Answer:
Explanation:
a )
This type of spectrum is called line emission spectrum . Because it consists of lines . It is emission spectrum because it is due to emission of radiation from a source .
b ) The wavelength of a photon is inversely proportional to its energy . Photon due to transition between n = 1 and n = 3 will have higher energy than
that due to transition between n = 2 and n = 5 . So the later photon ( B) will have greater wavelength or photon due to transition between n = 2 and n = 5 will have greater wavelength .
Consider an electromagnetic wave where the electric field of an electromagnetic wave is oscillating along the z-axis and the magnetic field is oscillating along the x-axis.
Required:
In what directions is it possible that the wave is traveling?
Answer:
The wave is traveling in the y axis direction
Explanation:
Because the wave will always travel in a direction 90° to the magnetic and electric components
Estimate the volume of a human heart (in mL) using the following measurements/assumptions:_______.
1. Blood flow through the aorta is approximately 11.2 cm/s
2. The diameter of the aorta is approximately 3.0 cm
3. Assume the heart pumps its own volume with each beat
4. Assume a pulse rate of 67 beats per minute.
Answer:
Explanation:
radius of aorta = 1.5 cm
cross sectional area = π r²
= 3.14 x 1.5²
= 7.065 cm²
volume of blood flowing out per second out of heart
= a x v , a is cross sectional area , v is velocity of flow
= 7.065 x 11.2
= 79.128 cm³
heart beat per second = 67 / 60
= 1.116666
If V be the volume of heart
1.116666 V = 79.128
V = 70.86 cm³.
How much heat is needed to melt 2.5 KG of water at its melting point? Use Q= mass x latent heat of fusion.
Answer:
Q = 832 kJ
Explanation:
It is given that,
Mass of the water, m = 2.5 kg
The latent heat of fusion, L = 333 kJ/kg
We need to find the heat needed to melt water at its melting point. The formula of heat needed to melt is given by :
Q = mL
[tex]Q=2.5\ kg\times 333\ kJ/kg\\\\Q=832.5\ kJ[/tex]
or
Q = 832 kJ
So, the heat needed to melt the water is 832 kJ.
two resistors of resistance 10 ohm's and 20 ohm's are connected in parallel to a batery of e.m.f 12V. Calculate the current passing through the 20hm's resister
A boat floating in fresh water displaces 16,000 N of water. How many newtons of salt water would it displace if it floats in salt water of specific gravity 1.10
Answer:
It will displace the same weight of fresh water i.e.16000N. The point is the body 'floats'- which is the underlying assumption here, and by Archimedes Principle, for this body or vessel or whatever it may be, to float it should displace an equal weight of water
Explanation:
select the example that best describes a renewable resource.
A.after a shuttle launch, you can smell the jet fuel for hours.
B.solar panels generate electricity that keeps the satellites running.
C.tractor trailers are large trucks that run on diesel fuel.
D. we use our barbeque every night; it cooks with propane.
Answer:
B.solar panels generate electricity that keeps the satellites running.
Explanation:
Solar panels are a renewable resource because they take energy from the sun.
A particle moves in a velocity field V(x, y) = x2, x + y2 . If it is at position (x, y) = (7, 2) at time t = 3, estimate its location at time t = 3.01.
Answer:
New location at time 3.01 is given by: (7.49, 2.11)
Explanation:
Let's start by understanding what is the particle's velocity (in component form) in that velocity field at time 3:
[tex]V_x=x^2=7^2=49\\V_y=x+y^2=7+2^2=11[/tex]
With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:
[tex]distance=v\,*\, t[/tex]
[tex]distance_x=49\,(0.01)=0.49\\distance_y=11\,(0.01)=0.11[/tex]
Therefore, adding these displacements in component form to the original particle's position, we get:
New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)
The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional toA) the particle's charge.B) the particle's momentum.C) the particle's energy.D) the flux density of the field.E)All of these are correct
Answer:
B) the particle's momentum.
Explanation:
We know that
The centripetal force on the particle when its moving in the radius R and velocity V
[tex]F_c=\dfrac{m\times V^2}{R}[/tex]
The magnetic force on the particle when the its moving with velocity V in the magnetic filed B and having charge q
[tex]F_m=q\times V\times B[/tex]
At the equilibrium condition
[tex]F_m=F_c[/tex]
[tex]q\times V\times B=\dfrac{m\times V^2}{R}[/tex]
[tex]R=\dfrac{m\times V}{q\times B}[/tex]
Momentum = m V
Therefore we can say that the radius of curvature is directly proportional to the particle momentum.
B) the particle's momentum.
Based on the graph below, what prediction can we make about the acceleration when the force is 0 newtons? A. It will be 0 meters per second per second. B. It will be 5 meters per second per second. C. It will be 10 meters per second per second. D. It will be 15 meters per second per second.
PLZ HURRY WILL MARK BRAINLIEST IF CORRECT
Answer:
Option A
Explanation:
Acceleration will be obviously zero when Force = 0
That is how:
Force = Mass * Acceleration
So, If force = 0
0 = Mass * Acceleration.
Dividing both sides by Mass
Acceleration = 0/Mass
Acceleration = 0 m/s²
Answer:
[tex]\boxed{\mathrm{A. \: It \: will \: be \: 0 \: meters \: per \: second \: per \: second. }}[/tex]
Explanation:
[tex]\mathrm{force=mass \times acceleration}[/tex]
The force is given 0 newtons.
[tex]\mathrm{force=0 \: N}[/tex]
Plug force as 0.
[tex]\mathrm{0=mass \times acceleration}[/tex]
Divide both sides by mass.
[tex]\mathrm{\frac{0}{mass} =acceleration}[/tex]
[tex]\mathrm{0 =acceleration}[/tex]
[tex]\mathrm{acceleration= 0\: m/s/s}[/tex]
A proton that is initially at rest is accelerated through an electric potential difference of magnitude 500 V. What speed does the proton gain? (e = 1.60 × 10-19 C , mproton = 1.67 × 10-27 kg)
Answer:
[tex]3.1\times 10^{5}m/s[/tex]
Explanation:
The computation of the speed does the proton gain is shown below:
The potential difference is the difference that reflects the work done as per the unit charged
So, the work done should be
= Potential difference × Charge
Given that
Charge on a proton is
= 1.6 × 10^-19 C
Potential difference = 500 V
[tex]v= \sqrt{\frac{2.q.\Delta V}{m_{p}}} \\\\\\= \sqrt{\frac{2\times 1.6\times 10^{-19}\times 5\times 10^{2}}{1.67\times 10^{-27}}}[/tex]
[tex]v= \sqrt{9.58\times 10^{10}}m/s \\\\= 3.095\times 10^{5}m/s\\\\\approx 3.1\times 10^{5}m/s[/tex]
Simply we applied the above formulas
A magnetic field is entering into a coil of wire with radius of 2(mm) and 200 turns. The direction of magnetic field makes an angle 25° in respect to normal to surface of coil. The magnetic field entering coil varies 0.02 (T) in every 2 seconds. The two ends of coil are connected to a resistor of 20 (Ω).
A) Calculate Emf induced in coil
B) Calculate the current in resistor
C) Calculate the power delivered to resistor by Emf
Answer:
a) 2.278 x 10^-5 volts
b) 1.139 x 10^-6 Ampere
c) 2.59 x 10^-11 W
Explanation:
The radius of the wire r = 2 mm = 0.002 m
the number of turns N = 200 turns
direction of the magnetic field ∅ = 25°
magnetic field strength B = 0.02 T
varying time = 2 sec
The cross sectional area of the wire = [tex]\pi r^{2}[/tex]
==> A = 3.142 x [tex]0.002^{2}[/tex] = 1.257 x 10^-5 m^2
Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°
==> Φ = 2.278 x 10^-7 Wb
The induced EMF is given as
E = NdΦ/dt
where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7
E = 200 x 1.139 x 10^-7 = 2.278 x 10^-5 volts
b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as
[tex]I[/tex] = E/R
where R is the resistor
[tex]I[/tex] = (2.278 x 10^-5)/20 = 1.139 x 10^-6 Ampere
c) power delivered to the resistor is given as
P = [tex]I[/tex]E
P = (1.139 x 10^-6) x (2.278 x 10^-5) = 2.59 x 10^-11 W
Uses of pressure and the uses of density
Answer:
Pressure is a scalar quantity defined as per unit area.
Density is the objects ,times its the acceleration due to gravity.
Explanation:
Pressure is the alternative object increases the area of contact decrease .
Pressure is the force component to the surface used to calculate pressure.
pressure is that collisions of the gas to container as the per unit time .
pressure is an physical important quantity to play the solid and fluid .
Pressure is the expressed in a number of units depend the context use, pressure exerted by the liquid alone.
Density is the objects, times, volume of the object that times acceleration objects.
Density is the used to the system complex objects and materials.
Density force is the weight of a region or objects static fluid.
Calculate the power of the eye in D when viewing an object 5.70 m away. (Assume the lens-to-retina distance is 2.00 cm. Enter your answer to at least one decimal place.)
Answer:
Power=50.17dioptre
Power=50.17D
Explanation:
P=1/f = 1/d₀ + 1/d₁
Where d₀ = the eye's lens and the object distance= 5.70m=
d₁= the eye's lens and the image distance= 0.02m
f= focal length of the lense of the eye
We know that the object can be viewed clearly by the person ,then image and lens of the eye's distance needs to be equal with the retinal and the eye lens distance and this distance is given as 0.02m
Therefore, we can calculate the power using above formula
P= 1/5.70 + 1/0.02
Power=50.17dioptre
Therefore, the power the eye's is using to see the object from distance is 5.70D
"A power of 200 kW is delivered by power lines with 48,000 V difference between them. Calculate the current, in amps, in these lines."
Answer:
9.6×10⁹ A
Explanation:
From the question above,
P = VI.................... Equation 1
Where P = Electric power, V = Voltage, I = current.
make I the subject of the equation
I = P/V............. Equation 2
Given: P = 200 kW = 200×10³ W, V = 48000 V.
Substitute these vales into equation 2
I = 200×10³×48000
I = 9.6×10⁹ A.
Hence the current in the line is 9.6×10⁹ A.
A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.
Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?
Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?
Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar.
Answer:
a) 6738.27 J
b) 61.908 J
c) [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]
Explanation:
The complete question is
A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.
Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?
Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?
Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.
moment of inertia is given as
[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex]
where m is the mass of the flywheel,
and r is the radius of the flywheel
for the flywheel with radius 1.1 m
and mass 11 kg
moment of inertia will be
[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]*11*1.1^{2}[/tex] = 6.655 kg-m^2
The maximum speed of the flywheel = 35 m/s
we know that v = ωr
where v is the linear speed = 35 m/s
ω = angular speed
r = radius
therefore,
ω = v/r = 35/1.1 = 31.82 rad/s
maximum rotational energy of the flywheel will be
E = [tex]Iw^{2}[/tex] = 6.655 x [tex]31.82^{2}[/tex] = 6738.27 J
b) second flywheel has
radius = 2.8 m
mass = 16 kg
moment of inertia is
[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex] = [tex]\frac{1}{2}[/tex][tex]*16*2.8^{2}[/tex] = 62.72 kg-m^2
According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels
for the first flywheel, rotational momentum = [tex]Iw[/tex] = 6.655 x 31.82 = 211.76 kg-m^2-rad/s
for their combination, the rotational momentum is
[tex](I_{1} +I_{2} )w[/tex]
where the subscripts 1 and 2 indicates the values first and second flywheels
[tex](I_{1} +I_{2} )w[/tex] = (6.655 + 62.72)ω
where ω here is their final angular momentum together
==> 69.375ω
Equating the two rotational momenta, we have
211.76 = 69.375ω
ω = 211.76/69.375 = 3.05 rad/s
Therefore, the energy stored in the first flywheel in this situation is
E = [tex]Iw^{2}[/tex] = 6.655 x [tex]3.05^{2}[/tex] = 61.908 J
c) one third of the initial energy of the flywheel is
6738.27/3 = 2246.09 J
For the car, the kinetic energy = [tex]\frac{1}{2}mv_{car} ^{2}[/tex]
where m is the mass of the car
[tex]v_{car}[/tex] is the velocity of the car
Equating the energy
2246.09 = [tex]\frac{1}{2}mv_{car} ^{2}[/tex]
making m the subject of the formula
mass of the car m = [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]
Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel's energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.5 m diameter and a mass of 250 kg. Its maximum angular velocity is 1200 rpm.
How long does it take the flywheel to reach top angular speed of 1200 rpm?
Answer:
t = 2.95 min
Explanation:
Given that,
The diameter of flywheeel, d = 1.5 m
Mass of flywheel, m = 250 kg
Initial angular velocity is 0
Final angular velocity, [tex]\omega_f=1200\ rpm = 126\ rad/s[/tex]
We need to find the time taken by the flywheel to each a speed of 1200 rpm if it starts from rest.
Firstly, we will find the angular acceleration of the flywheel.
The moment of inertia of the flywheel,
[tex]I=\dfrac{1}{2}mr^2\\\\I=\dfrac{1}{2}\times 250\times (0.75)^2\\\\I=70.31\ kg-m^2[/tex]
Now,
Let the torque is 50 N-m. So,
[tex]\alpha =\dfrac{\tau}{I}\\\\\alpha =\dfrac{50}{70.31}\\\\\alpha =0.711\ rad/s^2[/tex]
So,
[tex]t=\dfrac{\omega_f-\omega_i}{\alpha }\\\\t=\dfrac{126-0}{0.711}\\\\t=177.21\ s[/tex]
or
t = 2.95 min
Tuning a guitar string, you play a pure 330 Hz note using a tuning device, and pluck the string. The combined notes produce a beat frequency of 5 Hz. You then play a pure 350 Hz note and pluck the string, finding a beat frequency of 25 Hz. What is the frequency of the string note?
Answer:
The frequency is [tex]F = 325 Hz[/tex]
Explanation:
From the question we are told that
The frequency for the first note is [tex]F_1 = 330 Hz[/tex]
The beat frequency of the first note is [tex]f_b = 5 \ Hz[/tex]
The frequency for the second note is [tex]F_2 = 350 \ H_z[/tex]
The beat frequency of the first note is [tex]f_a = 25 \ Hz[/tex]
Generally beat frequency is mathematically represented as
[tex]F_{beat} = | F_a - F_b |[/tex]
Where [tex]F_a \ and \ F_b[/tex] are frequencies of two sound source
Now in the case of this question
For the first note
[tex]f_b = F_1 - F \ \ \ \ \ ...(1)[/tex]
Where F is the frequency of the string note
For the second note
[tex]f_a = F_2 - F \ \ \ \ \ ...(2)[/tex]
Adding equation 1 from 2
[tex]f_b + f_a = F_1 + F_2 + ( - F) + (-F) )[/tex]
[tex]f_b + f_a = F_1 + F_2 -2F[/tex]
substituting values
[tex]5 +25 = 330 + 350 -2F[/tex]
=> [tex]F = 325 Hz[/tex]
A 4g bullet, travelling at 589m/s embeds itself in a 2.3kg block of wood that is initially at rest, and together they travel at the same velocity. Calculate the percentage of the kinetic energy that is left in the system after collision to that before.
Answer:
The percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %
Explanation:
Given;
mass of bullet, m₁ = 4g = 0.004kg
initial velocity of bullet, u₁ = 589 m/s
mass of block of wood, m₂ = 2.3 kg
initial velocity of the block of wood, u₂ = 0
let the final velocity of the system after collision = v
Apply the principle of conservation of linear momentum
m₁u₁ + m₂u₂ = v(m₁+m₂)
0.004(589) + 2.3(0) = v(0.004 + 2.3)
2.356 = 2.304v
v = 2.356 / 2.304
v = 1.0226 m/s
Initial kinetic energy of the system
K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²
K.E₁ = ¹/₂(0.004)(589)² = 693.842 J
Final kinetic energy of the system
K.E₂ = ¹/₂v²(m₁ + m₂)
K.E₂ = ¹/₂ x 1.0226² x (0.004 + 2.3)
K.E₂ = 1.209 J
The kinetic energy left in the system = final kinetic energy of the system
The percentage of the kinetic energy that is left in the system after collision to that before = (K.E₂ / K.E₁) x 100%
= (1.209 / 693.842) x 100%
= 0.174 %
Therefore, the percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %
•• A metal sphere carrying an evenly distributed charge will have spherical equipotential surfaces surrounding it. Suppose the sphere’s radius is 50.0 cm and it carries a total charge of (a) Calculate the potential of the sphere’s surface. (b)You want to draw equipotential surfaces at intervals of 500 V outside the sphere’s surface. Calculate the distance between the first and the second equipotential surfaces, and between the 20th and 21st equipotential surfaces. (c) What does the changing spacing of the surfaces tell you about the electric field?
Answer:
Explanation:
For this exercise we will use that the potential is created by the charge inside the equinoctial surface and just like in Gauss's law we can consider all the charge concentrated in the center.
Therefore the potential on the ferric surface is
V = k Q / r
where k is the Coulomb constant, Q the charge of the sphere and r the distance from the center to the point of interest
a) On the surface the potential
V = 9 10⁹ Q / 0.5
V = 18 10⁹ Q
Unfortunately you did not write the value of the load, suppose a value to complete the calculations Q = 1 10⁻⁷ C, with this value the potential on the surfaces V = 1800 V
b) The equipotential surfaces are concentric spheres, let's look for the radii for some potentials
for V = 1300V let's find the radius
r = k Q / V
r = 9 109 1 10-7 / 1300
r = 0.69 m
other values are shown in the following table
V (V) r (m)
1800 0.5
1300 0.69
800 1,125
300 3.0
In other words, we draw concentric spheres with these radii and each one has a potential difference of 500V
C) The spacing of the spheres corresponds to lines of radii of the electric field that have the shape
E = k Q / r²
Astronauts increased in height by an average of approximately 40 mm (about an inch and a half) during the Apollo-Soyuz missions, due to the absence of gravity compressing their spines during their time in space. Does something similar happen here on Earth
Answer:
Yes. Something similar occurs here on Earth.
Explanation:
Gravity tends to pull objects perpendicularly to the ground. In space, the absence of this force means there is no compression on the spine due to gravity trying to pull it down. This means that astronauts undergo an increase in height in space.
Here on Earth, we experience gravity pull on our spine during the day. At night when we sleep, we lie down with our spine parallel to the ground, which means that our spine is no longer under compression from gravity force. The result is that we are a few centimetres taller in the morning when we wake up, than we are before going to bed at night. The increase is not much pronounced here on Earth because there is a repeated cycle of compression and decompression of our spine due to gravity, unlike when compared to that of astronauts that spend long duration in space, all the while without gravity forces on their spine
If an astronomer wants to find and identify as many stars as possible in a star cluster that has recently formed near the surface of a giant molecular cloud (such as the Trapezium cluster in the Orion Nebula), what instrument would be best for her to use
Answer:
Infrared telescope and camera
Explanation:
An infrared telescope uses infrared light to detect celestial bodies. The infrared radiation is one of the known forms of electromagnetic radiation. Infrared radiation is given off by a body possessing some form of heat. All bodies above the absolute zero temperature in the universe radiates some form of heat, which can then be detected by an infrared telescope, and infrared radiation can be used to study or look into a system that is void of detectable visible light.
Stars are celestial bodies that are constantly radiating heat. In order to see a clearer picture of the these bodies, Infrared images is better used, since they are able to penetrate the surrounding clouds of dust, and have located many more stellar components than any other types of telescope, especially in dusty regions of star clusters like the Trapezium cluster.
A piece of electronic equipment that is surrounded by packing material is dropped so that it hits the ground with a speed of 4 m/s. After impact, the equipment experiences an acceleration of a = 2kx, where k is a constant and x is the compression of the packing material. If the packing material experiences a maximum compression of 20 mm, determine the maximum acceleration of the equipment.
Answer:
Maximum acceleration is 800m/s^2
Explanation:
See attached file
Two buses are moving in opposite directions with velocities of 36 km/hr and 108
km/hr. Find the distance between them after 20 minutes.
Explanation:
It is given that,
Speed of bus 1 is 36 km/h and speed of bus 2 is 108 km/h. We need to find the distance between bus 1 and 2 after 20 minutes.
Time = 20 minutes = [tex]\dfrac{20}{60}\ h=\dfrac{1}{3}\ h[/tex]
As the buses are moving in opposite direction, then the concept of relative velocity is used. So,
Distance, [tex]d=v\times t[/tex]
v is relative velocity, v = 108 + 36 = 144 km/h
So,
[tex]d=144\ km/h \times \dfrac{1}{3}\ h\\\\d=48\ km[/tex]
So, the distance between them is 48 km after 20 minutes.
Determine the position in the oscillation where an object in simple harmonic motion: (Be very specific, and give some reasoning to your answer.) has the greatest speed has the greatest acceleration experiences the greatest restoring force experiences zero restoring force g
Answer:
Explanation:
The greatest speed is attained at middle point or equilibrium point or where displacement from equilibrium point is zero .
When the object remains at one of the extreme point it experiences greatest acceleration but at that point velocity is zero . Due to acceleration , its velocity goes on increasing till it come to equilibrium point . At this point acceleration becomes zero . After that its velocity starts decreasing because of negative acceleration . Hence at middle point velocity is maximum .
The greatest acceleration is attained at maximum displacement or at one of the two extreme end .
Greatest restoring force too will be at position where acceleration is maximum because acceleration is produced by restoring force .
Restoring force is proportional to displacement or extension against restoring force . So it will be maximum when displacement is maximum .
Zero restoring force exists at equilibrium position or middle point or at point where displacement is zero . It is so because acceleration at that point is zero .
Suppose that a 117.5 kg football player running at 6.5 m/s catches a 0.43 kg ball moving at a speed of 26.5 m/s with his feet off the ground, while both of them are moving horizontally.
(a) Calculate the final speed of the player, in meters per second, if the ball and player are initially moving in the same direction.
(b) Calculate the change in kinetic energy of the system, in joules, after the player catches the ball.
(c) Calculate the final speed of the player, in meters per second, if the ball and player are initially moving in opposite directions.
(d) Calculate the change in kinetic energy of the system, in joules, in this case.
Answer:
a) 6.57 m/s
b) 53.75 J
c) 6.37 m/s
d) -98.297 J
Explanation:
mass of player = [tex]m_{p}[/tex] = 117.5 kg
speed of player = [tex]v_{p}[/tex] = 6.5 m/s
mass of ball = [tex]m_{b}[/tex] = 0.43 kg
velocity of ball = [tex]v_{b}[/tex] = 26.5 m/s
Recall that momentum of a body = mass x velocity = mv
initial momentum of the player = mv = 117.5 x 6.5 = 763.75 kg-m/s
initial momentum of the ball = mv = 0.43 x 26.5 = 11.395 kg-m/s
initial kinetic energy of the player = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.5^{2}[/tex] = 2482.187 J
a) according to conservation of momentum, the initial momentum of the system before collision must equate the final momentum of the system.
for this first case that they travel in the same direction, their momenta carry the same sign
[tex]m_{p}[/tex][tex]v_{p}[/tex] + [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v
where v is the final velocity of the player.
inserting calculated momenta of ball and player from above, we have
763.75 + 11.395 = (117.5 + 0.43)v
775.145 = 117.93v
v = 775.145/117.93 = 6.57 m/s
b) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.57^{2}[/tex] = 2535.94 J
change in kinetic energy = 2535.94 - 2482.187 = 53.75 J gained
c) if they travel in opposite direction, equation becomes
[tex]m_{p}[/tex][tex]v_{p}[/tex] - [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v
763.75 - 11.395 = (117.5 + 0.43)v
752.355 = 117.93v
v = 752.355/117.93 = 6.37 m/s
d) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.37^{2}[/tex] = 2383.89 J
change in kinetic energy = 2383.89 - 2482.187 = -98.297 J
that is 98.297 J lost
Suppose Young's double-slit experiment is performed in air using red light and then the apparatus is immersed in water. What happens to the interference pattern on the screen?
Answer:
The bright fringes will appear much closer together
Explanation:
Because λn = λ/n ,
And the wavelength of light in water is smaller than the wavelength of light in air. Given that the distance between bright fringes is proportional to the wavelength
Sally who weighs 450 N, stands on a skate board while roger pushes it forward 13.0 m at constant velocity on a level straight street. He applies a constant 100 N force.
Work done on the skateboard
a. Rodger Work= 0J
b. Rodger work= 1300J
c. sally work= 1300J
d. sally work= 5850J
e. rodger work= 5850J
Answer:
b. Rodger work = 1300 J
Explanation:
Work done: This can be defined as the product of force and distance along the direction of the force.
From the question,
Work is done by Rodger using a force of 100 N in pushing the skateboard through a distance of 13.0 m.
W = F×d............. Equation 1
Where W = work done, F = force, d = distance.
Given: F = 100 N, d = 13 m
Substitute these values into equation 1
W = 100(13)
W = 1300 J.
Hence the right option is b. Rodger work = 1300 J
Suppose a 185 kg motorcycle is heading toward a hill at a speed of 29 m/s. The two wheels weigh 12 kg each and are each annular rings with an inner radius of 0.280 m and an outer radius of 0.330 m.
Randomized Variables
m = 185 kg
v = 29 m/s
h = 32 m
A. How high can it coast up the hill. if you neglect friction in m?
B. How much energy is lost to friction if the motorcycle only gains an altitude of 33 m before coming to rest?
Answer:
a) Height reached before coming to rest is 42.86 m
b) Energy lost to friction is 17902.45 J
Explanation:
mass of the motorcycle = 185 kg
speed of the towards the hill = 29 m/s
The wheels weigh 12 kg each
Wheels are annular rings with an inner radius of 0.280 m and outer radius of 0.330 m
a) To go up the hill, the kinetic energy of motion of the motorcycle will be converted to the potential energy it will gain in going up a given height
the kinetic energy of the motorcycle is given as
[tex]KE[/tex] = [tex]\frac{1}{2}mv^{2}[/tex]
where m is the mass of the motorcycle
v is the velocity of the motorcycle
[tex]KE[/tex] = [tex]\frac{1}{2}*185*29^{2}[/tex] = 77792.5 J
This will be converted to potential energy
The potential energy up the hill will be
[tex]PE[/tex] = mgh
where m is the mass
g is acceleration due to gravity 9.81 m/s^2
h is the height reached before coming to rest
[tex]PE[/tex] = 185 x 9.81 x m = 1814.85h
equating the kinetic energy to the potential energy for energy conservation, we'll have
77792.5 = 1814.85h
height reached before coming to rest = 77792.5/1814.85 = 42.86 m
b) if an altitude of 33 m was reached before coming to rest, then the potential energy at this height is
[tex]PE[/tex] = mgh
[tex]PE[/tex] = 185 x 9.81 x 33 = 59890.05 J
The energy lost to friction will be the kinetic energy minus this potential energy.
energy lost = 77792.5 - 59890.05 = 17902.45 J
A) The motorcycle can coast up the hill by ; 42.86m
B) The amount of energy lost to friction : 17902.45 J
A) Determine how high the motorcycle can coast up the hill when friction is neglected
apply the formula for kinetic and potential energies
K.E = 1/2 mv² ---- ( 1 )
P.E = mgH ---- ( 2 )
As the motorcycle goes uphiLl the kinetic energy is converted to potential energy.
∴ K.E = P.E
1/2 * mv² = mgH
∴ H = ( 1/2 * mv² ) / mg ---- ( 3 )
where ; m = 185 kg , v = 29 m/s , g = 9.81
Insert values into equation ( 3 )
H ( height travelled by motorcycle neglecting friction ) = 42.86m
B) Determine how much energy is lost to friction if the motorcycle attains 33m before coming to rest
P.E = mgh = 185 * 9.81 * 33 = 59890.05 J
where : h = 33 m , g = 9.81
K.E = 1/2 * mv² = 77792.5 J ( question A )
∴ Energy lost ( ΔE ) = ( 77792.5 - 59890.05 ) = 17902.45 J
Hence we can conclude that The motorcycle can coast up the hill by ; 42.86m , The amount of energy lost to friction : 17902.45 J.
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Three resistors, 6.0-W, 9.0-W, 15-W, are connected in parallel in a circuit. What is the equivalent resistance of this combination of resistors?
Answer:
2.9Ω
Explanation:
Resistors are said to be in parallel when they are arranged side by side such that their corresponding ends are joined together at two common junctions. The combined resistance in such arrangement of resistors is given by;
1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn
Where;
Req refers to the equivalent resistance and R1, R2, R3 .......Rn refers to resistance of individual resistors connected in parallel.
Note that;
R1= 6.0Ω
R2 = 9.0Ω
R3= 15.0 Ω
Therefore;
1/Req = 1/6 + 1/9 + 1/15
1/Req= 0.167 + 0.11 + 0.067
1/Req= 0.344
Req= (0.344)^-1
Req= 2.9Ω
The equivalent resistance of this combination of resistors is 2.9Ω.
Calculation of the equivalent resistance:The combined resistance in such arrangement of resistors is provided by;
1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn
here.
Req means the equivalent resistance and R1, R2, R3
.Rn means the resistance of individual resistors interlinked in parallel.
Also,
R1= 6.0Ω
R2 = 9.0Ω
R3= 15.0 Ω
So,
1/Req = 1/6 + 1/9 + 1/15
1/Req= 0.167 + 0.11 + 0.067
1/Req= 0.344
Req= (0.344)^-1
Req= 2.9Ω
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A person is nearsighted with a far point of 75.0 cm. a. What focal length contact lens is needed to give him normal vision
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]f= -75 \ cm = - 0.75 \ m[/tex]
b
[tex]P = -1.33 \ diopters[/tex]
Explanation:
From the question we are told that
The image distance is [tex]d_i = -75 cm[/tex]
The value of the image is negative because it is on the same side with the corrective glasses
The object distance is [tex]d_o = \infty[/tex]
The reason object distance is because the object father than it being picture by the eye
General focal length is mathematically represented as
[tex]\frac{1}{f} = \frac{1}{d_i} - \frac{1}{d_o}[/tex]
substituting values
[tex]\frac{1}{f} = \frac{1}{-75} - \frac{1}{\infty}[/tex]
=> [tex]f= -75 \ cm = - 0.75 \ m[/tex]
Generally the power of the corrective lens is mathematically represented as
[tex]P = \frac{1}{f}[/tex]
substituting values
[tex]P = \frac{1}{-0.75}[/tex]
[tex]P = -1.33 \ diopters[/tex]