Answer:
The string wasn't taut when he released the bob, causing the bob to move erratically.
The student only timed one cycle, introducing a significant timing error.
The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case.
Explanation:
The period of a simple pendulum is given by the formula
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]
where
T is the period of oscillation
L is the length of the pendulum
g is the acceleration due to gravity
Therefore, it is possible to measure the value of [tex]g[/tex] in an experiment, by taking a pendulum, measuring its length L, and measuring its period of oscillation T. Re-arranging the equation above, we get the value of g as:
[tex]g=(\frac{2\pi}{T})^2 L[/tex]
Here the value of g measured in the experiment is [tex]8 m/s^2[/tex] instead of [tex]9.8 m/s^2[/tex]. Let's now analyze the different options:
The length of the pendulum string was too long, so the equation for the period of a pendulum was no longer valid in this case. --> FALSE. There is no constraint on the length of the pendulum.
The string wasn't taut when he released the bob, causing the bob to move erratically. --> TRUE. This is possible, as if the string is not taut, the pendulum would not start immediately its oscillation, so the period would be larger causing a smaller value measured for g.
The student only timed one cycle, introducing a significant timing error. --> TRUE. This is also impossible: in fact, we can get a more accurate measurement of the period if we measure several oscillations (let's say 10), and then we divide the total time by 10.
The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case. --> TRUE. The formula written above for the period of the pendulum is valid only for small angles.
The mass of the bob was too large, so the equation for the period of a pendulum was no longer valid in this case. --> FALSE. The equation that gives the period of the pendulum does not depend on the mass.
Instead of releasing the bob from rest, the student threw the bob downward. --> FALSE. In fact, this force would have been applied only at the very first moment, but then later the only force acting on the pendulum is the force of gravity, so the formula of the period would still be valid.
Answer:
The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case.
The string wasn't taut when he released the bob, causing the bob to move erratically.
The student only timed one cycle, introducing a significant timing error.
Instead of releasing the bob from rest, the student threw the bob downward.
Explanation:
The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case. your equation only works for angles > 20 degrees
The string wasn't taut when he released the bob, causing the bob to move erratically. this could lead to errors in either direction
The student only timed one cycle, introducing a significant timing error. unforseen events like wind could heavily influence one cycle.
Instead of releasing the bob from rest, the student threw the bob downward. this would artificially decrease your period (T) and therefor throw the other variables in your equation off. since L is a constant g would be forced to change.
A rotating viscometer consists of two concentric cylinders –an inner cylinder of radius Rirotating at angular velocity (rotation rate) ωi, and a stationary outer cylinder of inside radius Ro. In the tiny gap between the two cylinders is the fluid of viscosity μ. The length of the cylinders (into the page) is L. L is large such that end effects are negligible (we can treat this as a two-dimensional problem). Torque (T) is required to rotate the inner cylinder at a constant speed. (a) Showing all of your work and algebra, generate an approximate expression for T as a function of the other variables.
(b) Explain why your solution is only an approximation. In particular, do you expect the velocity prole in the gap to remain linear as the gap becomes larger and larger (i.e., if the outer radiusR0 were to increase, all else staying the same)?
Answer:
b) the result we got can be termed approximation because we are neglecting the shear stress acting on the two ends of the cylinder. Here we have considered only the share stress acting on the curved surface area only.
Explanation:
check attachment for solution to A
The torque required to rotate the inner cylinder of a rotating viscometer with a fluid of viscosity μ can be calculated using fluid mechanics. However, this solution is approximated and based on a linear velocity profile in the gap, which can change with a significant increase in the gap.
Explanation:In a rotating viscometer with a fluid of viscosity μ between two concentric cylinders, the torque required to rotate the inner cylinder at a constant speed can be computed using fluid mechanics principles. The shear stress on the fluid due to the rotating cylinder is the product of the viscosity and the velocity gradient across the gap, given by (ωi*ri)/ (ro - ri). This shear stress relates to the force acting on a unit area of the fluid as F/A, and the torque T is the product of the force and the radius of the inner cylinder. Following the equations, we'll obtain T = 2π * L * μ * ωi * (((ro^2 - ri^2) / ln (ro/ri)).
This solution is an approximation as it's based on the assumption that the velocity profile in the gap between the cylinders is linear, which holds true when the gap is thin. If the gap increases disproportionately (i.e., the outer radius R0 were to increase), the linear approximation of the velocity profile may no longer hold. In this scenario, edge effects and non-linear velocity profiles have to be considered, potentially modifying the expected torque value.
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A smooth circular hoop with a radius of 0.600 m is placed flat on the floor. A 0.350-kg particle slides around the inside edge of the hoop. The particle is given an initial speed of 10.00 m/s. After one revolution, its speed has dropped to 5.50 m/s because of friction with the floor.
(a) Find the energy transformed from mechanical to internal in the particle—hoop—floor system as a result of friction in one revolution.
(b) What is the total number of revolutions the particle makes before stopping? Assume the friction force remains constant during the entire motion.
Answer:
a. The energy transformed from mechanical to internal in the particle—hoop is 12.21 Joules
b. The total number of revolutions the particle makes before stopping is 1.43 revolutions
Explanation:
a.
Given
m = mass of particle = 0.350-kg
u = initial speed of 10.00 m/s
v = final speed = 5.50 m/s
r = radius = 0.600 m
We assume that the floor is horizontal;
This means that F = mg.
We also assume the rotational kinetic energy to be negligable.
Having listed the assumptions, we proceed as follows;
Let ∆E represent The energy transformed from mechanical to internal in the particle hoop.
This is given by
∆E = KE1 - KE2
Where KE1 = ½mu²
KE2 = ½mv²
So, ∆E = KE1 - KE2 becomes
∆E = ½mu² - ½mv²
∆E = ½m(u² - v²)
∆E = ½ * 0.350 * (10² - 5.5²)
∆E = 12.20625
∆E = 12.21J (Approximated)
Hence, the energy transformed from mechanical to internal in the particle—hoop is 12.21 Joules
b.
Let n = number of revolutions
The relationship between n and the energy is
1/n = (KE1 - KE2)/KE1
Make n the subject of formula
n = KE1 / (KE1 - KE2)
n = ½mu² / (½mu² - ½mv²) --- Simplify
n = ½mu² / (½m(u² - v²)) ---- Divide through by ½m
n = u² / (u² - v²)
n = 10² / (10² - 5.5²)
n = 1.433691756272401
n = 1.43 rev
Hence, the total number of revolutions the particle makes before stopping is 1.43 revolutions
Final answer:
The energy lost to friction in one revolution is 11.9875 J, and the particle makes a total of 2 revolutions before coming to a stop.
Explanation:
Energy Loss Due to Friction
The energy transformed from mechanical to internal in the particle—hoop—floor system as a result of friction in one revolution (a) is determined by the change in kinetic energy of the particle. The initial kinetic energy is given by ½ mv², where ‘m’ is the mass of the particle and ‘v’ is the initial velocity. The final kinetic energy is likewise given by ½ mv² using the final velocity. The difference in these energies gives the energy lost to friction.
Kinetic Energy Initial = ⅓ (0.350 kg)(10.00 m/s)² = 17.5 J
Kinetic Energy Final = ⅓ (0.350 kg)(5.50 m/s)² = 5.5125 J
Energy Transformed = Kinetic Energy Initial - Kinetic Energy Final = 17.5 J - 5.5125 J = 11.9875 J
Total Number of Revolutions
For part (b), to find the total number of revolutions the particle makes before stopping, assuming a constant frictional force, we can use a deceleration approach or an energy approach. The work done against friction for each subsequent revolution will be the same; thus each revolution results in the same amount of energy loss until the particle comes to a stop. By dividing the total initial kinetic energy by the energy lost per revolution, we find the number of revolutions.
Revolutions = Total Initial Kinetic Energy / Energy Transform per Revolution = 17.5 J / 11.9875 J ≈ 1.46 revolutions
Since it cannot complete a partial revolution after losing all kinetic energy, we round down to get 1 full revolution. Therefore, the particle makes a total of 1 + 1 = 2 revolutions before stopping.
g A 48.7 g clay, moving at 9.8 m/s, crashes into 1.1 kg brick that is sitting at rest on a frictionless surface. After the collision, the clay is stuck to the brick. This is an inelastic collision, where some kinetic energy (for the system of both masses) is lost. How much kinetic energy is lost (so your answer will be negative!)
Answer:
-2.2394181 J amount of kinetic energy is lost.
Explanation:
Let Vf be the final speed
(48.7) (9.8) = (48.7 + 1100) Vf
477.26 = 1148.7 Vf
=> Vf = 0.4155 m/s
KE initiali = 1/2 (0.0487) (9.8)²
KE initial = 1/2 (0.0487) (96.04)
KE initial = 1/2 (4.677148)
KE initial = 2.338574 J
KE final = 1/2 (1.1 + 0.0487) (0.4155)²
KE final = 1/2 (1.1487) (0.17264025)
KE final = 1/2 (0.198311855175)
KE final = 0.0991559 J
KE (lost) = KEf - KEi
= 0.0991559 J - 2.338574 J
= -2.2394181 J
An object is thrown with a force of 30 Newtons and ends up with an acceleration of 3 m/s ^ 2 due to that throw. What is the mass of that ball?
Answer:
F=30N
a= 3m/s^2
m=?
F=ma
30=m(3)
30/3=m
m=10kg
The mass of the ball is 10kg
Final answer:
The mass of an object thrown with a force of 30 Newtons resulting in an acceleration of 3 m/s² is calculated using the formula m = F/a, resulting in a mass of 10 kg.
Explanation:
The question is regarding the calculation of the mass of an object when a known force is applied to it and an acceleration is observed as a result. According to Newton's Second Law of Motion, which is typically expressed as F = ma, where F is the force applied to an object in newtons (N), m is the mass of the object in kilograms (kg), and a is the acceleration of the object in meters per second squared (m/s²). To find the mass of the object, we rearrange this formula to get m = F / a. Given the applied force of 30 N and the acceleration of 3 m/s² due to that force, we can calculate the mass of the object.
Mass Calculation:Force (F) = 30 NAcceleration (a) = 3 m/s²Mass (m) = F / a = 30 N / 3 m/s² = 10 kg
Therefore, the mass of the object that was thrown with a force of 30 Newtons, resulting in an acceleration of 3 m/s², is 10 kg.
An object with mass 3.7 kg is executing simple harmonic motion, attached to a spring with spring constant 320 N/m . When the object is 0.017 m from its equilibrium position, it is moving with a speed of 0.60 m/s . Part A Calculate the amplitude of the motion. Express your answer to two significant figures and include the appropriate units. A = nothing nothing
Answer:
The amplitude of the motion of the spring is 1 m.
Explanation:
From the conservation of energy;
[tex](PE_{spring})_{max}= PE+KE[/tex]
which is
[tex]\frac{1}{2} kx^2_{max}=\frac{1}{2} kx^2+\frac{1}{2} mv^2[/tex]
Make [tex]x_{max}[/tex] the subject of formula
[tex]x^2_{max}= \frac{kx^2\times mv^2}{k}[/tex]
[tex]x_{max}= \sqrt{\frac{kx^2\times mv^2}{k} }[/tex]
Substitute 320 N/m for k, 0.017 m for x, 3.7 kg for m, and 0.60 m/s for v.
[tex]x_{max}= \sqrt{\frac{(320)(0.017)^2+ (3.7)(0.60)^2}{320} } \\\\=1m[/tex]
Thus, The amplitude of the motion of the spring is 1 m.
Answer: 0.0667m
Explanation:
given
Mass of the object, m = 3.7kg
Force constant of the spring, k = 320 N/m
Speed of the object, v = 0.6 m/s
Distance from equilibrium position, x = 0.017 m
Using the law of conservation of energy. The total energy conserved is
= 1/2kx² + 1/2mv²
= 1/2 * 320 * 0.017² + 1/2 * 3.7 * 0.6²
= 0.046 + 0.666
= 0.712 J
When v = 0, the maximum deflection is Xmas. This Xmas, is also the amplitude. Such that,
Energy = 1/2kx²
0.712 = 1/2 * 320 * x²
1.424 = 320 x²
x² = 1.424 / 320
x² = 0.00445
x = 0.0667
x = 6.67*10^-2 m
Thus, the amplitude is 0.0667 m
According to Newton’s second law of motion what is force equals to
Answer:force equals to rate of change of momentum
Explanation:
F=force
t=time
m=mass
v=final velocity
u=initial velocity
(mv-mu)/t=rate of change of momentum
Force=rate of change of momentum
F=(mv-mu)/t
Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the netforce, and inversely proportional to the mass of the object.
A force of 40 N is applied in a direction perpendicular to the end of a 9 m long bar that pivots about its other end. Find the torque that this force produces about the pivot point. magnitude
Answer:
360 Nm
Explanation:
Torque: This is the force that tend to cause a body to rotate or twist. The S.I unit of torque is Newton- meter (Nm).
From the question,
The expression of torque is given as
τ = F×d......................... Equation 1
Where, τ = Torque, F = force, d = distance of the bar perpendicular to the force.
Given: F = 40 N, d = 9 m
Substitute into equation 1
τ = 40(9)
τ = 360 Nm.
Answer:
360Nm
Explanation:
Torque is defined as the rotational effect of a force. The magnitude of a torque τ, is given by;
τ = r F sin θ
Where;
r = distance from the pivot point to the point where the force is applied
F = magnitude of the force applied
θ = the angle between the force and the vector directed from the point of application to the pivot point.
From the question;
r = 9m
F = 40N
θ = 90° (since the force is applied perpendicular to the end of the bar)
Substitute these values into equation (i) as follows;
τ = 9 x 40 sin 90°
τ = 360Nm
Therefore the torque is 360Nm
An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavier fragment slides 6.80 m before stopping. How far does the lighter fragment slide
Answer:
Explanation:
Given that,
One fragment is 7 times heavier than the other
Let one fragment mass be M
Let this has a velocity v
And the other 7M
And this a velocity V
Initially the fragment is at rest u = 0
Applying conservation of momentum
Momentum is given as p=mv
Initial momentum = final momentum
Po = Pf
(M+7M) × 0 = 7M •V − Mv
0 = 7M•V - Mv
Divide both sides by M
0 = 7V -v
v = 7V
Since friction decelerates the masses to zero speed, we can calculate the NET work on the individual blocks and relate this quantity to the change in kinetic energy of each block
The workdone by the 7M mass is
Distance moved by 7M mass is 6.8m, Then, d =6.8m
W = fr × d
Where fr = µkN
When N=W =mg, where m=7M
N= 7Mg
fr = −µk × 7mg
Then, W(7m) = −7µk•Mg×d
W(7m) = −7µk•Mg×6.8
W(7m) = −47.6 µk•Mg
Then, same procedure,
Let distance move by the small mass be m
Work done by M mass
W(m) = −µk•Mg×d'
Since it is a wordone by friction, that is why we have a negative sign.
Using conservation of energy
Work done by 7M mass is equal to work done by M mass
W(7m) = W(m)
−47.6 µk•Mg = −µk•Mg×d
Then, M, g and µk cancels out
We are left with
-46.7 = -d
Then, d = 46.7m
Answer:
Distance the lighter fragment slides before stopping= 333.2 m
Explanation:
Gravitational acceleration = g
Mass of the lighter fragment = m
Mass of the heavier fragment = 7m
Velocity of the lighter fragment after the explosion = V1
Velocity of the heavier fragment after the explosion = V2
The object is at rest before the explosion hence the total momentum of the system is zero.
mV1 + 7mV2 = 0
V1 = -7V2
Coefficient of friction = [tex]\mu[/tex]
Friction force on the lighter fragment = f1 =[tex]\mu mg[/tex]
Friction force on the heavier fragment = f2 =[tex]7\mu mg[/tex]
Deceleration of the lighter fragment due to friction = a1
ma1 = -f1
ma1 = [tex]-\mu mg[/tex]
a1 =[tex]-\mu g[/tex]
Deceleration of the heavier fragment due to friction = a2
7ma2 = -f2
7ma2 = -7[tex]\mu mg[/tex]
a2 = [tex]-\mu g[/tex]
Final velocity of the lighter fragment = V3 = 0 m/s
Final velocity of the heavier fragment = V4 = 0 m/s
Distance traveled by the lighter fragment before coming to rest = d1
Distance traveled by the heavier fragment before coming to rest= d2 =6.8 m
V4² = V2² + 2a₂d₂
[tex],[/tex]
[tex](0)^2 = V_2^2 + (2)(-\mu g)d_2[/tex]
[tex]d_2 = \frac{V_2^2}{2 \mu g}[/tex]
V₃² = V₁² + 2a₁d₁
[tex](0) = (-7V_2)^2 + 2(-\mu g)d_1[/tex]
[tex]d_1 = \frac{49V_2^2}{2 \mu g}[/tex]
d₁ = 49d₂
d₁ = (49)(6.8)
d₁ = 333.2 m
Distance the lighter fragment slides before stopping = 333.2 m
Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of north. The resultant vector A + B is given by:
(a) 9.8 m at an angle of 26° north of east,
(b) 3.3 m at an angle of 26° north of east,
(c) 7.2 m at an angle of 26° east of north,
(d) 3.3 m at an angle of 64° east of north or
(e) 9.8 m at an angle of 64° east of north
Answer:
The resultant vector A + B is given by 7.2 m at an angle of 26° east of north,
C
Explanation:
Resolving the vectors to vertical and horizontal component;
Vertical;
Vector A = 6sin30
Vector B = 4sin60
Resultant vertical = 6sin30 + 4sin60 = 6.464m
Horizontal;
Vector A = 6cos30
Vector B = -4cos60
Resultant horizontal = 6cos30 - 4cos60 = 3.196m
Resultant R = √(6.464^2 + 3.196^2) = 7.2m
Tanθ = 6.464/3.196
θ = taninverse (6.464/3.196) BN
θ = 64° north of East.
Or
26° east of north
The resultant vector A + B is given by 7.2 m at an angle of 26° east of north,
The resultant vector A + B is given by 7.2 m at an angle of 26° east of north. Hence, option (c) is correct.
Given data:
The magnitude of vector A is, A = 6.0 m.
The direction of vector A is, 30° north of east.
The magnitude of vector B is, B = 4.0 m.
The direction of vector B is, 30° west of north.
The quantity having both the magnitude as well as the magnitude are known as vector quantities.
Resolving the vectors to vertical and horizontal component;
Along the Vertical direction;
Vector A = 6sin30
Vector B = 4sin60
Resultant vertical vector = 6sin30 + 4sin60 = 6.464 m
Along the Horizontal direction;
Vector A = 6cos30
Vector B = -4cos60
Resultant horizontal vector = 6cos30 - 4cos60 = 3.196 m
Now, the resultant vector is calculated as,
[tex]R=\sqrt{6.464^{2}+3.196^{2}}\\\\R = 7.2 \;\rm m[/tex]
And the resultant direction of vectors is,
[tex]tan \theta = 6.464/3.196\\\\\theta = tan^{-1} (6.464/3.196) \\\theta =64 ^{\circ}[/tex]( North of East)
θ = 64° north of East.
Or
26° east of north
Thus, we can conclude that the resultant vector A + B is given by 7.2 m at an angle of 26° east of north. Hence, option (c) is correct.
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What is equilibrium????
Answer:
Explanation:
An equilibrium is a state in which opposing forces or influences are banned.
An example of equilibrium is in economics when supply and demand are equal. An example of equilibrium is when you are calm and steady. An example of equilibrium is when hot air and cold air are entering the room at the same time so that the overall temperature of the room does not change at all.
Answer:
a state in which opposing forces or influences are balanced.
Explanation:
In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time, so that there is no observable change in the properties of the system.
The diagram below shows a light ray from a pencil hitting a mirror. An upright pencil sits to the left of a mirror, which is convex facing left. An arrow from the tip of the pencil runs to the center of the mirror. Line 1 runs from the point reflection at an acute angle below the first arrow. Arrow 2 runs at an acute angle above the first arrow. Arrow 3 runs perpendicularly up from the point reflection. Arrow 4 runs through the mirror down at an obtuse angle pointing toward a distant point on a line running from the base of the pencil. Which ray shows the correct direction of the reflected ray? 1 2 3 4
the answer would be 2, I just took the test..hope this helps :))
The ray 2 shows the correct direction of the reflected ray. Therefore, the correct option is b.
What is reflected ray?The ray that depicts the light that is reflected by the surface is the reflected ray that corresponds to a certain incident ray. The angle of reflection is the angle formed between the surface normal and the reflected beam. According to the Law of Reflection, the angle of reflection for a specular surface is always the same as the angle of incidence.
In the given problem, the second ray is the ray that is depicted correctly as a reflected ray as it is bounced back after striking the mirror at a specified angle above the horizontal. Rest all the other rays are not depicted correctly as the ray 1 is reflected below normal, ray 3 is reflected horizontally and the ray 4 is refracted.Therefore, the correct option is b.
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The question is incomplete, but most probably the complete question is,
The diagram below shows a light ray from a pencil hitting a mirror. An upright pencil sits to the left of a mirror, which is convex facing left. An arrow from the tip of the pencil runs to the center of the mirror. Line 1 runs from the point of reflection at an acute angle below the first arrow. Arrow 2 runs at an acute angle above the first arrow. Arrow 3 runs perpendicularly up from the point reflection. Arrow 4 runs through the mirror down at an obtuse angle, pointing toward a distant point on a line running from the base of the pencil. Which ray shows the correct direction of the reflected ray?
a.1
b.2
c.3
d.4
At t1, Car A and car B are each located at position xo moving forward at speed v. At t2, car A is located at position 2xo moving forward at speed 3v, while car B is located at position 2xo but is moving backward at speed v. Is the average velocity of car A between t1 and t2greater than, less than, or equal to the average velocity of car B between t1 and t2?
The average velocity of car A between t₁ and t₂ is greater than the average velocity of car B between t₁ and t₂.
What is average velocity?The average velocity is the ratio of the total displacement traveled to the total time taken by the body. Its unit is m/sec.
The given data in the problem is;
t₁ is the time at position 1
xo is the position of car A and car B
v is the speed of cars A and B
t₂ is the time at position 2
3v is the speed of car A
v is the speed of car B
The average velocity is the ratio of the total distance traveled to the total time taken by the body. Its unit is m/sec.
At time t₁ A moved from Xo to 2Xo, resulting in a displacement of 2Xo.
At time t2, A travels at a speed of 3V, therefore his new position is 3Xo, as opposed to 2Xo, which is at 5Xo. A's displacement from the starting location is 5Xo.
B moved from Xo to 2Xo at time ti, and the displacement is 2Xo.
At time t₂ proceeds with speed V in the opposite direction, returning to his starting location, and therefore his new position is at Xo. A's the displacement from his starting location is zero.
[tex]\rm V_{avg}= \frac{total \ displacement}{total \ time}[/tex]
The displacement of car A between t₁ and t₂ is greater than the displacement of car B between t₁ and t₂.
Hence the average velocity of car A between t₁ and t₂ is greater than the average velocity of car B between t₁ and t₂.
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A helium balloon can just lift a load of 790 N. The skin of the balloon has a mass of 1.80 kg. What is the volume of the balloon?(the density of air is 1.29 kg/m³ and the density of helium is 0.179 kg/m³)
Answer:
74.10 cubic meters
Explanation:
The weught of load, balloon and heilium equals the weight of air displaced.
We know that density is mass per unit volume hence mass is product of density and volume, m=dv where d is density and v is volume
Weight is product of mass and acceleration due to gravity, W=mg=mgv where g is acceleration due to gravity and m is mass. We take g to be 9.81 m/s2
Weight of load=790 N
Weight of ballon= mg= 1.8*9.81
Weight of helium=0.179*9.81*v
Weight of air displaced =1.29*9.81*v
790+(1.8*9.81)+(9.81*0.179*v)=1.29*9.81*v
(1.29-0.179)v*9.81=807.658
10.89891v=807.658
V=74.104474667650
Rounding off, the volume is 74.10 cubic meters
A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 20.5 m above the river, whereas the opposite side is a mere 2.4 m above the river. The river itself is a raging torrent 57.0 m wide.
A) How fast should the car be traveling just as it leaves the cliff in order just to clear the river and land safely on the opposite side?
B) What is the speed of the car just before it lands safely on the other side?
Answer:
(A) Velocity of car in order just to clear the river and land safely on the opposite side is [tex]29.68 \frac{m}{s}[/tex]
(B) The speed of the car is [tex]35.14 \frac{m}{s}[/tex]
Explanation:
Given:
Car distance from river [tex]y = 20.5[/tex] m
Mere distance from river [tex]y_{o} = 2.4[/tex] m
River length [tex]x= 57[/tex] m
(A)
For finding the velocity of car,
Using kinematics equation we find velocity of car
[tex]y - y_{o} = v_{o}t + \frac{1}{2} gt^{2}[/tex]
Where [tex]v_{o} = 0[/tex], [tex]g = 9.8[/tex] [tex]\frac{m}{s^{2} }[/tex] ( given in question )
[tex]20.5 - 2.4 = \frac{1}{2} \times 9.8 \times t^{2}[/tex]
[tex]t= 1.92[/tex] sec
The speed of the car before it lands safely on the opposite side of the river is given by,
[tex]v_{x} = \frac{x-x_{o} }{t}[/tex]
[tex]v_{x} = \frac{57-0}{1.92}[/tex] ( [tex]x_{o} = 0[/tex] )
[tex]v_{x} = 29.68[/tex] [tex]\frac{m}{s}[/tex]
(B)
For finding the speed,
The horizontal distance travel by car,
[tex]v_{y} = v_{o} + at[/tex]
Where [tex]a = 9.8[/tex] [tex]\frac{m}{s^{2} }[/tex]
[tex]v_{y} = 9.8 \times 1.92[/tex]
[tex]v_{y} = 18.81[/tex] [tex]\frac{m}{s}[/tex]
For finding the speed,
[tex]v = \sqrt{v_{x}^{2} + v_{y} ^{2} }[/tex]
[tex]v = \sqrt{(29.68)^{2}+ (18.81)^{2} }[/tex]
[tex]v = 35.14[/tex] [tex]\frac{m}{s}[/tex]
Therefore, the speed of the car is [tex]35.14 \frac{m}{s}[/tex]
1. A series RC circuit has a time constant of 1.0 s. The battery has a voltage of 50 V and the maximum current just after closing the switch is 500 mA. The capacitor is initially uncharged. What is the charge on the capacitor 2.0 s after the switch is closed?
The charge on the capacitor after 2 s is 0.43 C.
Explanation:
The formula for finding the voltage while charging a capacitor is
[tex]V = V_{0}(1-e^{\frac{-t}{RC} })[/tex]
Here , V₀ is the initial potential before charging and t is the time at which we have to determine the voltage, R is the resistance and C is the capacitance for the given circuit.
The given problem have given us the values for V₀ = 50 V and maximum current I is given as 500 mA.
Then, resistance can be determined using Ohm's law: [tex]R = \frac{V}{I} =\frac{50}{500*10^{-3} } = 100 Ohm[/tex]
The capacitance is defined as the ratio of charge to the unit voltage.
[tex]C = \frac{Q}{V} = \frac{IT}{IR} = \frac{T}{R}[/tex]
Here T is the time constant which is given as 1 s and R is found to be 100 ohm, then capacitance will be [tex]\frac{1}{100} = 10 mF[/tex]
So, the values for parameters like V₀ = 50 V, R = 100 Ω, C = 10 mF and t = 2 s.
Then, [tex]V =50*(1-e^{\frac{-2}{100*10*10^{-3} } }) =50*(1-e^{-2}) = 50*(1-0.1353)[/tex]
V= 43 V.
Then, [tex]Q = CV = 10*10^{-3} * 43 V = 0.43 C[/tex]
Thus, the charge on the capacitor after 2 s is 0.43 C.
The unit of charge is Coulomb.
The charge on the capacitor after 2 second is 0.43 Coulomb.
To finding the voltage while charging a capacitor is given as,
[tex]V=V_{0}(1-e^{-\frac{t}{RC} } )[/tex]
Here , V₀ is the initial potential, R is the resistance and C is the capacitance
It is given that V₀ = 50 V and maximum current I is given as 500 mA.
So, [tex]R=\frac{V}{I}=\frac{50}{500*10^{-3} } =100ohm[/tex]
Time constant = RC = 1
So, [tex]C=\frac{1}{R} =\frac{1}{100}F[/tex]
Now, we have V₀ = 50 V, R = 100 Ω, C = 1/100 F and t = 2 s.
[tex]V=50(1-e^{-\frac{2}{100*\frac{1}{100} } } )=50(1-e^{-2} )=50(1-0.1353)=43V[/tex]
We know that, [tex]Q=CV=\frac{1}{100}*43=0.43C[/tex]
Thus, the charge on the capacitor after 2 s is 0.43 C.
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A portable x-ray unit has a step-up transformer. The 120 V input is transformed to the 100 kV output needed by the x-ray tube. The primary has 73 loops and draws a current of 14.1 A when in use. What is the number of loops in the secondary
Answer:
60833 loops
Explanation:
Given parameters:
Input voltage = 120V
Output voltage = 100kV
Primary loops = 73 loops
Current = 14.1A
Unknown:
Number of loops in the secondary coil = ?
Solution:
A step up transformer of this nature increases the output voltage. Here, the voltage is raised on the output side.
One thing about a transformer is that the output and input power are always the same.
To find the number of loops in the secondary coil;
[tex]\frac{N_{p} }{N_{s} } = \frac{V_{p} }{V_{s} }[/tex]
Where N is the number of loops and V is the voltage
s and p are the secondary and primary parts
Note: 100kV = 100000V
Now input the parameters;
[tex]\frac{73}{N_{s} } = \frac{120}{100000}[/tex]
120N[tex]_{s}[/tex] = 73 x 100000
N[tex]_{s}[/tex] = 60833 loops
An 8 foot metal guy wire is attached to a broken stop sign to secure its position until repairs can be made. Attached to a stake in the ground, the guy wire makes an angle of 51º with the ground. How far from the foot of the stop sign is the stake, to the nearest tenth of a foot?
Answer:
Explanation:
The guy wire is making a right angled triangle with the ground and stop sign . It makes an angle of 51 degree with the ground. In this triangle stop sign is the perpendicular and distance from the base on the ground forms the base of the triangle . Wire forms the hypotenuse.
base / hypotenuse = cos51
base = hypotenuse x cos51
= 8 x cos51
= 5.03 ft .
The distance of the stake with which guy wire was attached from the foot of the stop sign is 5.03 ft .
Suppose that the resistive force of the air on a skydiver can be approximated by f = −bv2. If the terminal velocity of an 82.0 kg skydiver is 33.4 m/s, what is the value of b (in kg/m)?
Answer:
The value of b 0.7351 kg/m
Explanation:
Given that;
Mass of sky diver = 82 kg
Velocity = 33.4 m/s
f = −bv²
It is a resistance force, therefore the negative sign is ignored.
since; f = mg
∴ mg = bv²
b = mg / v ² ........ (1)
At terminal velocity a = 0
Put parameters in (1)
b = (82 × 10) / (33.4)²
b = 820 / 1,115.56
b = 0.7351
The chart shows the voltage of four electric currents.
Which is best supported by the data in the chart?
Voltage of Currents
Current
Volts (V)
19.0
1.5
W
X
Current w flows at a higher rate than Current Z.
Current Y flows at a lower rate than Current X.
© Current X has a lower potential difference than Current
3.0
Z
4.5
Current Z has a greater potential difference than Current
W.
The question is incomplete. The complete question is as follows.
The chart shows the voltage of four electric currents. A two-column table with 2 row titled Voltage of Currents. The first column labeled Current has entries W, X, Y, Z. The second column labeled Volts (volts) has entries 9.0, 1.5, 3.0, 4.5.
Which is best supported by the data in the chart?
A. Current W flows at a higher rate than Current Z.
B. Current Y flows at a lower rate than Current X.
C. Current X has a lower potential difference than Current Y.
D. Current Z has a greater potential difference than Current W.
Answer: C. Current X has a lower potential difference than Current Y.
Explanation: The table described is show below:
Current Volts (volts)
W 9.0
X 1.5
Y 3.0
Z 4.5
Current is defined by the flow of charge that passes through a material, called conductor in an amount of time:
I = [tex]\frac{Q}{t}[/tex]
where:
Q is charge in Coulomb
t is time in seconds
I is the current in Ampere: A = C/s
According to the question, the currents are not defined by charge or numbers, so it's difficult to determine its value. So, alternatives A and B aren't correct.
Voltage is the difference in electrical potential between two points. It is because of the voltage that the charges flow.
From the table, Current W has the biggest voltage of all the others. So, alternative D is incorrect.
Analysing the voltages of Current X and Y, we observe that they are 1.5 and 3.0, respectively. So, volts or the potential diference of current X is lower than Y, making alternative C the correct one.
Answer:
Current X has a lower potential difference than Current Y.
Explanation:
3. an observer is at the midpoint between two coherent in-phase sources of sound of frequency 780 hz. what is the minimum distance the observer must move toward one of the sources in order to be at a point were destructive interference occurs?
Answer:
Explanation:
wavelength of sound = velocity / frequency
= 340 / 780
= .4359 m
Let the observer be at equal distance d from in phase source .
let it moves by distance x for destructive interference .
path difference from source
= d + x - (d - x )
= 2x
for destructive interference
path difference = wave length / 2
2x = .4359 / 2 m
x = .4359 / 4
= .108975 m
10.9 cm
so observer must move by distance 10.9 cm towards on of the centers.
Does light travel faster or slower when it has to pass through air or water
Answer:
Light travels fast in air but i dont know about water
Answer:
light slower in water than in Air
Explanation:
Air's refractive index is about 1.0003, while water's is about 1.3. This means that light is “slower” in water than air. This is because it's more likely to hit a molecule and then get re-emitted, lengthening the amount of time the light takes to get through a certain distance of the medium.
An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s ). If a particular disk is spun at 235.6 rad/s while it is being read, and then is allowed to come to rest over 0.167 seconds , what is the magnitude of the average angular acceleration of the disk? average angular acceleration: rad s 2 If the disk is 0.12 m in diameter, what is the magnitude of the tangential acceleration of a point 1 / 3 of the way out from the center of the disk? tangential acceleration
Answer:
Part(a): The angular acceleration is [tex]\bf{1410.78~rad/s^{2}}[/tex].
Part(b): The linear acceleration is [tex]\bf{28.21~m/s^{2}}[/tex].
Explanation:
Given:
The angular velocity, [tex]\omega = 235.6~rad/s[/tex]
Time taken, [tex]t = 0.167~s[/tex]
Diameter of the disk, [tex]d = 0.12~m[/tex]
Radius of the concerned point, [tex]r_{p} = \dfrac{1}{3}(d/2)[/tex].
Part(a):
The angular acceleration ([tex]\alpha[/tex]) is given by
[tex]\alpha &=& \dfrac{\omega}{t}\\&=& \dfrac{235.6~rad/s}{0.167~s}\\&=& 1410.78~rad/s^{2}[/tex]
Part(b):
The radius ([tex]r_{p}[/tex]) of the concerned point is given by
[tex]r_{p} &=& \dfrac{1}{3}\dfrac{0.12~m}{2}\\&=& 0.02~m[/tex]
Linear acceleration ([tex]a[/tex]) is give by
[tex]a &=& \alpha \times r_{p}\\&=& (1410.78~rad/sec)(0.02~m)\\&=& 28.21~m/s^{2}[/tex]
This question involves the concepts of equations of motion for angular motion and tangential acceleration.
a) The magnitude of the average angular acceleration is "1410.78 rad/s²".
b) The magnitude of the tangential acceleration of a point 1/3 of the way out from the center of the disk is "28.22 m/s²".
a)
We will use the first equation of motion for the angular motion to find out the angular acceleration:
[tex]\alpha = \frac{\omega_f-\omega_i}{t}[/tex]
where,
[tex]\alpha[/tex] = angular acceleration = ?
[tex]\omega _f[/tex] = final angular speed = 0 rad/s
[tex]\omega_i[/tex] = initial angular acceleration = 235.6 rad/s
t = time = 0.167 s
Therefore,
[tex]\alpha=\frac{0\ rad/s-235.6\ rad/s}{0.167\ s}\\\\\alpha = 1410.78\ rad/s^2[/tex]
b)
The tangential acceleration can be found using the following formula:
[tex]a_T=(r)(\alpha)[/tex]
where,
[tex]a_T[/tex] = tangential acceleration = ?
r = distance of point from center = [tex]\frac{1}{3}(\frac{d}{2}) = \frac{1}{3}(\frac{0.12\ m}{2}) = 0.02\ m[/tex]
Therefore,
[tex]a_T=(0.02\ m)(1410.78\ rad/s^2)\\a_T=28.22\ m/s^2[/tex]
Learn more about the angular motion here:
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The attached picture shows the angular equations of motion.
One end of a steel rod of radius R = 10 mm and length L = 80cm is held in a vise. A force of magnitude F = 60kN is then applied perpendicularly to the end face (uniformly across the area) at the other end, pulling directly away from the vise. What is the stress on the rod?
Answer: 1.91*10^8 N/m²
Explanation:
Given
Radius of the steel, R = 10 mm = 0.01 m
Length of the steel, L = 80 cm = 0.8 m
Force applied on the steel, F = 60 kN
Stress on the rod, = ?
Area of the rod, A = πr²
A = 3.142 * 0.01²
A = 0.0003142
Stress = Force applied on the steel/Area of the steel
Stress = F/A
Stress = 60*10^3 / 0.0003142
Stress = 1.91*10^8 N/m²
From the calculations above, we can therefore say, the stress on the rod is 1.91*10^8 N/m²
By considering the magnetic force in the second region, develop a mathematical expression that relates the mass of the particle to the other variables. Do not include the velocity in your expression. You can use the condition that the particle passed through the region of electric and magnetic fields undeflected to eliminate v from your expression. Your expression will also contain the radius of the circular path. i. Your expression for m should depend on B, E, r, and q
Answer:
M = [tex]\frac{qrB^{2} }{E}[/tex]
Explanation:
considering the magnetic force in the second region derive a mathematical expression that equates the mass of the particle to other variables
In a magnetic field
q = charge, M = mass of particle, E = electric field,B= magnetic field
qvb = [tex]\frac{mv^{2} }{r}[/tex] = therefore m = [tex]\frac{qrb}{v}[/tex] (equation 1)
note : the particle passes through the region undeflected
therefore : qvb = qE therefore (E = VB)
hence v = [tex]\frac{E}{B}[/tex] ( equation 2 )
insert equation 2 into equation 1
m = [tex]\frac{qrB^{2} }{E}[/tex]
A traveling sinusoidal electromagnetic wave in vacuum has an electric field amplitude of 96.9 V/m. Find the intensity of this wave and calculate the energy flowing during 14.9 s through an area of 0.0227 m2 that is perpendicular to the wave\'s direction of propagation.
Answer:
the intensity of this wave, I = 12.42 W/m²the energy of this wave, U = 4.2 JExplanation:
Given;
peak electric field, E₀ = 96.9 V/m
time of flow, t = 14.9s
area through which the energy flows, A = 0.0227 m²
The intensity of this wave is calculated using the following formula;
[tex]I = \frac{E_{rms}^2}{c \mu_o}[/tex]
where;
root-mean-square electric field, [tex]E_{rms} = \frac{E_o}{\sqrt{2}} = \frac{96.9}{\sqrt{2} } = 68.5187 \ V/m[/tex]
c is speed of light, c = 3 x 10⁸ m/s
μ₀ is permeability of free space (constant), μ₀ = 1.26 x 10⁻⁶
Substitute these values and calculate the intensity of the wave;
[tex]I = \frac{E_{rms}^2}{c \mu_o} = \frac{(68.5187)^2}{(3*10^8)(1.26*10^{-6})} = 12.42 \ W/m^2[/tex]
Thus, intensity of this wave is 12.42 W/m²
The energy of the wave is calculated as follows;
U = IAt
U = 12.42 x 0.0227 x 14.9
U = 4.2 J
Thus, the energy of this wave is 4.2 J
A 1.3 kgkg block slides along a frictionless surface at 1.3 m/sm/s . A second block, sliding at a faster 5.0 m/sm/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.5 m/sm/s . What was the mass of the second block?
Answer:
0.624 kg
Explanation:
We are given that
Mass of one block,[tex]m_1=1.3 kg[/tex]
[tex]v_1=1.3 m/s[/tex]
[tex]v_2=5 m/s[/tex]
[tex]V=2.5 m/s[/tex]
We have to find the mass of second block.
Mass of second block,[tex]m_2=\frac{m_1V-m_1v_1}{v_2-V}[/tex]
Substitute the values
[tex]m_2=\frac{1.3\times 2.5-1.3\times 1.3}{5-2.5}[/tex]
[tex]m_2=0.624 kg[/tex]
Hence, the mass of second block=0.624Kg
Answer:
The mass of the second block is 0.624 kg.
Explanation:
Given that,
Mass of the block 1, m₁ = 1.3 kg
Speed of block 1, u₁ = 1.3 m/s
Speed of block 2, u₂ = 5 m/s
The final velocity of the combined blocks is 2.5 m/s, V = 2.5 m/s
It is a case of inelastic collision. Using the conservation of linear momentum as :
[tex]m_1u_1+m_2u_2=(m_1+m_2)V\\\\1.3\times 1.3+5m_2=(1.3+m_2)2.5\\\\1.69+5m_2=3.25+2.5m_2\\\\m_2=0.624\ kg[/tex]
So, the mass of the second block is 0.624 kg.
If given a device that has unknown circuitry, and you measure that the voltage across the device leads the voltage across a resistor at 500 Hz, can you know what type of elements are in the device?If so, how?
Final answer:
A leading voltage across a device at 500 Hz compared to a resistor suggests inductive elements within the device. Voltage leads current in inductors, which can be determined by comparing phase differences using a voltmeter parallel to the device.
Explanation:
When measuring the voltage across a device and comparing it to the voltage across a resistor at a specific frequency, such as 500 Hz, the phase difference provides information about the type of elements in the device. If the voltage across the device leads the voltage across the resistor, this indicates the presence of inductive reactance, suggesting that there are inductive elements within the unknown circuitry.
Reactance and impedance play key roles in AC circuits. The voltage drop across a resistor is in phase with the current, meaning it neither leads nor lags the current. However, in an inductor, the voltage leads the current, and in a capacitor, the voltage lags the current. Therefore, measuring a leading voltage implies inductive characteristics. To evaluate this, a voltmeter would be placed in parallel with the device, ensuring minimal disturbance to the circuit as high resistance in the voltmeter minimizes current flow through it.
For components like inductors and capacitors, the relationship between voltage and current is not linear; thus, they are not ohmic devices like resistors. This is evident in an RLC series circuit, where the impedance at a given frequency depends on the resistance (R), inductance (L), and capacitance (C) of the circuit.
The voltage leading the resistor's voltage at 500 Hz suggests the device contains capacitive elements. A phase difference measurement can confirm this. Capacitors cause the voltage to lead the current in AC circuits.
If you observe that the voltage across a device leads the voltage across a resistor at 500 Hz, it is likely that the device includes a capacitive element. In AC circuits, capacitors cause the voltage to lead the current, which means that the voltage across the capacitor will peak before the voltage across a purely resistive element.
To confirm the presence of a capacitive component, you can perform a more detailed analysis:
Measure the phase difference between the voltage across the device and the voltage across the resistor.If the phase difference is close to 90 degrees, it indicates the presence of a pure capacitor.If the phase difference is less than 90 degrees but still significant, it may indicate a combination of resistive and capacitive elements, known as an RC circuit.This phase difference can be analyzed using an oscilloscope to visualize the waveform and confirm the capacitive nature of the device.
A long straight wire carrying a 2.5 A current passes through the centre of a rectangle of dimensions 2.5 m by 7.5 m at the angle of 45 degrees to the surface normal. What is the path integral of ∮B⇀ ∙ ds⇀ around the outside of the rectangle? Express your answer using two significant figures. The answer should be in μT∙m.
Answer:
The value of path integral is 2.2 [tex]\mu T.m[/tex]
Explanation:
Given:
Current carrying by long wire [tex]I = 2.5[/tex] A
Area of rectangle [tex]A = 18.75[/tex] [tex]m^{2}[/tex]
Angle with surface normal [tex]\theta =[/tex] 45°
According to the ampere's circuital law,
[tex]\int\limits {B} \, ds = \mu _{o} I_{net}[/tex]
Where [tex]\mu _{o} = 4\pi \times 10^{-7}[/tex] [tex]ds=[/tex] area element
Here, [tex]I _{net} = I \cos 45[/tex]
[tex]I_{net} = 2.5 \times \frac{1}{\sqrt{2} }[/tex]
Put value of current in above equation,
[tex]\int\limits {B} \, ds = 4\pi \times 10^{-7} \times 2.5 \times \frac{1}{\sqrt{2} }[/tex]
[tex]\int\limits {B} \, ds = 22.2 \times 10^{-7}[/tex]
[tex]\int\limits {B} \, ds = 2.2 \times 10^{-6}[/tex]
[tex]\int\limits {B} \, ds = 2.2[/tex] [tex]\mu T.m[/tex]
Therefore, the value of path integral is 2.2 [tex]\mu T.m[/tex]
A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total mass of the wagon, rider, and rock is 95.8 kg. The mass of the rock is 0.325 kg. Initially the wagon is rolling forward at a speed of 0.530 m/s. Then the person throws the rock with a speed of 15.1 m/s. Both speeds are relative to the ground. Find the speed of the wagon after the rock is thrown (a) directly forward in one case and (b) directly backward in another.
Answer:
a)0.48 m/s
b) 0.583 m/s
Explanation:
As the wagon rolls,
momentum'p'= m x v => 95.8 x 0.530 = 50.774 Kgm/s
(a)Rock is thrown forward,
momentum of rock = 0.325 x 15.1 = 4.9075 Kgm/s
Conservation of momentum says momentum of wagon is given by
50.774 - 4.9075 = 45.8665
Therefore, Speed of wagon = 45.8665 / (95.8-0.325) = 0.48 m/s
(b) Rock is thrown backward,
momentum of wagon = 50.774 + 4.9075 = 55.68 Kgm/s
Therefore, speed of wagon = 55.68 / (95.8-0.325) = 0.583 m/s
Final answer:
The speed of the wagon after the rock is thrown directly forward is 0.529 m/s, and after the rock is thrown directly backward is 1.63 m/s.
Explanation:
To find the speed of the wagon after the rock is thrown, we can use the principle of conservation of momentum. Since there is no external force acting on the system, the total momentum before and after the rock is thrown will be the same.
For part (a) when the rock is thrown directly forward, the momentum of the wagon and rider will remain the same, but the momentum of the rock will change to zero. So, the final momentum of the system will be the same as the initial momentum.
Using the equation: initial momentum = final momentum
(mass of wagon + rider + rock) x initial velocity of wagon = (mass of wagon + rider) x final velocity of wagon
(95.8 kg) x (0.530 m/s) = (95.8 kg + 0.325 kg) x vf
vf = 0.529 m/s
For part (b) when the rock is thrown directly backward, the momentum of the wagon and rider will change to zero, but the momentum of the rock will remain the same. So, the final momentum of the system will be the same as the initial momentum.
Using the equation: initial momentum = final momentum
(mass of wagon + rider + rock) x initial velocity of wagon = (mass of rock) x final velocity of rock
(95.8 kg) x (0.530 m/s) = (0.325 kg) x vf
vf = 1.63 m/s
An elastic conducting material is stretched into a circular loop of 10.7 cm radius. It is placed with its plane perpendicular to a uniform 0.803 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 81.0 cm/s. What emf is induced in volts in the loop at that instant?
Answer:
0.44 V
Explanation:
Parameters given:
Radius of loop, r = 10.7 cm = 0.107 m
Magnetic field strength, B = 0.803 T
Rate of shrinkage of Radius, dr/dt = 81 cm/s = 0.81m/s
EMF induced is given in terms of magnetic Flux, Φ, as:
EMF = dΦ/dt
Magnetic Flux, Φ, is given as:
Φ = B * A (where A is area of loop)
Therefore, EMF is:
EMF = d(B*A)/dt
The area of the loop is given as A = πr²
EMF = d(Bπr²) / dt = Bπ*d(r²)/dt
=> EMF = Bπ*2r*(dr/dt)
EMF = 0.803 * π * 2 * 0.107 * 0.81
EMF = 0.44 V