Answer:
Vx' = (Vx - u) / (1 - Vx *u / c^2) velocity transformation formula
In both cases we wish to measure the velocity in the frame of the earth which is moving at speed u = -.99 c relative to the spaceship
VA' = (c + .99c) / (1 - (-.99 c * c) / c^2) = 1.99c / 1.99 = c
VB' = (-c + .99c) / (1 - (-c * -.99c) / c^2) = .01 c / .01 = c
In both cases an observer on earth will observe the light traveling at speed c.
A 800 kg safe is 2.1 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 46 cm. What is the spring constant of the spring?
Answer:
k = 17043.5 N/m = 17.04 KN/m
Explanation:
First we need to find the force applied by safe pn the spring:
F = Weight of Safe
F = mg
where,
F = Force Applied by the safe on the spring = ?
m = mass of safe = 800 kg
g = 9.8 m/s²
Therefore,
F = (800 kg)(9.8 m/s²)
F = 7840 N
Now, using Hooke's Law:
F = kΔx
where,
K = Spring Constant = ?
Δx = compression = 46 cm = 0.46 m
Therefore,
7840 N = k (0.46 m)
k = 7840 N/0.46 m
k = 17043.5 N/m = 17.04 KN/m
A spring balance is attached with string to the piece of aluminum in the preceding problem. What reading will the balance register when the metal is submerged
A toboggan is sliding down an icy slope. As it goes down, _________ does work on the toboggan and ends up converting __________ energy to _________ energy.
Answer:
As it goes down, weight does work on the toboggan and it ends up converting gravitational potential energy to kinetic energy.
1. weight
2. gravitational potential energy to kinetic energy.
Explanation:
As it goes down, weight does work on the toboggan and it ends up converting gravitational potential energy to kinetic energy.
work done by toboggan = weight × distance
W = mg and the distance is down the icy slope
By using law of conservation of energy, energy can neither be created nor destroyed, but can be conserve from one form to another in a closed system.
Toboggan converts gravitational potential energy (mgh) to kinetic energy(¹/₂mv²)
g How many rpm would a 25 m diameter Ferris wheel need to travel if a 75 kg person were to experience an effective weight of 810 N at the lower-most point of the ride
Answer:
2.52 rpm
Explanation:
given that
diameter of the wheel, d = 25 m
Mass of the person, m = 75 kg
Weight experienced, N = 810 N
Since diameter is 25, radius then is 25/2 = 12.5 m
We all know that,
v = rw
Also, the passengers weight is equal to the centripetal acceleration, and thus
mg = mv²/r
Substitute for v, we have
mg = m/r * (rw)²
mg = mr²w²/r
g = rw²
If we make w the subject of formula, we have
w² = g/r
w = √(g/r)
mg = 810
75 * g = 810
g = 810 / 75
g = 1.08 m/s²
w = √(g/r)
w = √(1.08 / 12.5)
w = √0.0864
w = 0.294 rad/s
Since the question asked us in rpm, we convert to rpm
0.294 * (60 / 2π)
2.52 revolution per minute.
In a high school swim competition, a student takes 1.6 s to complete 2.0 somersaults. Determine the average angular speed of the diver, in rad/s, during this time interval.
Answer:
9.82 rads/sec
Explanation:
We are given;
Time taken; t = 1.6 secs
Number of somersaults = 2
Now, we know that,
1 revolution = 2π radians
And number of somersaults is the same thing as number of revolutions
So,
Total radians = 2π × 2 = 5π
Angular velocity = total number of revolutions/time period = 5π/1.6 = 9.82 rads/sec
7. How many 1.00 µF capacitors must be connected in parallel to store a charge of 1.00 C with a potential of 110 V across the capacitors?
Answer:
q = C V charge on 1 capacitor
q = 1 * 10E-6 * 110 = 1.1 * 10E-4 C per capacitor
N = Q / q = 1 / 1.1 * 10E-4 = 9091 capacitors
9.09 × 10³ capacitors must be connected in parallel.
How to calculate the number of capacitors connected in parallel?
Given C = 1.00μF = 1 × 10⁻⁶ F
q = 1.00 C
V = 110 V
The equivalent capacitance is given by
Ceq = q/V
where q = total charge on all the capacitors
V = potential difference
For N number of identical capacitors in parallel,
Ceq = NC
Therefore,
NC = q/V
N = q/VC
Putting on the values in the above formula,
N = 1/ (110)(1 × 10⁻⁶)
= 1 / 110 × 10⁻⁶
= 9.09 × 10³
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Use Kepler's third law to determine how many days it takes a spacecraft to travel in an elliptical orbit from a point 6 965 km from the Earth's center to the Moon, 385 000 km from the Earth's center.
Answer:
0.0665 days
Explanation:
We are given;
The mean distance from the Earth's center to the moon;a1 = 385000 km
The mean distance from the Earth's center to the space craft;a2 = 6965 km
Formula for kepplers third law is;
T² = 4π²a³/GM
However, the proportion of both distances would be;
(T1)²/(T2)² = (a1)³/(a2)³
Where;
T1 is the period of orbit of the moon around the earth. T1 has a standard value of 27.322 days
T2 is the period of the space craft orbit.
Making T2 the subject, we have;
T2 = √((T1)²×(a2)³)/(a1)³)
Thus, plugging in the relevant values;
T2 = √(27.322² × 6965³)/(385000)³
T2 = 0.0665 days
You set two parallel slits 0.1 mm apart at a distance of 2 m from a screen and illuminate them with light of wavelength 450 nm. The distance between a bright spot in the interference pattern and the dark spot adjacent to it is Group of answer choices
Answer:
Δx = 9 x 10⁻³ m = 9 mm
Explanation:
The formula for fringe spacing in Young's Double Slit Experiment is given as follows:
Δx = λL/d
where,
Δx = fringe spacing = ?
λ = wavelength of light = 450 nm = 450 x 10⁻⁹ m
L = Distance between slits and screen = 2 m
d = distance between slits = 0.1 mm = 0.1 x 10⁻³ m
Therefore,
Δx = (450 x 10⁻⁹ m)(2 m)/(0.1 x 10⁻³ m)
Δx = 9 x 10⁻³ m = 9 mm
A cylindrical shell of radius 7.00 cm and length 2.59 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 20.1 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C.
A) Use approximate relationships to find the net charge on the shell.
B) Use approximate relationships to find the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell.
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 1.8 kg microphone is connected to a spring and is oscillating in simple harmonic motion up and down with a period of 3s. Below the microphone is 1.8 hz, calculate the spring constant
Answer:
230N/m
Explanation:
Pls see attached file
Comparing helium atoms with nitrogen molecules at the same temperature, the helium atoms on average are moving _______ and have _______ kinetic energy.
Answer:
Helium atoms compared to nitrogen atoms are moving faster and have a greater kinetic energy.
Explanation:
The molecular velocity of a gas at room temperature is inverse proportional to the square root of its molecular mass.
The greater the molecular mass of the gas the lesser the average speed of its molecules. Comparing the molecular masses of nitrogen and helium, helium is found to have a lower molecular mass and a corresponding greater velocity.
Hence helium moves faster than nitrogen and has a higher kinetic energy than nitrogen
The maximum velocity of a particle executing simple harmonic motion with amplitude 7.0 mm is 4.4 m/s. The period of oscillation is? A. 0.5 s B. 0.1 s C. 100 s D. 0.01 s E. 10 s Reset Selection
Answer:
D. 0.01 s
Explanation:
The maximum speed is the amplitude times the angular frequency.
v = Aω
4.4 m/s = (0.0070 m) ω
ω = 628.6 rad/s
The period is:
ω = 2π / T
T = 2π / ω
T = 2π / (628.6 rad/s)
T = 0.01 s
To test the resiliency of its bumper during low-speed collisions, a 2 010-kg automobile is driven into a brick wall. The car's bumper behaves like a spring with a force constant 4.00 106 N/m and compresses 3.18 cm as the car is brought to rest. What was the speed of the car before impact, assuming no mechanical energy is transformed or transferred away during impact with the wall?
Answer:
Vi = 2 m/s
Explanation:
First we find the force applied to the car by wall to stop it. We use Hooke's Law:
F = kx
where,
F = Force = ?
k = spring constant = 4 x 10⁶ N/m
x = compression = 3.18 cm = 0.0318 m
Therefore,
F = (4 x 10⁶ N/m)(0.0318 m)
F = 127200 N
but, from Newton's Second Law:
F = ma
a = F/m
where,
m = mass of car = 2010 kg
a = deceleration = ?
Therefore,
a = 127200 N/2010 kg
a = 63.28 m/s²
a = - 63.28 m/s²
negative sign due to deceleration.
Now, we use 3rd equation of motion:
2as = Vf² - Vi²
where,
s = distance traveled = 3.18 cm = 0.0318 m
Vf = Final Speed = 0 m/s
Vi = Initial Speed = ?
Therefore,
2(- 63.28 m/s²)(0.0318 m) = (0 m/s)² - Vi²
Vi = √4.02 m²/s²
Vi = 2 m/s
An electron is released from rest at a distance of 9.00 cm from a proton. If the proton is 11) held in place, how fast will the electron be moving when it is 3.00 cm from the proton?
Answer:
Vf = 1.43 m/s
Explanation:
From Coulomb's Law, the electrostatic force between electron and proton is given as:
F = kq₁q₂/r²
F = Electrostatic force = ?
k = Coulomb's Constant = 9 x 10⁹ N.m²/C²
q₁ = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
q₂ = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C
r = distance between electron and proton = 9 cm = 0.09 m
Therefore,
F = (9 x 10⁹ N.m²/C²)(1.6 x 10⁻¹⁹ C)(1.6 x 10⁻¹⁹ C)/(0.09 m)²
F = 2.84 x 10⁻²⁶ N
but, from Newton's second law:
F = 2.84 x 10⁻²⁶ N = ma
where,
m = mass of electron = 9.1 x 10⁻³¹ kg
a = acceleration of electron = ?
Therefore,
2.84 x 10⁻²⁶ N = (1.67 x 10⁻²⁷ kg)(a)
a = 2.84 x 10⁻²⁶ N/1.67 x 10⁻²⁷ kg
a = 17.03 m/s²
Now, we apply 3rd equation of motion to the motion of electron from a distance of 9 cm to 3 cm near to the proton:
2as = Vf² - Vi²
where,
s = distance traveled = 9 cm - 3 cm = 6 cm = 0.06 m
Vf = speed of electron when it is 3 cm from proton = ?
Vi = Initial speed of electron = 0 m/s
Therefore,
2(17.03 m/s²)(0.06 m) = Vf² - (0 m/s)²
Vf = √2.04 m²/s²
Vf = 1.43 m/s
Unpolarized light is incident upon two polarization filters in sequence. The two filters transmission axes are not aligned. If 18% of the incident light passes through this combination of filters, what is the angle between the transmission axes of the filters
Answer:
53°
Explanation:
I/Io*2= 0.18
0.18= cos²theta
Cos^-1(0.36) = 53°
The primary of an ideal transformer has 100 turns and its secondary has 200 turns. If the input current at the primary is 100 A, we can expect the output current at the secondary to be
Answer:
Explanation:
For current in ideal transformer the formula is
I₁ / I₂ = N₂ / N₁
I₁ and I₂ are current in primary and secondary coil respectively and N₁ and N₂ are no of turns in primary and secondary coil .
Putting the given values
100 / I₂ = 200 / 100 = 2
I₂ = 50 A .
output current = 50 A .
a football is kicked toward a goal keeper with an initial speed of 20m/s at an angle of 45 degrees with the horizontal .at the moment the ball is kicked the goal keeper is 50m from the player .at what speed and in what direction must the goalkeeper run in order to catch the ball at the same height at which it was kicked
Answer:
3.18 m/s
Explanation:
Given that
Initial speed of the ball, u = 20 m/s
Angle of inclination, θ = 45°
Distance from the ball, h = 50 m
Using equations of projectile to solve this, we have
We start by finding the time of flight, T
T = 2Usinθ/g
T = (2 * 20 * sin45)/9.8
T = (40 * 0.7071) / 9.8
T = 28.284/9.8
T = 2.89 s
Next we find the Range, R
R = u²sin2θ/g
R = (20² * sin 90) / 9.8
R = (400 * 1) / 9.8
R = 400/9.8 = 40.82 m
Distance the gk must cover
40.82 - 50 m
-9.18 m or 9.18 m in the opposite direction.
Speed of the GK = d/t
9.18 / 2.89 = 3.18 m/s
differences between
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and horse
What is the relationship between the surface area of a parachute and the amount of air resistance it builds up when it is deployed by a sky diver?
Answer:
An open parachute increases the cross-sectional area of the falling skydiver and thus increases the amount of air resistance which he encounters. Once the parachute is opened, the air resistance overwhelms the downward force of gravity.
Explanation:
The larger a parachute, the greater the force.
Hope it helps you in a little way.
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.
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.
In what directions is it possible that the wave is traveling?
A. The-z direction.
B. The ty direction
C. The +x direction.
D. The -y direction
E. The -x direction.
F. The +z direction.
Answer:
The wave will be travelling in the y-axis
Explanation:
An e-m wave has a spatially varying electric field that is always associated with a magnetic field that changes over time and vice versa. The electric field and the magnetic field oscillates perpendicularly to each other, and together form a wave that travels in a perpendicular direction to the magnetic and the electric field in space. The movement of the e-m wave through space is usually away from the source where it is generated. So, if the electric field travels in the z-axis, and the magnetic field travels through along the x-axis, then the e-m wave generated will travel in the y-axis direction.
In a single-slit diffraction experiment, the width of the slit through which light passes is reduced. What happens to the width of the central bright fringe
Answer:
It becomes wider
Explanation:
Because The bigger the object the wave interacts with, the more spread there is in the interference pattern. Decreasing the size of the opening increases the spread in the pattern.
A positively charged particle has a velocity in the negative z direction at point P. The magnetic force on the particle at this point is in the negative y direction. Which one of the following statements about the magnetic field at point P can be determined from this data?
A. Bx is positive.
B. Bz is positive.
C. By is negative.
D. By is positive.
E. Bx is negative.
Answer:
When reviewing the correct answer is A
Explanation:
The magnetic force is given by the expression
F = qv xB
where the bold letters indicate vectors, from this expression the module can be calculated
F = = q v b sin θ
the direction of the force is given by the rule of the right hand, for a positive charge the speed held by the thumb, the extended fingers point in the direction of the magnetic field and the palm points the direction of the force
in this case
the speed is in the negative part of the z axis
the force is in the negative direction of the axis and
consequently the magnetizing field is in the positive direction of the x axis
When reviewing the correct answer is A
what is transmission of heat?
Answer:
Heat transfer is the transmission of heat energy from a body at higher temperature to lower temperature. The three mechanisms of heat transfer are
Conduction ConvectionRadiation.Example of Conduction:
Heating a metal
Example of Convection:
Sea Breeze
Example of Radiation:
Sun
Hope this helps ;) ❤❤❤
Answer:
Transmission of heat is the movement of thermal energy from one thing to another thing of different temperature.
There are three(3) different ways heat can transfer and they are:
a) Conduction (through direct contact).
b) Convection (through fluid movement).
c) Radiation (through electromagnetic waves).
Examples: 1.Heating a saucepan of water using a coalpot.(conduction&convection).
2. Baking a pie in an oven(radiation).
Hope it helps!!Please mark me as the brainliest!!!Thanks!!!!❤❤❤
Consider a long rod of mass, m, and length, l, which is thin enough that its width can be ignored compared to its length. The rod is connected at its end to frictionless pivot.
a) Find the angular frequency of small oscillations, w, for this physical pendulum.
b) Suppose at t=0 it pointing down (0 = 0) and has an angular velocity of 120 (that is '(t = 0) = 20) Note that 20 and w both have dimensions of time-1. Find an expression for maximum angular displacement for the pendulum during its oscillation (i.e. the amplitude of the oscillation) in terms of 20 and w assuming that the angular displacement is small.
Answer:
Explanation:
The rod will act as pendulum for small oscillation .
Time period of oscillation
[tex]T=2\pi\sqrt{\frac{l}{g} }[/tex]
angular frequency ω = 2π / T
= [tex]\omega=\sqrt{\frac{g}{l} }[/tex]
b )
ω = 20( given )
velocity = ω r = ω l
Let the maximum angular displacement in terms of degree be θ .
1/2 m v ² = mgl ( 1 - cosθ ) ,
[ l-lcosθ is loss of height . we have applied law of conservation of mechanical energy .]
.5 ( ω l )² = gl( 1 - cos θ )
.5 ω² l = g ( 1 - cosθ )
1 - cosθ = .5 ω² l /g
cosθ = 1 - .5 ω² l /g
θ can be calculated , if value of l is given .
A metal ring 4.00 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.200 T/s.
(a) What is the magnitude of the electric field induced in the ring?
(b) In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?
Explanation:
a) d[phi]/dt = (dB/dt)*Acos(0) = (-0.20)*(pi(2.25*10^-2)^2) = -3.98*10^-4 Wb
E = (1/2r*pi)*(d[phi]/dt) = -2.8*10^-3 N/C
b) Clockwise because The induced magnetic field will be in the direction to oppose the change. Since the magnetic flux from the magnets is decreasing, the induced magnetic field will be in the same direction as the magnet's field.
What is the minimum magnitude of an electric field that balances the weight of a plasticsphere of mass 5.4 g that has been charged to -3.0 nC
Answer:
E = 17.64 x 10⁶ N/C = 17.64 MN/C
Explanation:
The electric field is given by the following formula:
E = F/q
E= W/q
E = mg/q
where,
E = magnitude of electric field = ?
m = mass of plastic sphere = 5.4 g = 5.4 x 10⁻³ kg
g = acceleration due to gravity = 9.8 m/s²
= charge = 3 nC = 3 x 10⁻⁹ C
Therefore,
E = (5.4 x 10⁻³ kg)(9.8 m/s²)/(3 x 10⁻⁹ C)
E = 17.64 x 10⁶ N/C = 17.64 MN/C
In your words, describe how momentum is related to energy.
Answer:
you need momentum in order to release energy. For example, if you need to push something heavy and you get a running head start, then it will be easier.
Explanation:
A body of mass 2.5kg is raised 4.0m above the ground.Calculate the potential energy if g=10m/s squared
Answer:
100 joulesExplanation:
[tex]mass = 2.5kg\\height = 4.0\\Acceleration \: due \:to\:gravity = 10m/s^2\\\\P.E = mgh\\P.E = 2.5kg\times10\times4\\\\Potential \: Energy = 100 joules[/tex]
The potential energy if g = 10m/s² is 98 J
What is Potential Energy ?Potential energy is the energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors.
Formula for potential energy :
P.E. = mgh
m = mass (kg)g = gravity (m/s²)h = height/distance (m)Given :
The book is held from the ground of a distance = 4.0 m,
so
h = 4.0 m
we know that the book weighs,
2.5 kg
so
m = 2.5 kg.
Now we just put it in the formula ;
PE = (2.5kg) × (9.8 m/s²) × ( 4.0 m)
P E = 98 J
Therefore, The potential energy if g = 10m/s² is 98 J
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