Answer:
nail will stop after traveling 1000 m
Explanation:
We have given mass m = 50 gram = 0.05 kg
Frictional force which is used to stop the mass F = 62.5 kN
Initial velocity is given u = 50 m/sec
From newton's law force is equal to F = ma, here m is mass and a is acceleration
So [tex]a=\frac{F}{m}=\frac{62.5\times 10^3}{0.05}=1.25\times 10^6m/sec^2[/tex]
As finally nail stops so final velocity v = 0 m/sec
From third equation of motion [tex]v^2=u^2+2as[/tex]
So [tex]0^2=50^2-2\times 1.25\times 10^6\times s[/tex]
s = 1000 m
So nail will stop after traveling 1000 m
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
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 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.
Radar uses radio waves of a wavelength of 2.4 \({\rm m}\) . The time interval for one radiation pulse is 100 times larger than the time of one oscillation; the time between pulses is 10 times larger than the time of one pulse. What is the shortest distance to an object that this radar can detect? Express your answer with the appropriate units.
Answer:
120 m
Explanation:
Given:
wavelength 'λ' = 2.4m
pulse width 'τ'= 100T ('T' is the time of one oscillation)
The below inequality express the range of distances to an object that radar can detect
τc/2 < x < Tc/2 ---->eq(1)
Where, τc/2 is the shortest distance
First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'
f = c/λ (c= speed of light i.e 3 x [tex]10^{8}[/tex] m/s)
f= 3 x [tex]10^{8}[/tex] / 2.4
f=1.25 x [tex]10^{8}[/tex] hz.
As, T= 1/f
time of one oscillation T= 1/1.25 x [tex]10^{8}[/tex]
T= 8 x [tex]10^{-9}[/tex] s
It was given that pulse width 'τ'= 100T
τ= 100 x 8 x [tex]10^{-9}[/tex] => 800 x [tex]10^{-9}[/tex] s
From eq(1), we can conclude that the shortest distance to an object that this radar can detect:
[tex]x_{min}[/tex]= τc/2 => (800 x [tex]10^{-9}[/tex] x 3 x [tex]10^{8}[/tex])/2
[tex]x_{min}[/tex]=120m
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.
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.
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 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
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
A fixed 16.1-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.23 s , the field is changed to 0.19 T pointing down. Part APart complete What is the average induced emf in the coil?
Answer:
Average induced emf will be equal to 0.00156 volt
Explanation:
We have given diameter of the wire d = 16.1 cm
So radius [tex]r=\frac{16.1}{2}=8.05cm=0.08m[/tex]
Cross sectional area of the wire [tex]A=\pi r^2=3.14\times 0.08^2=0.02m^2[/tex]
Magnetic field is changing from 0.53 T to 0.19 T
So change in magnetic field dB = 0.19-0.53 = -0.34 T
Time taken to change in magnetic field dt = 0.23 sec
Induced emf is given by [tex]e=\frac{-d\Phi }{dt}=-A\frac{dB}{dt}=-0.02\times \frac{0.34}{0.23}=0.00156volt[/tex]
So average induced emf will be equal to 0.00156 volt
A charged particle is accelerated from rest through a potential difference of magnitude |ΔV|. After exiting the potential difference at an emission point, the particle enters a region of uniform magnetic field. The magnetic field is perpendicular to the particle's velocity, and the particle travels along a complete circular path. The particle's mass is 2.10 ✕ 10−16 kg, its charge is 26.0 nC, and the magnetic field magnitude is 0.600 T. The particle's circular path, as it returns to the emission point, encloses a magnetic flux of 15.0 µWb.
a. What is the speed in (m/s) of the particle when it is in the region of the magnetic field?
b. What is the magnitude of the potential difference through which the particle was accelerated?
Answer:
a
The speed of the particle is [tex]v = 209485.71 m/s[/tex]
b
The potential difference is [tex]\Delta V = 177.2 \ V[/tex]
Explanation:
From the question we are told that
The mass of the particle is [tex]m = 2.10 *10^{-16} kg[/tex]
The charge on the particle is [tex]q = 26.0 nC = 26.0 *10^{-9} C[/tex]
The magnitude of the magnetic field is [tex]B = 0.600 T[/tex]
The magnetic flux is [tex]\O = 15.0 \mu Wb = 15.0 *10^{-6} Wb[/tex]
The magnetic flux is mathematically represented as
[tex]\O = B *A[/tex]
Where A is the the area mathematically represented as
[tex]A = \pi r^2[/tex]
Substituting this into the equation w have
[tex]\O = B (\pi r^2 )[/tex]
Making r the subject of the formula
[tex]r = \sqrt{\frac{\O}{B \pi} }[/tex]
Substituting value
[tex]r = \sqrt{\frac{15 *10^{-6}}{3.142 * 0.6} }[/tex]
[tex]r = 2.82 *10^{-3}m[/tex]
For the particle to form a circular path the magnetic force the partial experience inside the magnetic must be equal to the centripetal force of the particle and this is mathematically represented as
[tex]F_q = F_c[/tex]
Where [tex]F_q = q B v[/tex]
and [tex]F_c = \frac{mv^2 }{r}[/tex]
Substituting this into the equation above
[tex]qBv = \frac{mv^2}{r}[/tex]
making v the subject
[tex]v = \frac{r q B}{m}[/tex]
substituting values
[tex]v = \frac{2.82 * 10^{-3} * 26 *10^{-9} 0.6}{2.10*10^{-16}}[/tex]
[tex]v = 209485.71 m/s[/tex]
The potential energy of the particle before entering the magnetic field is equal to the kinetic energy in the magnetic field
This is mathematically represented as
[tex]PE = KE[/tex]
Where [tex]PE = q * \Delta V[/tex]
and [tex]KE = \frac{1}{2} mv^2[/tex]
Substituting into the equation above
[tex]q \Delta V = \frac{1}{2} mv^2[/tex]
Making the potential difference the subject
[tex]\Delta V = \frac{mv^2}{2 q}[/tex]
[tex]\Delta V = \frac{2.16 * 10^ {-16} * (209485.71)^2 }{2 * 26 *10^{-9}}[/tex]
[tex]\Delta V = 177.2 \ V[/tex]
Final answer:
The speed of the particle in the magnetic field is found using the force equation for a moving charge in a magnetic field and the work-energy principle is used to determine the magnitude of the potential difference.
Explanation:
To answer the student's question:
The speed of the particle in the magnetic field can be calculated using the formula for the magnetic force acting on a moving charge, which is F = qvB. Since this force provides the centripetal force required for circular motion, F = mv2/r, where m is the mass of the particle, q is the charge of the particle, v is the speed of the particle, B is the magnetic field, and r is the radius of the circular path. Combining these equations, we can solve for the speed: v = qBr/m. To determine the radius, we can use the relationship between magnetic flux (ΦB), magnetic field strength (B), and area of the circular path (A): ΦB = BA or r = sqrt(ΦB / πB). The mass, charge, and magnetic field are given, and we can calculate the radius from the magnetic flux to then find the speed.
The magnitude of the potential difference the particle was initially accelerated through can be determined using the work-energy principle. The work done by the electric field on the charge is equal to the change in kinetic energy of the particle. Using the formula W = ΔKE = qΔV and knowing that the kinetic energy is KE = 1/2 mv2, we can solve for the potential difference: ΔV = KE/q, where KE is the kinetic energy of the particle right before entering the magnetic field.
The gold-like mineral in the metamorphic rock pictured below is known as pyrite. When pyrite is exposed to air and water it can react
to form sulfate and an acid. The acid can discolor and eat away the surrounding rock.
Image courtesy of David Usher, USGS
Pyrite breaks down rocks in which of the following processes?
A
deposition
B. physical weathering
C.
erosion
D.
chemical weathering
Answer: C. Chemical Weathering
Explanation:
Answer:
Chemical Weathering
Explanation:
Starting from rest, a 0.0367 kg steel ball sinks into a vat of corn syrup. The thick syrup exerts a viscous drag force that is proportional to the ball's velocity. → F drag = − C → v where C = 0.270 N ⋅ s/m is a constant related to the size and composition of the ball as well as the viscosity of the syrup. Find the rate at which gravitational energy is converted to thermal energy once the ball reaches terminal velocity.
Answer:
0.48 W
Explanation:
Given that:
the mass of the steel ball = 0.0367 kg
C = 0.270 N
g (acceleration due to gravity) = 9.8
Now;
At Terminal Velocity Weight is balance by drag force
mg =Cv
Making v the subject of the formula:we have:
[tex]v = \frac {mg}{C}[/tex]
[tex]v = \frac {0.0367*9.8}{0.270}[/tex]
v = 1.332 m/s
Thus, the Rate at which gravitational energy is converted to thermal energy once the ball reaches terminal velocity is:
P=Fv
P = mgv
[tex]P = (0.0367*9.8)*1.332[/tex]
P=0.48 W
Final answer:
To find the rate at which gravitational energy is converted to thermal energy at terminal velocity, equate the drag force to the gravitational force on the ball and solve for power as the product of force and velocity.
Explanation:
The question asks for the rate at which gravitational energy is converted to thermal energy once the ball reaches terminal velocity in a vat of corn syrup. To find this rate, we must recognize that at terminal velocity, all the gravitational force acting on the ball (minus any buoyant force, which is ignored here) is converted into thermal energy due to viscous drag. The viscous drag force [tex]F_{drag}[/tex] is given by [tex]F_{drag[/tex] = -Cv, where C is the drag coefficient, and v is the velocity of the ball. The gravitational force acting on the ball can be calculated using [tex]F_{gravity[/tex] = mg, where m is the mass of the ball, and g is the acceleration due to gravity. At terminal velocity, [tex]F_{drag[/tex] = [tex]F_{gravity[/tex], hence the power (rate of energy conversion) P can be found as P = Fv = mgv. Assuming standard gravity (9.81 m/s2), the mass of the ball (0.0367 kg), and solving for v using [tex]F_{drag[/tex], one can determine the rate of gravitational energy conversion to thermal energy.
Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocity of 1700 ft/s. For steady-state operation, and neglecting potential energy effects, determine the exit temperature, in F
Answer:
[tex]386.2^{\circ}F[/tex]
Explanation:
We are given that
[tex]P_1=200lbf/in^2[/tex]
[tex]P_2=60lbf/in^2[/tex]
[tex]v_1=200ft/s[/tex]
[tex]v_2=1700ft/s[/tex]
[tex]T_1=500^{\circ}F[/tex]
[tex]Q=0[/tex]
[tex]C_p=1BTU/lb^{\circ}F[/tex]
We have to find the exit temperature.
By steady energy flow equation
[tex]h_1+v^2_1+Q=h_2+v^2_2[/tex]
[tex]C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}[/tex]
[tex]1BTU/lb=25037ft^2/s^2[/tex]
Substitute the values
[tex]1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}[/tex]
[tex]500+1.598=T_2+115.4[/tex]
[tex]T_2=500+1.598-115.4[/tex]
[tex]T_2=386.2^{\circ}F[/tex]
We can use the First Law of Thermodynamics and the provided initial conditions from the problem statement to calculate the exit temperature of steam in a steady-flow system, such as a nozzle. The calculation requires knowledge of thermal dynamics principles and the use of steam tables.
Explanation:This question is an application of the First Law of Thermodynamics (also known as the Law of Energy Conservation) and it pertains specifically to the flow of steam. In this case, we're working with a steady-flow system (a nozzle) where only the kinetic energy changes are important. There is no mention of the incoming steam doing work on the nozzle, nor of any heat is transferred to or from the nozzle, so, in this case, the work done (W) and heat transfer (Q) are both zero.
Furthermore, the exit pressure of the steam is given, as is the exit velocity; what we need to ascertain is the exit temperature, T2. To do this, we can use the enthalpies for steam at the given initial state, and then use the first law of thermodynamics to determine the exit enthalpy. The corresponding temperature can then be found in the steam tables associated with the exit enthalpy and pressure.
Therefore, the calculation requires an understanding of Thermal Dynamics principles and the use of Steam Tables.
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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.
A mass M is attached to an ideal massless spring. When the system is set in motion, it oscillates with a frequency f. What is the new oscillation frequency if the mass is doubled to 2M? g
Answer:
[tex]\frac {f_i}{\sqrt 2}[/tex]
Explanation:
Frequency is given by [tex]\frac {1}{2\pi}\sqrt{{\frac {k}{m}}[/tex]
Let the initial frequency be denoted by [tex]f_i[/tex]
[tex]f_i=\frac {1}{2\pi}\sqrt{{\frac {k}{M}}[/tex]
When M is 2M then
[tex]f_f=\frac {1}{2\pi}\sqrt{{\frac {k}{2M}}=\frac {f_i}{\sqrt 2}[/tex]
Therefore, the final frequency is
[tex]\frac {f_i}{\sqrt 2}[/tex]
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]
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 electron is accelerated through a potential difference of 2.8 kV and directed into a region between two parallel plates separated by 17 mm with a potential difference of 100 V between them. The electron is moving perpendicular to the electric field when it enters the region between the plates. What magnetic field is necessary perpendicular to both the electron path and the electric field so that the electron travels in a straight line
Answer:
B=1.89*10^{-4} T
Explanation:
First we have to calculate the electric force on the electron, and then we have to take into account that this force is equal to the force generated by the magnetic field.
The formula is
[tex]F=q_eE\\\\E=\frac{V}{d}[/tex]
q: charge of the electron = 1.6*10^{-19}C
V: potential
d: separation between plates
[tex]E=\frac{100V}{17*10^{-3}m}=5882.35N/C[/tex]
[tex]F=(1.6*10^{-19}C)(5882.35N/C)=9.4*10^{-16}N[/tex]
This force must equal the Lorentz's force
[tex]F_E=F_B\\\\F_B=qvB\\\\B=\frac{F_B}{q_ev}[/tex]
But before we have to calculate the speed of the electron by using (me=9.1*10^{-31}kg)
[tex]E_e=\frac{1}{2}m_ev^2=q_eV\\\\v=\sqrt{\frac{2q_eV}{m_e}}=\sqrt{\frac{2(1.6*10^{-19}C)(2.8*10^3V)}{9.1*10^{-31}kg}}\\\\v=3.13*10^7\frac{m}{s}[/tex]
Hence, we have
[tex]B=\frac{9.4*10^{-16}N}{(1.6*10^{-19}C)(3.1*10^{7})\frac{m}{s}}=1.89*10^{-4}T[/tex]
hope this helps!!
Answer:
The magnetic field that is necessary is equal to 1.88x10⁻⁴ T
Explanation:
If the electron is accelerated:
[tex]e*V=\frac{1}{2} mv^{2} \\v=\sqrt{\frac{2eV}{m} }[/tex]
Where
e = 1.6x10⁻¹⁹C
V = 2.8 kV = 2800 V
m = 9.1x10⁻³¹kg
Replacing:
[tex]v=\sqrt{\frac{2*1.6x10^{-19}*2800 }{9.1x10^{-31} } } =3.13x10^{7} m/s[/tex]
When the electron is moving in straight line, the magnetic force is balanced with the electric force, thus:
V = E * d
Where V = 100 V
d = 17 mm = 0.017 m
E = V/d = 100/0.017 = 5882.35 N/C
The magnetic field that is necessary is equal to:
B = E/v = 5882.35/3.13x10⁷ = 1.88x10⁻⁴ T
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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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
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
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
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
Consider the standing wave pattern below created by a string fixed at both ends. The string is under a tension of 0.98 N and has a mass of 2.0 grams. What is the frequency of the standing wave?
Answer:
11.07Hz
Explanation:
Check the attachment for diagram of the standing wave in question.
Formula for calculating the fundamental frequency Fo in strings is V/2L where;
V is the velocity of the wave in string
L is the length of the string which is expressed as a function of its wavelength.
The wavelength of the string given is 1.5λ(one loop is equivalent to 0.5 wavelength)
Therefore L = 1.5λ
If L = 3.0m
1.5λ = 3.0m
λ = 3/1.5
λ = 2m
Also;
V = √T/m where;
T is the tension = 0.98N
m is the mass per unit length = 2.0g = 0.002kg
V = √0.98/0.002
V = √490
V = 22.14m/s
Fo = V/2L (for string)
Fo = 22.14/2(3)
Fo = 22.14/6
Fo = 3.69Hz
Harmonics are multiple integrals of the fundamental frequency. The string in question resonates in 2nd harmonics F2 = 3Fo
Frequency of the wave = 3×3.69
Frequency of the wave = 11.07Hz
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 battery with an emf of 12.0 V shows a terminal voltage of 11.7 V when operating in a circuit with two lightbulbs, each rated at 4.0 W (at 12.0 V), which are connected in parallel.
What is the battery's internal resistance?
0.46Ω
Explanation:The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;
E = V + Ir --------------------(a)
Where;
I = current flowing through the circuit
But;
V = I x Rₓ ---------------------(b)
Where;
Rₓ = effective or total resistance in the circuit.
First, let's calculate the effective resistance in the circuit:
The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.
Let;
R₁ = resistance in the first bulb
R₂ = resistance in the second bulb
Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;
P = [tex]\frac{V^{2} }{R}[/tex]
=> R = [tex]\frac{V^{2} }{P}[/tex] -------------------(ii)
Where;
P = Power of the bulb
V = voltage across the bulb
R = resistance of the bulb
To get R₁, equation (ii) can be written as;
R₁ = [tex]\frac{V^{2} }{P}[/tex] --------------------------------(iii)
Where;
V = 12.0V
P = 4.0W
Substitute these values into equation (iii) as follows;
R₁ = [tex]\frac{12.0^{2} }{4}[/tex]
R₁ = [tex]\frac{144}{4}[/tex]
R₁ = 36Ω
Following the same approach, to get R₂, equation (ii) can be written as;
R₂ = [tex]\frac{V^{2} }{P}[/tex] --------------------------------(iv)
Where;
V = 12.0V
P = 4.0W
Substitute these values into equation (iv) as follows;
R₂ = [tex]\frac{12.0^{2} }{4}[/tex]
R₂ = [tex]\frac{144}{4}[/tex]
R₂ = 36Ω
Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;
[tex]\frac{1}{R_{X} }[/tex] = [tex]\frac{1}{R_1}[/tex] + [tex]\frac{1}{R_2}[/tex] -----------------(v)
Substitute the values of R₁ and R₂ into equation (v) as follows;
[tex]\frac{1}{R_X}[/tex] = [tex]\frac{1}{36}[/tex] + [tex]\frac{1}{36}[/tex]
[tex]\frac{1}{R_X}[/tex] = [tex]\frac{2}{36}[/tex]
Rₓ = [tex]\frac{36}{2}[/tex]
Rₓ = 18Ω
The effective resistance (Rₓ) is therefore, 18Ω
Now calculate the current I, flowing in the circuit:
Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;
11.7 = I x 18
I = [tex]\frac{11.7}{18}[/tex]
I = 0.65A
Now calculate the battery's internal resistance:
Substitute the values of E = 12.0, V = 11.7V and I = 0.65A into equation (a) as follows;
12.0 = 11.7 + 0.65r
0.65r = 12.0 - 11.7
0.65r = 0.3
r = [tex]\frac{0.3}{0.65}[/tex]
r = 0.46Ω
Therefore, the internal resistance of the battery is 0.46Ω
Answer:
[tex]R_i_n_t=0.45 \Omega[/tex]
Explanation:
Internal resistance is a concept that helps model the electrical consequences of the complex chemical reactions that occur within a battery. When a charge is applied to a battery, the internal resistance can be calculated using the following equation:
[tex]R_i_n_t=(\frac{V_N_L}{V_F_L} -1)R_L[/tex]
Where:
[tex]V_F_L=Load\hspace{3}voltage=11.7V\\V_N_L= O pen\hspace{3}circuit\hspace{3}voltage=12V\\R_L=Load\hspace{3}resistance[/tex]
As you can see, we don't know the exactly value of the [tex]R_L[/tex]. However we can calculated that value using the next simple operations:
The problem tell us that the power of each lightbulb is 4.0 W at 12.0 V, hence let's calculated the power at 11.7V using Cross-multiplication:
[tex]\frac{12}{11.7} =\frac{4}{P}[/tex]
Solving for [tex]P[/tex] :
[tex]P=\frac{11.7*4}{12} =3.9W[/tex]
Now, the electric power is given by:
[tex]P=\frac{V^2}{R_b}[/tex]
Where:
[tex]R_b=Resistance\hspace{3}of\hspace{3}each\hspace{3}lightbulb[/tex]
So:
[tex]R_b=\frac{V^2}{P} =\frac{11.7^2}{3.9} =35.1\Omega[/tex]
Now, because of the lightbulbs are connected in parallel the equivalent resistance is given by:
[tex]\frac{1}{R_L} =\frac{1}{R_b} +\frac{1}{R_b} =\frac{2}{R_b} \\\\ R_L= \frac{R_b}{2} =\frac{35.1}{2}=17.55\Omega[/tex]
Finally, now we have all the data, let's replace it into the internal resistance equation:
[tex]R_i_n_t=(\frac{12}{11.7} -1)17.55=0.45\Omega[/tex]
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 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.
1. Is the collision between the ball and the pendulum elastic or inelastic? Justify your answer by calculating the kinetic energy of the system before collision using the value of vxo in the experiment and the kinetic energy just after collision using the experimental value of h in Eq. 9.2.
Answer:
So energy is not conserved and inelastic shock
Explanation:
In the collision between a bullet and a ballistic pendulum, characterized in that the bullet is embedded in the block, if the kinetic energy is conserved the shock is elastic and if it is not inelastic.
Let's find the kinetic energy just before the crash
K₀ = ½ m vₓ₀²
After the crash we can use the law of conservation of energy
Starting point. Right after the crash, before starting to climb
Em₀ = K = ½ (m + M) v₂²
Final point. At the maximum height of the pendulum
[tex]Em_{f}[/tex] = U = (m + M) g h
Where m is the mass of the bullet and M is the mass of the pendulum
Em₀ = Em_{f}
½ (m + M) v₂² = (m + M) g h
v₂ = √ 2g h
Now we can calculate the final kinetic energy
K_{f} = ½ (m + M) v₂²²
K_{f} = ½ (m + M) (2gh)
The relationship between these two kinetic energies is
K₀ / K_{f} = ½ m vₓ₀² / (½ (m + M) 2 g h)
K₀ / K_{f} = m / (m + M) vₓ₀² / 2 g h
We can see that in this relationship the Ko> Kf
So energy is not conserved and inelastic shock
To determine if the collision between the ball and the pendulum is elastic or inelastic, we need to calculate the kinetic energy of the system before and after the collision.
Explanation:An elastic collision is one that conserves kinetic energy, while an inelastic collision does not conserve kinetic energy. To determine if the collision between the ball and the pendulum is elastic or inelastic, we need to calculate the kinetic energy of the system before and after the collision.
In this case, we can calculate the kinetic energy of the system before the collision using the value of vxo and the kinetic energy just after the collision using the experimental value of h in Eq. 9.2.
If the kinetic energy of the system is the same before and after the collision, then it is an elastic collision. If the kinetic energy decreases after the collision, then it is an inelastic collision.
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