
Req = R1 + ( R2 * R3 / R2 + R3 )
= 766.66 Ω
Total current ( I ) = V / Req = 1.6 / 766.66 = 2.08 * 10^-3 amps
magnitude of currents
I₁ = ( I * R3 ) / ( R2 + R3 ) = 1.43 * 10^-3 amps
I₂ = ( I * R2 ) / ( R2 + R3 ) = 0.65 * 10^-3 amps
An electric eel can generate a 180-V, 0.1-A shock for stunning its prey. What is the eel's power output
Power output = volts x amps
Power output = 170 volts x 0.1 amps
Power output = 18 watts
A simple pendulum takes 2.00 s to make one compete swing. If we now triple the length, how long will it take for one complete swing
Answer:
3.464 seconds.
Explanation:
We know that we can write the period (the time for a complete swing) of a pendulum as:
[tex]T = 2*\pi*\sqrt{\frac{L}{g} }[/tex]
Where:
[tex]\pi = 3.14[/tex]
L is the length of the pendulum
g is the gravitational acceleration:
g = 9.8m/s^2
We know that the original period is of 2.00 s, then:
T = 2.00s
We can solve that for L, the original length:
[tex]2.00s = 2*3.14*\sqrt{\frac{L}{9.8m/s^2} }\\\\\frac{2s}{2*3.14} = \sqrt{\frac{L}{9.8m/s^2}}\\\\(\frac{2s}{2*3.14})^2*9.8m/s^2 = L = 0.994m[/tex]
So if we triple the length of the pendulum, we will have:
L' = 3*0.994m = 2.982m
The new period will be:
[tex]T = 2*3.14*\sqrt{\frac{2.982m}{9.8 m/s^2} } = 3.464s[/tex]
The new period will be 3.464 seconds.
If an electrical component with a resistance of 53 Q is connected to a 128-V source, how much current flows through the component?
Answer:
the current that flows through the component is 2.42 A
Explanation:
Given;
resistance of the electrical component, r = 53 Ω
the voltage of the source, V = 128 V
The current that flows through the component is calculated using Ohm's Law as demonstrated below;
[tex]V = IR\\\\I = \frac{V}{R} = \frac{128 \ V}{53 \ ohms} = 2.42 \ A[/tex]
Therefore, the current that flows through the component is 2.42 A
~~~NEED HELP ASAP~~~
Please solve each section and show all work for each section.
Explanation:
Forces on Block A:
Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass [tex]m_A[/tex] as
[tex]x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)[/tex]
[tex]y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)[/tex]
Substituting (2) into (1), we get
[tex]\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)[/tex]
where [tex]f_N= \mu_kN[/tex], the frictional force on [tex]m_A.[/tex] Set this aside for now and let's look at the forces on [tex]m_B[/tex]
Forces on Block B:
Let the x-axis be (+) up along the inclined plane. We can write the forces on [tex]m_B[/tex] as
[tex]x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)[/tex]
[tex]y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)[/tex]
From (5), we can solve for N as
[tex]N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)[/tex]
Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by
[tex]T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)[/tex]
Substituting (7) into (4) we get
[tex]m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba[/tex]
Collecting similar terms together, we get
[tex](m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag[/tex]
or
[tex]a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)[/tex]
Putting in the numbers, we find that [tex]a = 1.4\:\text{m/s}[/tex]. To find the tension T, put the value for the acceleration into (7) and we'll get [tex]T = 21.3\:\text{N}[/tex]. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get [tex]N = 50.9\:\text{N}[/tex]
When you hammer a nail into wood, the nail heats up. 30 Joules of energy was absorbed by a 5-g nail as it was hammered into place. How much does the nail's temperature increase (in °C) during this process? (The specific heat capacity of the nail is 450 J/kg-°C, and round to 3 significant digits.
Answer:
13.33 K
Explanation:
Given that,
Heat absorbed, Q = 30 J
Mass of nail, m = 5 g = 0.005 kg
The specific heat capacity of the nail is 450 J/kg-°C.
We need to find the increase in the temperature during the process. The heat absorbed in a process is as follows:
[tex]Q=mc\Delta T\\\\\Delta T=\dfrac{Q}{mc}\\\\\Delta T=\dfrac{30}{0.005\times 450}\\\\=13.33\ K[/tex]
So, the increase in temperature is 13.33 K.
a nano second is what
Answer:
one thousand-millionth of a second.
A nanosecond is an SI unit of time equal to one billionth of a second, that is, ¹⁄₁ ₀₀₀ ₀₀₀ ₀₀₀ of a second, or 10⁻⁹ seconds. The term combines the prefix nano- with the basic unit for one-sixtieth of a minute. A nanosecond is equal to 1000 picoseconds or ¹⁄₁₀₀₀ microsecond.
d. On the afternoon of January 15, 1919, an unusually warm day in Boston, a 17.7-m-high, 27.4-m-diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded into the streets in a 5-m-deep stream, killing pedestrians and horses and knocking down buildings. The molasses had a density of 1600 kg>m3 . If the tank was full before the accident, what was the total outward force the molasses exerted on its sides
Answer:
F = 1.638 x 10⁸ N = 163.8 MN
Explanation:
The total force exerted by the molasses is given as:
F = PA
where,
F = Force exerted by the molasses = ?
P = Pressure = ρgh
ρ = density of molasses = 1600 kg/m³
g = acceleration due to gravity = 9.81 m/s²
h = height of tank = 17.7 m
A = cross-sectional area of tank = πr²
r = radius of tank = 27.4 m/2 = 13.7 m
Therefore,
[tex]F = \rho ghA = \rho gh(\pi r^2)\\\\F = (1600\ kg/m^3)(9.81\ m/s^2)(17.7\ m)(\pi)(13.7\ m)^2[/tex]
F = 1.638 x 10⁸ N = 163.8 MN
A block of mass M is connected by a string and pulley to a hanging mass m.The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg. Find the acceleration of the system and tensions on the string.
The free body diagram for the block of mass M consists of four forces:
• the block's weight, Mg, pointing downward
• the normal force of the table pushing upward on the block, also with magnitude Mg
• kinetic friction with magnitude µMg = 0.2 Mg, pointing to the left
• tension of magnitude T pulling the block to the right
For the block of mass m, there are only two forces:
• its weight, mg, pulling downward
• tension T pulling upward
The m-block will pull the M-block toward the edge of the table, so we take the right direction to be positive for the M-block, and downward to be positive for the m-block.
Newton's second law gives us
T - 0.2Mg = Ma
mg - T = ma
where a is the acceleration of either block/the system. Adding these equations together eliminates T and we can solve for a :
mg - 0.2 Mg = (m + M) a
a = (m - 0.2M) / (m + M) g
a = 1.96 m/s²
Then the tension in the string is
T = m (g - a)
T = 78.4 N