Lindsay should fly the plane in the direction [W 12.5° S] to get Hamilton.
Using Sine rule to solve this question
Sine rule => SinA/a = SinB/b = SinC/c = constant
The magnitude of wind is 50 with an angle of 60 degrees.
The magnitude of plane is 200 and the angle at which it should fly is unknown and should be θ.
One side is 50 km/hr at an angle of 60 degrees.
sin 60°/200 = sin θ / 50
50 × sin 60° = 200 × sin θ
√3/2 = 4 × sin θ
√3/8 = sin θ
sin θ = 0.2165
θ = sin⁻¹(0.2165)
θ = 12.5°
So Lindsay have to fly the plane in the direction of [W 12.5° S].
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A concave mirror
light, whereas a concave lens
light.
Answer: Answer.
1)Concave lens is transparent whereas the concave mirror is opaque. 2)concave mirrors cause reflection whereas concave lense cause refraction of light.
3)concave mirror forms both real, inverted images of various sizes and virtual, erect, and enlarged images whereas a concave lens form only virtual images.
Explanation:
. In a common test for cardiac function (the “stress test”), the patient walks on an
inclined treadmill. Estimate the power required from a 75-kg patient when the
treadmill is sloping at an angle of 12° and the velocity is 3.1 km/h.
The power required from a 75-kg patient when the treadmill is sloping at required speed is 131.58 W.
Parallel force on the patient
The parallel force on the patient is calculated as follows;
F = mg sinθ
F = (75 x 9.8) x sin(12)
F = 152.82 N
Average power requiredP = FV
where;
V is speed = 3.1 km/h = 3.1/3.6 = 0.861 m/sF = 152.82 x 0.861
F = 131.58 W
Thus, the power required from a 75-kg patient when the treadmill is sloping at required speed is 131.58 W.
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two protons are released from rest when they are 0.750 nm apart (a)what is the maximum speed they will reach ? when does this speed occur? (b)what is the maximum acceleration they will achieve?when does this acceleration occur?
Answer:
And so calculating we get the maximum speed at each proton will reach To be 1.36 Times 10 to the four m the second
A bowling ball of mass 7 kg is dropped from the top of a tall building. It safely lands on the ground 4.7 seconds later. Neglecting air friction, what is the height of the building in meters? (Give the answer without a unit and round it to the nearest whole number)
The height of the building in meters, given the data is 108 m
Data obtained from the questionMass (m) = 7 KgTime (t) = 4.7 sAcceleration due to gravity (g) = 9.8 m/s²Height (h) = ?How to determine the height of the buildingThe height of the building can be obtained as illustrated below:
h = ½gt²
h = ½ × 9.8 × 4.7²
h = 108 m
Thus, the height of the building is 108 m
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If the pitch of the sound coming out of a speaker increases, which statement is true about the sound wave?
Answer:
frequency and amplitude increases
[ b) The time of reverberation of an empty hall without and with 500 audiences is 1.5 sec and 1.4 sec respectively. Find the reverberation time with 800 audiences. Wait
The reverberation time with 800 audiences is 0.875 seconds.
Reverberation time with 800 audience
R₁V₁ = R₂V₂
where;
R₁ is the reverberation time with 400 audienceR₂ is the reverberation time with 800 audienceV₁ is initial volume V₂ is final volumeR₂ = R₁V₁/V₂
R₂ = (1.4 x 500) / 800
R₂ = 0.875 seconds
Thus, the reverberation time with 800 audiences is 0.875 seconds.
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A basketball player passes the ball to a teammate. The ball reaches a velocity of 5.5 m/s, 2 seconds after the ball leaves the player’s hands. What is the acceleration of the ball during those two seconds?
Answer:
acceleration = 2.25 m/s²
Explanation:
[tex]a = \frac{v - u}{t}[/tex]
where:
a = acceleration (?)
v = final velocity (5.5 m/s)
u = initial velocity (0 m/s, because the ball starts from rest)
t = time taken (2s).
Using the formula:
acceleration = [tex]\frac{5.5 - 0 }{2}[/tex]
= 2.75 m/s²
Set local ground level to 700 ft, and record the inches of mercury.
What is the altimeter reading in inches of mercury (from the Kollsman window)?
The altimeter reading is 29.17 from the Kollsman window is 29.17 in Hg
How to determine the altimeter reading?The given parameters are:
Ground level = 700 ft i.e. the field elevationPressure altitude = 1450 ftThe formula to calculate the pressure altitude is:
Altitude = (29.92 – Altimeter reading) * 1,000 + Ground level
Substitute the known values
1450 = (29.92 - Altimeter reading) * 1000 + 700
Evaluate the like terms
750 = (29.92 - Altimeter reading) * 1000
Divide both sides by 1000
0.75 = 29.92 - Altimeter reading
Evaluate the like terms
Altimeter reading = 29.17
Hence, the altimeter reading is 29.17 from the Kollsman window is 29.17 in Hg
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What percentage of the area under the normal distribution is shaded?
The middle ear provides impedance matching between air and fluid by?
The middle ear provides impedance matching between air and fluid by the reduction in area between the tympanic membrane and the stapes footplate.
What is the middle ear?Middle ear is part of ear that separates the outer ear from the inner ear.
The middle ear consists of the following bones:
hammer (malleus), anvil (incus), and stirrup (stapes).The middle ear provides impedance matching between air and fluid by
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While buying a hot plate you notice the resistance of the hot
plate is 22.02. When connected to a 120. V source, how
much current will the hot plate carry? How much power does
it consume?
Answer:
Current = 5.45amps
Power = 654 watts
Explanation:
E =120V
I =?
R = 22.02 Ohms
I= E/R
I= 120/22.02
I = 5.45AmPs
P = ? P= E x R
E = 120V P= 120x5.45
I = 5.45 AmPs P= 654W
This is a scenario common with many physics students: you push a heavy car by hand. The car, in turn, pushes back with an opposite but equal force on you. Doesn't this mean that the forces cancel each other, making acceleration impossible? Resolve the misunderstanding underlying this question.
Considering that the moment you provide the action force, the car will certainly apply the response force. You will consequently come into an opposing force.
What is Newton's third law of motion?
According to Newton's third law of motion, every action has an equal and opposite reaction. As well as the action and reaction are always acted in pairs.
You manually move a big automobile. You are then pushed back by the automobile with an opposing but equivalent force.
Because the automobile will undoubtedly apply the response force when you apply the action force. As a result, you will encounter an opposing force.
Hence, when you push a heavy car by hand. The car, in turn, pushes back due to the action-reaction analogy.
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A net force of 16 N [40° W of N] is caused by two applied forces acting on the same
object. These two forces are:
A) 10 N [W] and 12 N [N]
B) 12 N [W] and 10 N [N]
C) 14 N [W] and 12 N [N]
D) 12 N [W] and 14 N [N]
E) 10 N [W] and 14 N [N]
The two forces are 12N and 10N (option B)
What is the net force?The net force is the force that has the same effect in magnitude and direction as the two forces acting together.
This problem can be solved using a right angled triangle;
Let the forces be P and Q
sin 40 = P/16
P= 16 sin 40
P = 10 N
Cos 40 = Q/16
Q = 16 cos 40
Q = 12 N
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A snowboarder on a slope starts from rest and reaches a speed of 1.6 m/s after 8.4 s.
(a)
What is the magnitude (in m/s2) of the snowboarder's average acceleration?
m/s2
(b)
How far (in m) does the snowboarder travel in this time?
m
The average acceleration of snowboarder's is 0.19 m/s² and distance travel by snowboarder is 6.70 metres.
We have given that snowboarder start moving from rest means
Initial speed (u) = 0 m/s
Final speed (v) = 1.6 m/s
time taken (t) = 8.4 s
Applying first equation of motion
v = u + at
1.6 = 0 + a×8.4
1.6 = 8.4a
a = 1.6/8.4 m/s²
a = 0.19 m/s²
So average acceleration is 0.19 m/s².
Now Applying second equation of motion
s = ut + 1/2at²
s = (0) × (8.4) + (1/2)(0.19)(8.4)² metre
s = (1/2)(0.19)(8.4)² metre
s = 6.70 metre
So that distance covered by snowboarder is 6.70 metre.
So we find out the average acceleration by using first equation of motion and the average acceleration came out to be 0.19 m/s² and to find out the distance covered by snowboarder we use second equation of motion and the distance came out to be 6.70 metre.
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You are trying to drag an object 5.0 m along a
straight path. You are pulling with a force of 40 N
along a rope that is inclined 30° to the horizontal.
How much work do you do?
The work done in dragging an object along a straight path is 173.2 J.
What is work done?Work done is equal to product of force applied and distance moved.
Given You are trying to drag an object 5.0 m along a straight path. You are pulling with a force of 40 N along a rope that is inclined 30° to the horizontal.
Work = Force x Distance x cos(angle)
W= 40 x 5 x cos 30°
W = 173.2 Joules
Thus, the work done is 173.2 Joules
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After washing lettuce, you can dry the water from it using a salad spinner. You put the lettuce in the bowl, and then crank the handle, spinning and drying the lettuce.
From the Earth frame of reference, what causes the water on the lettuce to come off the lettuce and go to the walls of the bowl?
The centrifugal force
The centripetal force
Newton's first law of motion
The force of gravity
A. The force that causes the water on the lettuce to come off the lettuce and go to the walls of the bowl is centrifugal force.
What is centrifugal force?
Centrifugal force is an inertial force that appears to act on all objects when viewed in a rotating frame of reference.
This force is directed away from the center around which the body is moving.
What is centripetal force?This is force that acts on a body moving in a circular path and is directed towards the center around which the body is moving.
While centripetal force is directed towards to the center, the centrifugal force is directed away.
Thus, the force that causes the water on the lettuce to come off the lettuce and go to the walls of the bowl is centrifugal force.
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Centripetal force causes the water on the lettuce to come off the lettuce and go to the walls of the bowl.
What is centripetal force?Centripetal force is the force that is necessary to keep an object moving in a curved path and that is directed inward toward the center of rotation. This force can move the body in the circular direction.
So we can conclude that Centripetal force causes the water on the lettuce to come off the lettuce and go to the walls of the bowl.
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A small ball is tied to a string and set rotating with negligible friction in a vertical circle. If the ball moves over the top of the circle at its slowest possible speed (so that the tension in the string is negligible), what is the tension in the string at the bottom of the circle, assuming there is no additional energy added to the ball during rotation?
An explanation would be very much appreciated as there are a few things I am confused about
- why is tension negligible at the top? Isn't there still a tension force pulling inward? I thought there would be two forces acting on the ball at the top of the circle, Fg and [tex]F_T[/tex] ?
- is finding a numerical value for this answer even possible? we don't have the mass of the ball
Since the tension in the string over the top of the circle is negligible, the tension in the string at the bottom of the circle will be zero.
What is Circular Motion ?Circular motion is the same as rotational motion. That is, motion moving in a circle.
Given that a small ball is tied to a string and set rotating with negligible friction in a vertical circle. If the ball moves over the top of the circle at its slowest possible speed (so that the tension in the string is negligible)
Since the tension in the string over the top of the circle is negligible, the tension in the string at the bottom of the circle, assuming there is no additional energy added to the ball during rotation will be zero.
The tension is negligible at the top in order to allow the ball rotate. And since the tension is negligible, there no tension force pulling inward
Therefore, the numerical value of the answer is Zero.
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2. A 1.54 kΩ resistor is connected to an AC voltage source with an rms voltage of 240 V.
(a)What is the maximum potential difference across the resistor (in V)?
(b)What is the maximum current through the resistor (in A)?
(c)What is the rms current through the resistor (in A)?
(d)What is the average power dissipated by the resistor (in W)?
(a) The maximum potential difference across the resistor is 339.41 V.
(b) The maximum current through the resistor is 0.23 A.
(c) The rms current through the resistor is 0.16 A.
(d) The average power dissipated by the resistor is 38.4 W.
Maximum potential differenceVrms = 0.7071V₀
where;
V₀ is peak voltageV₀ = Vrms/0.7071
V₀ = 240/0.7071
V₀ = 339.41 V
rms current through the resistorI(rms) = V(rms)/R
I(rms) = (240)/(1,540)
I(rms) = 0.16 A
maximum current through the resistorI₀ = I(rms)/0.7071
I₀ = (0.16)/0.7071
I₀ = 0.23 A
Average power dissipated by the resistorP = I(rms) x V(rms)
P = 0.16 x 240
P = 38.4 W
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perpetual motion machines have fascinated people especially inventors for hundreds of years before the laws of thermodynamics became known. Ideally once energy is added to start a perpetual motion machine in motion, the machine would continue to operate indefinitely. Using a law of thermodynamics, explain why such a machine cannot be produced
HELP HELP HELP
Given the fact that energy conversion is not entirely efficient, it is impossible to produce a perpetual motion machine.
What is a perpetual motion machine?The perpetual motion machine in one that is able to work continuously without stopping. This would mean that the efficiency of this machine must that the machine is 100% efficient which violates the second law of thermodynamics.
Thus, given the fact that energy conversion is not entirely efficient and energy looses cause machines not function effectively, it is impossible to produce a perpetual motion machine.
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4
Select the correct answer from each DROP-DOWN MENU.
A graph plots time in seconds versus position (m). A line rises from a point A (0, 0) to a point B (4, unit 6) extends linearly through a point C (11, unit 6), and rises to a point D (12, unit 7 point 5).
The graph describes the motion of an object.
The object moves with
from A to B. It
from B to C. It moves with
from C to D.
Reset Next
(a) The object moves with uniform velocity from A to B.
(b) The object moves with constant velocity from B to C.
(c) The object moves with increasing velocity from C to D.
Velocity of the object from point A to B
V(A to B) = (6 - 0)/(4 - 0) = 1.5 m/s
Velocity of the object from point B to CV(B to C) = (6 - 6)/(11 - 4) = 0 m/s
Velocity of the object from point C to DV(C to D) = (7 - 6)/(12 - 11) = 1 m/s
final velocity = 1 + 1.5 m/s = 2.5 m/s
Thus, we can conclude the following;
The object moves with uniform velocity from A to B.
The object moves with constant velocity from B to C.
The object moves with increasing velocity from C to D.
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Answer:The object moves with
positive accelaration
from A to B. It
speeds up
from B to C. It moves with
constant velocity
from C to D.
Explanation:I took the test
Why are sedimentation pond useful for reducing the affects of soil erosion?
- They provide a water source that can be used for irrigation of crops?
-They help transport soil sediments to rivers, ponds and lake.
-They provide a place to raise fish that can be use to replace fish harmed by sedimentation
-They provide a place to collect soil sediments so that it doesn't reach ponds, lakes and rivers
Answer: D (They provide a place to collect soil sediments so that it doesn't reach ponds, lakes and rivers)
Explanation: I just took the test on accelerate. :)
Sedimentation ponds are useful for reducing the effects of soil erosion because : ( D ) They provide a place to collect soil sediments so that it doesn't reach ponds, lakes and rivers
What are sedimentation ponds ?Sedimentation ponds are ponds that hold sediments and suspended solids discharged by runoff waters thereby preventing the soil sediments from reaching water bodies such as ponds, lakes and rivers.
The sediment ponds hold the water while allowing the debris and solids contained in the water to settle down.
Hence we can conclude that sedimentation ponds provide a place to collect soil sediments so that it doesn't reach ponds, lakes and rivers
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Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 56.3 N. Calculate the magnitude of the tension in the string marked A.
The tension in the string marked A is determined as 331.35 N.
Angle made by A with respect to verticaltanθ = Δy/Δx
tanθ = (6 - 1)/(7 - 1) = 0.8333
θ = arc tan(0.8333) = 39.81⁰
with respect to horizontal = 90 - 39.81 = 50.19⁰
Angle made by B with respect to horizontaltanθ = Δy/Δx
tanθ = (9 - 6)/(7 - 5) = 1.5
θ = arc tan(1.5) = 56.31 ⁰
Angle made by C with respect to horizontaltanθ = Δy/Δx
tanθ = (6 - 4)/(14 - 7) = 0.2857
θ = arc tan(0.2857) = 15.95 ⁰
Bcosθ + Aosθ = Ccosθ
Bcos(56.31) + A[cos(50.19)] = 56.3cos(15.95)
0.55B + 0.64A = 54.13 ----- (1)
Bsinθ + Asinθ = Csinθ
Bsin(56.31) + A[sin(50.19)] = 56.3sin(15.95)
0.832B + 0.77A = 15.47---- (2)
Solve (1) and (2)
0.2A = 66.27
A = 66.27/0.2
A = 331.35 N
Thus, the tension in the string marked A is determined as 331.35 N.
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Which pair of quantities includes one quantity that increases as the other decreases during simple harmonic motion? acceleration and displacement acceleration and net force velocity and displacement displacement and net force
Velocity and displacement are the two quantities that increases as the other decreases during simple harmonic motion.
In a basic harmonic motion, how are velocity and displacement related?If a body's velocity at any given moment is inversely proportional to its displacement from the mean position, its motion is said to be simple harmonic.
Angular harmonic frequency and time t are functions that can be used to explain simple harmonic motion:
Displacement from equilibrium has amplitude A:
x = Asin(ωt)
and velocity has amplitude ωA
v = dx/dt = ωAcos(ωt)
At ωt = 0, x = 0 while v = Aω (max).
At ωt = π/2, x = A (max) while v = 0.
As a result, v reaches it's maximum /2 (90°) before x does. Therefore, displacement follows velocity at a 90° angle.
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your manger wants everyine to make setting up seasonal aisle display their top priority you noticed one if associate benn focusing on other task mostly likly to do or least likly to do
The manager is both most and least likely to lead seminars and provide everyone skills.
What is seasonal aisle?When a store has a “seasonal aisle,” it suggests it has extra money to spend on items that are in season. Festivals, special days, seasons, and other things are examples of seasonal components.
An operational strategy focused on creating profit. Seasonal goods are required by a supply chain. A manager wants everyone to prioritize setting up the seasonal aisle displays because of this.
Hence, the significance of the seasonal aisle is aforementioned.
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A bungee cord is essentially a very long rubber band that can stretch up to four times its unstretched length. However, its spring constant varies over its stretch. Take the length of the cord to be along the x-direction and define the stretch x as the length of the cord l minus its un-stretched length [tex]l_0[/tex]; that is, [tex]x=l-l_0[/tex] (see below). Suppose a particular bungee cord has a spring constant, for 0 ≤ x ≤ 4.88m , of [tex]k_1=204 N/m[/tex] and for x ≥ 4.88m , of [tex]k_2=111N/m[/tex]. (Recall that the spring constant is the slope of the force F(x) versus its stretch x.) (a) What is the tension in the cord when the stretch is 16.7 m (the maximum desired for a given jump)? (b) How much work must be done against the elastic force of the bungee cord to stretch it 16.7 m?
please explain if possible!
A) When the cord is stretched 16.7 m, the tension in the cord is 18537 N.
b) The work that must be done against the elastic force of the bungee cord to stretch it 16.7 m is 15.48 kJ.
What is the tension in the Bungee cord when stretched 16.7 m?
The tension or force in an elastic string is given by the formula:
Force = spring constant × extensiona) The force or tension in the cord = 111 N/m × 16.7
Tension = 18537 N
b) Work done in a string = ke²/2
where k = 111 N/m, e = 16.7
Work done = 111 × 16.7²/2 = 15,478.395 J
Work done = 15.48 kJ
In conclusion, the force constant of the cord is required to calculate the tension and work done against the elastic force of the cord.
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A 12kg object is pushed with a horizontal force of 6N due east across a horizontal table. If the force of friction between the two surfaces is 2N, what is the acceleration of the object?
The acceleration of the body is obtained as 0.33 m/s^2.
What is the acceleration of the body?Now we know that the object is acted upon by a system of forces. The net force is the effective force that acts on the body.
Hence;
ma = F -Ff
The applied force is F while the frictional force is Ff. The net force is represented as ma
Thus;
12 * a = 6 - 2
a = 4/12
a = 0.33 m/s^2
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A woman exerts a horizontal force of 145 N on a crate with a mass of 33.2 kg.
(a)
If the crate doesn't move, what's the magnitude of the static friction force (in N)?
N
(b)
What is the minimum possible value of the coefficient of static friction between the crate and the floor? (Assume the crate remains stationary.)
(a) The magnitude of the static friction force is 145 N
(b) The minimum possible value of static friction is 0.44.
What is the friction force?It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
it is defined as the product of the coefficient of friction and normal reaction.
Given data;
The horizontal force exerted, F =145 N
Mass of crate,m =33.2 kg.
The velocity of the crate,v =0 m/sedc
The magnitude of the static friction force, [tex]\rm f_s[/tex] =? N
The minimum possible value of the coefficient of static friction between the crate and the floor is,
Normal reaction,N = mg
a)
The forces acting on the crate are balanced since it is at rest; as a result, the horizontal force is equal to the frictional force, f:
F = [tex]\rm f_s[/tex] = 145 N
(b)
The following equations provide the greatest allowable force of friction between the floor and the crate:
[tex]\rm f_{min}= \mu_s N \\\\ \rm f_{min}= \mu_s mg \\\\[/tex]
The force applied to the box must be less than or equal to the maximum force of friction in order for it to stay at rest.
[tex]\rm f \leq f_{min} \\\\ f \leq \mu_s mg \\\\ 145 \leq \mu_s (33.2 \times 9.81 ) \\\\ \mu_s \geq 0.44[/tex]
Hence, the magnitude of the static friction force and minimum possible value of the coefficient of static friction between the crate and the floor will be 145 N and 0.44.
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A section of a high speed test track is circular with a radius of curvature R = 1860 m.
At what angle of θ should the track be inclined so that a car traveling at 61.0 m/s (136 mph) would keep moving in a circle if there is oil on that section of the track, i.e., it would not slip sideways even with zero friction on that section. (Hint: The car's vertical acceleration is zero.)
The banking angle of curved road to allow to move without slipping is 11.53 ⁰.
Banking angle of curved roadThe banking angle of curved road is calculated as follows;
V(max) = √(rg tanθ)
where;
r is radius of the curveg is acceleration due to gravityV² = rg tanθ
tanθ = V²/rg
tanθ = (61²)/(1860 x 9.8)
tanθ = 0.204
θ = arc tan(0.204)
θ = 11.53 ⁰
Thus, the banking angle of curved road to allow to move without slipping is 11.53 ⁰.
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A yoyo with a mass of m = 150 g is released from rest as shown in the figure.
The inner radius of the yoyo is r = 2.14 cm, and the outer radius is R = 4.00 cm, and the moment of inertia about the axis perpendicular to the plane of the yoyo and passing through the center of mass is ICM = 1.01×10-4 kgm2.
1. Determine the linear acceleration of the yoyo.
2. Determine the angular acceleration of the yoyo.
3. What is the weight of the yoyo?
4. What is the tension in the rope?
5. If a 1.27 m long section of the rope unwinds from the yoyo, then what will be the angular speed of the yoyo?
(1) The linear acceleration of the yoyo is 3.21 m/s².
(2) The angular acceleration of the yoyo is 80.25 rad/s²
(3) The weight of the yoyo is 1.47 N
(4) The tension in the rope is 1.47 N.
(5) The angular speed of the yoyo is 71.385 rad/s.
Linear acceleration of the yoyoThe linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.
∑τ = Iα
rT - Rf = Iα
where;
I is moment of inertiaα is angular accelerationT is tension in the roper is inner radiusR is outer radiusf is frictional forcerT - Rf = Iα ----- (1)
T - f = Ma -------- (2)
a = Rα
where;
a is the linear acceleration of the yoyoTorque equation for frictional force;
[tex]f = (\frac{r}{R} T) - (\frac{I}{R^2} )a[/tex]
solve (1) and (2)
[tex]a = \frac{TR(R - r)}{I + MR^2}[/tex]
since the yoyo is pulled in vertical direction, T = mg [tex]a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2[/tex]
Angular acceleration of the yoyoα = a/R
α = 3.21/0.04
α = 80.25 rad/s²
Weight of the yoyoW = mg
W = 0.15 x 9.8 = 1.47 N
Tension in the ropeT = mg = 1.47 N
Angular speed of the yoyov² = u² + 2as
v² = 0 + 2(3.21)(1.27)
v² = 8.1534
v = √8.1534
v = 2.855 m/s
ω = v/R
ω = 2.855/0.04
ω = 71.385 rad/s
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Two blocks are connected by a massless rope over a massless, frictionless pulley, as shown in the figure. The mass of block 2 is 2=11.3 kg , and the coefficient of kinetic friction between block 2 and the incline is =0.200 . The angle of the incline is 31.5° . If block 2 is moving up the incline at constant speed, what is the mass 1 of block 1?
The mass of block 1 is 1.30176 kg.
What is tension in physics?A tension force in physics is a force developed in a rope, string, or cable when stretched under an applied force.
Tension is acted along the length of the rope/cable in a direction that is opposite to the force applied on it.
According to the question,
Mass of block 2= 11.3 kg
Kinetic friction μ=0.200
The component of the weight in the direction of the incline
W = m₂gsinθ
T = [tex]F_{f} - W[/tex]........(1)
When the force of friction
[tex]F_{f}[/tex] = μ[tex]_k[/tex]N
[tex]F_{f}[/tex] = μ[tex]_k[/tex](m₂gcosθ)
Here , Put the value in equation 1
T = m₂gcos 31.5° - μ[tex]_k[/tex]m₂gcos 31.5°
We know that ,
T - m₁g = 0
T = m₁g
So,
m₁g = m₂g (Sin 31.5° - 0.200 x cos 31.5°)
m₁ = 11.3 x (0.0840 - 0.2 x 0.996)
m₁ = 11.3 x (0.1152)
m₁ = 1.30176 kg
Hence, the mass of block 1 is 1.30176 kg.
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