Assuming constant acceleration, the requisite magnitude is [tex]a[/tex] such that
[tex]\left(50\dfrac{\rm m}{\rm s}\right)^2 - 0^2 = 2a (2000\,\mathrm m)[/tex]
Solve for [tex]a[/tex] :
[tex]a = \dfrac{\left(50\frac{\rm m}{\rm s}\right)^2}{4000\,\mathrm m} = \boxed{0.625 \dfrac{\rm m}{\mathrm s^2}}[/tex]
[ 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 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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Set the Coleman window to 29.92 in Hg and record the pressure altitude:
What is the altimeter reading in ft above sea level?
Answer:
The long pointer measures altitude in intervals of 10,000 feet (2 = 20,000 feet). The short, wide pointer measures altitude in intervals of 1,000 feet (2 = 2,000 feet). The medium, thin pointer measures altitude in intervals of 100 feet (2 = 200 feet).
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 crate with a mass of M = 60.5 kg is suspended by a rope from the midpoint of a uniform boom. The boom has a mass of m = 119 kg and a length of l = 9.64 m. The end of the boom is supported by another rope which is horizontal and is attached to the wall as shown in the figure.
1. The boom makes an angle of θ = 60.2° with the vertical wall. Calculate the tension in the vertical rope.
2. What is the tension in the horizontal rope?
The boom makes an angle of θ = 60.2° with the vertical wall and the tension in the horizontal rope is mathematically given as
[tex]T_1=730.85 \mathrm{~N}[/tex]
[tex]T_1'=\frac{1980.51 \mathrm{~N}}{}[/tex]
What is the tension in the vertical rope.?Generally, the equation for is mathematically given as
1) Tension in vertical rope,
[tex]T_{1}=74.5 \times 9.81\\T_1=730.85 \mathrm{~N}[/tex]
2) Tension in horizontal rope,
[tex]\sum{Mg} =0\\Mg\frac{l}{2} \sin \theta+T_{1} \frac{l}{2} \sin \theta &=T_{2} l \cos \theta \\[/tex]
[tex]T_{2} &=\frac{M g(l / 2) \sin \theta+T_{1}(l / 2) \sin \theta}{l \cos \theta} \\\\&=\frac{M g \tan \theta}{2}+\frac{T_{1} \tan \theta}{2} \\\\T_{2} &=\frac{142 \times 9.81 \tan 61.8}{2}+\frac{730.85 \times \tan 61.8}{2} \\\\=& 1299+681.51[/tex]
[tex]T_1=\frac{1980.51 \mathrm{~N}}{}[/tex]
In conclusion, the tension in the horizontal rope is
[tex]T_1=\frac{1980.51 \mathrm{~N}}{}[/tex]
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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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What would be the direction of the vector given a horizontal component of 4 and a vertical component of
3?
Answer:
5
Explanation:
Use the Pythagorean theorem. 3^2 + 4^2 = 25, which is the square of 5.
Brainliest, please :)
The direction of the vector would eventually depend on the horizontal component and vertical component which are 4 and 3 respectively. So, it would be calculated as 5 according to the Pythagoras theorem.
What do you mean by the Direction of the vector?In physics, the direction of the vector may be defined as the orientation of the vector, that is, the angle it makes with the X-axis. A vector is drawn by a line with an arrow on the top and a fixed point at the other end.
The direction in which the arrowhead of the vector is directed gives the direction of the vector. The direction of the vector measures the angle it makes with a horizontal line.
According to the question,
The horizontal component of the vector = 4
The vertical component of the vector = 3
The direction of the vector = square of the horizontal component + square of the vertical component.
= [tex]4^2+3^2[/tex] = 16 + 9 = 25 i.e. = [tex]5^2.[/tex]
Therefore, the direction of the vector would eventually depend on the horizontal component and vertical component which are 4 and 3 respectively. So, it would be calculated as 5 according to the Pythagoras theorem.
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which charateristic of the observant function?
Answer:
All people with the observant trait are often a steadying force that always wants to get things done. Their energy is very elegant in the sense of working on real things in real time.
Explanation:
Hope I helped
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
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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When working with two-dimensional motion, choose the direction of _________________ to be a positive direction.
Rightward and upward
What is One-dimensional motion and two-Dimensional Motion?Whenever any object moves in one direction only, the motion is known as one-dimensional motion.
Two-dimensional motion is the movement of the object in two directions simultaneously.
Example: A ball thrown at an angle is a two-dimensional motion.
In two-dimensional motion, there are two axes used, generally the x-axis and the y-axis.
Generally, the x-axis is positive towards the rightward direction and negative towards the leftward direction.
Similarly, the y- axis is positive towards the upward direction and negative toward the downward direction.
So, the rightward and upward directions are chosen as positive.
Therefore:
When working with two-dimensional motion, choose the direction of , the rightward and upward to be a positive direction.
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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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A student holding a 324Hz tuning fork approaches a wall at a speed of 6ms^(-1). The speed of sound in air is 336ms^(-1). What frequency will the student detect from waves omitted from the fork and waves coming from the wall?
The frequency the student detect from waves omitted from the fork and waves coming from the wall is 348.1 Hz.
Frequency detected by the student
The observed frequency is determined by applying Doppler effect;
f' = f₀(v + v₀)/(v)
where;
f' is the observed frequencyv is speed of soundv₀ is the source speedf₀ is the original frequencyf' = 342(6 + 336)/(336)
f' = 348.1 Hz
Thus, the frequency the student detect from waves omitted from the fork and waves coming from the wall is 348.1 Hz.
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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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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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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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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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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
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
Car A starts from rest at t = 0 and travels along a straight road with a constant acceleration of 6 ft/s2 until it reaches a speed of 80 ft/s. Afterwards it maintains this speed. Also, when t = 0, car B located 6000 ft down the road is traveling towards A at a constant speed of 60 ft/s. Determine the distance traveled by car A when they pass each other.
The distance traveled by car A when they pass each other is 3,428.56 ft.
Time when the two cars pass each other(Va)t - (Vb)t = d
where;
Va is speed of car AVb is speed of car Bt is time when they pass each otherd is the distance between them80t - (-60)t = 6000
80t + 60t = 6000
140t = 6000
t = 6000/140
t = 42.857 s
Distance traveled by car at the calculated times = vt
s = 80 x 42.857
s = 3,428.56 ft
Thus, the distance traveled by car A when they pass each other is 3,428.56 ft.
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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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An owl is carrying a mouse to the chicks in its nest. It is 4.00 m west and 12.0 m above the center of the 30 cm diameter nest and is flying east at 3.50 m/s at an angle 32° below the horizontal when it accidentally drops the mouse. Will it fall into the nest? Find out by solving for the horizontal position of the mouse (measured from the point of release) when it has fallen the 12.0 m.
m (from the point of release)
The mouse will be 4.61 meters distant horizontally. and the owl won't fall into the nest.
What is velocity?The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
The given data in the problem is
Diameter of nest,d = 30 cm
The velocity of flying, V = 3.50 m/s toward the east
The angle of flying below the horizontal, Θ = 32°
Position of owl,12.0 m above the middle of the 30 cm-diameter nests, at 4.00 m west.
Height of owl above the ground,h = 12 m
Distance from the west direction, a = 4.00 m
The vertical component of velocity,vₐ
The horizontal component of velocity,vₓ
After covering a vertical distance of 12 m, the vertical component of the mouse's velocity, v, is given by
vₐ-u²=2gh
u = asinΘ
Substitute the above value;
vₐ²- (a sin Θ )² = 2gh
Substitute the given value;
vₐ² - (4 sin 32)² = 2 ×9.81 ×12
vₐ² - (4 × 0.53)² = 235.44
vₐ² - 4.494 = 235.44
vₐ² = 235.44 + 4.494
vₐ² = 239.934
vₐ = 15.49 m/s
The time required to descend the aforementioned 12 meters;
[tex]\rm t = \frac{v_{rel}}{g}[/tex]
t = (vₐ - u )/g
t =( vₐ - a sin Θ )/g
Substitute the given value;
t =(15.49 - 4 ×sin 32°) / 9.81
t= (15.49 - 4 * 0.53) / 9.81
t =(15.49 - 2.12) / 9.81
t = 13.37 / 9.81
t= 1.36 s
Horizontal distance traveled for the given period;
x=uₓ × t
x =a cosΘ × t
Substitute the given value;
x = 4×(cos 32°)×1.36
x= 4×0.848×1.36
x= 4.61 m
Hence, the mouse will be 4.61 meters distant in the horizontal direction. and the owl won't fall into the nest,
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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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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
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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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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A shell is launched with an initial velocity at an angle of 40.0° above horizontal
from ground level. The shell needs to hit 1.5 km away. There is no appreciable air resistance,
and g = 9.80 m/s2
.
a. What should the initial velocity be?
b. What are the components of the shell’s velocity when it hits the ground? (horizontal
and vertical components)
Answer:
122.17 m/s
Explanation:
x cos 40 = horizontal velocity
1500 m / x cos 40 = time in the air = 1958.11 / x
x sin 40 = vertical velocity
find when shell vertical velocity = 0 (this is max height....1/2 way through its flight) , the time when it hits the ground will be twice this...
0 = x sin 40 - 9.8 t
t = x sin40 / 9.8 time in the air is twice this = .13118 x
Equate the two times from above to solve for x
1958.11/ x = .13118 x
x = 122.17 m/s
A uniform rod with a mass of m = 1.94 kg and a length of l = 2.10 m is attached to a horizontal surface with a hi=nge. The rod can rotate around the hi=nge without friction. (See figure.)
Initially the rod is held at rest at an angle of θ = 70.4° with respect to the horizontal surface. Then the rod is released.
1. What is the angular speed of the rod, when it lands on the horizontal surface?
2. What is the angular acceleration of the rod, just before it touches the horizontal surface?
Answer:
H = L sin θ / 2 height of center of mass
E = m g H = m g L sin θ / 2
I = m L^2 / 3 moment of inertia about end of rod
1/2 I ω^2 = m g L sin θ / 2 KE of rod when it hits surface
1/2 m L^2 / 3 ω^2 = m g L sin θ / 2
ω^2 = 3 g sin θ / L
ω = (3 g sin θ / L)^1/2 angular speed of rod striking surface
P = I α if P = torque
α = m g L/2 / (m L^2 / 3) = 3 g / (2 L) angular acceleration of rod
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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A force of 42 N changes the length of Elastic X by 1.2 m while a force of 24 N changes the length of Elastic Y by 2.1
m. Calculate the work done on both Elastic X and Elastic Y. Solve this problem by showing your work in an
organized step-by-step fashion and include proper significant digits. (3 marks)
The work done on both Elastic X and Elastic Y is 25.2 J.
Work done on elastic XThe work done on the Elastic X is calculated as follows;
W = ¹/₂fx
where;
f is the applied force to material Xx is the extension of material XW = ¹/₂(42)(1.2)
W = 25.2 J
Work done on elastic YThe work done on the Elastic Y is calculated as follows;
W = ¹/₂fx
where;
f is the applied force to material Yx is the extension of material YW = ¹/₂(24)(2.1)
W = 25.2 J
Thus, the work done on both Elastic X and Elastic Y is 25.2 J.
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