When you ride a bicycle, the direction of the angular velocity of the wheels is; Option A; to your left
Complete question is;
When you ride a bicycle, in what direction is the angular velocity of the wheels? A) to your left B) to your right C) forwards D) backwards
While an object rotates, each particle will have a different velocity:
the 'Speed' component will vary with radius while the 'Direction' component will vary with angle.
Now, all of the velocity vectors are aligned in the same plane and as such we can be solve this by choosing a single vector normal to ALL of the possible velocity vectors of the rotating object in that plane.
The convention that will be used to answer this question is known as "Right-hand rule". The angular velocity vector points along the wheel's axle.
For instance, if you Imagine wrapping your right hand around the axle so that your fingers point in the direction of rotation, with your thumb sticking out. You will notice that your thumb points to the left.
Thus;
In conclusion, by right-hand rule, a wheel rotating on a forward - moving bicycle has an angular velocity vector pointing to the rider's left.
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A particle with a charge of 5.1 μC is 3.02 cm from a particle with a charge of 2.51 μC . The potential energy of this two-particle system, relative to the potential energy at infinite separation, is
Answer:
U = 3.806 J
Explanation:
The potential energy between the two charges q1 and q2, is given by the following formula:
[tex]U=k\frac{q_1q_2}{r}[/tex] (1)
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
q1 = 5.1*10^-6 C
q2 = 2.51*10^-6 C
r: distance of separation between particles = 3.02cm = 3.02*10^-2 m
You replace the values of all parameters in the equation (1):
[tex]U=(8.98*10^9Nm^2/C^2)\frac{(5.1*10^{-6}C)(2.51*10^{-6}C)}{3.02*10^{-2}m}\\\\U=3.806J[/tex]
The potential energy of the two particle system is 3.806 J
Mark Watney (Matt Damon in the Martian movie) and Marvin the Martian (Looney Tunes cartoon character) are having an argument on the surface of Mars (negligible air resistance). They are testing out their new potato launcher that fires projectiles at a constant speed. Mark launches his potato at an angle of 60◦ and Marvin launches his identical potato at an angle of 30◦ . Without any calculations try to answer the following questions, and justify each answer.
(A) Which potato lands farther away from the launcher (potatoes are launched from ground level)?
(B) Which potato spends more time in the air before hitting the ground
(C) Which potato has a greater speed just before it hits the ground?
Answer:
A) The two potatoes cover the same horizontal distance from the launcher.
B) Mark's potato spends more time in the air than Marvin's potato before hitting the ground.
C) Marvin's potato hits the ground with a greater speed than Mark's potato
Explanation:
A) For projectile motion, the final horizontal distance of the projectile from where it was initially launched (its range) is given as
R = (u² sin 2θ)/g
where
u = initial velocity of the projectile
θ = angle above the horizontal at which the projectile was launched = 30°, 60°
g = acceleration due to gravity on Mars
Since, u and g are the same for Mark and Marvin, sin 2θ would determine which range is higher.
Sin (2×60°) = sin 120°
Sin (2×30°) = sin 60°
Sin 120° = Sin 60°
Hence, the two potatoes cover the same horizontal distance from the launcher.
B) Time spent in the air for a projectile is given as
T = (2u sin θ)/g
Again, since u and g are the same for Mark and Marvin on Mars, sin θ will give the required idea of whose potato spends more time in the air.
Sin 60° = 0.866
Sin 30° = 0.50
Sin 60° > Sin 30°
Hence, Mark's potato spends more time in the air than Marvin's potato.
C) The horizontal velocity for projectile motion is constant all through the motion and is equal to u cos θ
u cos 60° < u cos 30°
And the initial vertical velocity is u sin θ
Final vertical velocity
= (initial vertical velocity) - gt
g = acceleration due to gravity on Mars
T = time of flight
For Mark,
initial vertical velocity = u sin 60°, greater than Marvin's u sin 30°
And Mark's potato's time of flight is greater as established in (B) above.
But for Marvin
initial vertical velocity = u sin 30°, less than Mark's u sin 60°
And Marvin's potato's time of flight is lesser as established in (B) above
So, at the end of the day, the final vertical velocity is almost the same for both Mark's and Marvin's potatoes.
Hence, the horizontal component of the final velocity edges the final speed of the potatoes just before hitting the ground in Marvin's favour.
Hope this Helps!!!
1. In 214 BC, Archimedes invented a large spherical-type mirror used to focus the sun's intense rays onto far away enemy boats, which would eventually light them on fire. If the boats were travelling in a nearby channel approximately 1,000 m from the river bank, what would the radius of curvature of his mirror need to be? Show your work.
Answer:
2000 m
Explanation:
since the boat is 1000 m from the river bank, the beam must be focused at this point. This indicates that the focal length is 1000 m
for a spherical mirror, the focal length is given by
f = R/2
where R is the radius of curvature
1000 = R/2
R = 2000 m
R = 2000 m
this means that the radius of curvature must be 2000 m
How much displacement will a spring with a constant of 120N / m achieve if it is stretched by a force of 60N?
Answer:
Explanation:
There's a formula for this:
[tex]F = k*displacement[/tex]
F being force, k being the spring constant, and displacement being the change in x
We are given the force and the spring constant, so this is essentially isolating the Δx term. Do 60N/120N per meter. The newtons cancel out and you get a final answer of Δx = 0.5 meters
A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 5.37x10-2 kg/s. The density of the gasoline is 739 kg/m3, and the radius of the fuel line is 3.37x10-3 m. What is the speed at which gasoline moves through the fuel line
Answer:
Speed v = 2.04 m/s
the speed at which gasoline moves through the fuel line is 2.04 m/s
Explanation:
Given;
Mass transfer rate m = 5.37x10^-2 kg/s.
Density d = 739 kg/m3
radius of pipe r = 3.37x10^-3 m
We know that;
Density = mass/volume
Volume = mass/density
Volumetric flow rate V = mass transfer rate/density
V = m/d
V = 5.37x10^-2 kg/s ÷ 739 kg/m3
V = 0.00007266576454 m^3/s
V = 7.267 × 10^-5 m^3/s
V = cross sectional area × speed
V = Av
Area A = πr^2
V = πr^2 × v
v = V/πr^2
Substituting the given values;
v = 7.267 × 10^-5 m^3/s/(π×(3.37x10^-3 m)^2))
v = 0.203678639672 × 10 m/s
v = 2.04 m/s
the speed at which gasoline moves through the fuel line is 2.04 m/s
1. Describe what must happen to an atom to make it
A. A cation
B. An anion
2. Describe why some acids are strong while other acids are weak
3. Compare protons, neutrons and electron, listing their similarities and differences
4. Explain why you breathe faster and deeper when exercising
Answer:
Explanation:
Atoms—and the protons, neutrons, and electrons that compose them—are extremely small. For example, a carbon atom weighs less than 2 × 10−23 g, and an electron ... The amu was originally defined based on hydrogen, the lightest element, ... but three-letter symbols have been used to describe some elements that have ...
Protons: Protons are positively charged particles that are also found in the nucleus. Like neutrons, protons give mass to the atom but do not participate in ... 3) Electrons: Electrons are negatively charged particles that are found in ... pair of electrons with 4 different hydrogen atoms, forming a molecule of CH4 (methane).Elements differ from each other in the number of protons they have, e.g. ... Atoms of an element that have differing numbers of neutrons (but a constant atomic ... Electrons, because they move so fast (approximately at the speed of light), ...toms are made up of particles called protons, neutrons, and electrons, which ... Therefore, they do not contribute much to an element's overall atomic mass. ... For instance, iron, Fe, can exist in its neutral state, or in the +2 and +3 ionic states. ... Isotopes of the same element will have the same atomic number but different ...
In Physics lab, a lab team places a cart on one of the horizontal, linear tracks with a fan attached to it. The cart is positioned at one end of the track, and the fan is turned on. Starting from rest, the cart takes 4.34 s to travel a distance of 1.62 m. The mass of the cart plus fan is 354 g. Assume that the cart travels with constant acceleration.
A) What is the net force exerted on the cart-fan combination?B) Mass is added to the cart until the total mass of the cart-fan combination is 762 g, and the experiment is repeated. How long does it take for the cart, starting from rest, to travel 1.62 m now?
Answer:
A. F = 0.06 N
B. t = 6.37 s
Explanation:
A)
First we need to find the constant acceleration of the cart. For this purpose, we use 2nd equation of motion:
s = (Vi)(t) + (0.5)at²
where,
s = distance traveled = 1.62 m
Vi = 0 m/s (Since, it starts from rest)
t = Time Taken = 4.34 s
a = acceleration = ?
Therefore,
1.62 m = (0 m/s)(4.34 s) + (0.5)(a)(4.34 s)²
1.62 m/9.4178 s² = a
a = 0.172 m/s²
Now, from Newton's Second law, we know that:
F = ma
where,
F = Net Force of the combination = ?
m = Mass pf combination = 354 g = 0.354 kg
Therefore,
F = (0.354 kg)(0.172 m/s²)
F = 0.06 N
B)
Now, for the same force, but changed mass = 762 g = 0.762 kg, we have the acceleration to be:
F = ma
a = F/m
a = 0.06 N/0.762 kg
a = 0.08 m/s²
Now, using 2nd equation of motion:
s = (Vi)(t) + (0.5)at²
1.62 m = (0 m/s)(t) + (0.5)(0.08 m/s²)t²
t² = 1.62 m/(0.04 m/s²)
t = √40.54 s²
t = 6.37 s
A 1.70 m tall woman stands 5.00 m in front of a camera with a 50.00 cm focal
length lens. Calculate the size of the image formed on flim
Answer:
18.89cm
Explanation:
As we know that the person is standing 5m in front of the camera
[tex]d_0=5m=500cm[/tex]
The focal length of the lens =50cm
f=50 cm
By Lens formula we have:
[tex]\dfrac{1}{f} = \dfrac{1}{d_i} + \dfrac{1}{d_o}\\\dfrac{1}{50} = \dfrac{1}{d_i} + \dfrac{1}{500}\\\dfrac{1}{d_i} =\dfrac{1}{50}-\dfrac{1}{500}\\\dfrac{1}{d_i}=0.018\\d_i=55.56cm[/tex]
By the formula of magnification
[tex]\dfrac{h_i}{h_o} = \dfrac{55.56}{500}\\\\h_i = \dfrac{55.56}{500} \times h_o\\\\ h_o=1.70m=170cm\\\\Therefore: h_i=\dfrac{55.56}{500} \times$ 170 cm\\\\h_i =18.89 cm[/tex]
The height of the image formed is 18.89cm.
g The potential energy of a pair of hydrogen atoms separated by a large distance x is given by U(x)=−C6/x6, where C6 is a positive constant. Part A What is the force that one atom exerts on the other? Express your answer in terms of C6 and x. Fx = nothing Request Answer Part B Is this force attractive or repulsive? Is this force attractive or repulsive? attractive repulsive
Answer:
[tex]F_x = -\frac{6 C_6}{2^7}[/tex]
Attractive
Explanation:
Data provided in the question
The potential energy of a pair of hydrogen atoms given by [tex]\frac{C_6}{X_6}[/tex]
Based on the given information, the force that one atom exerts on the other is
Potential energy μ = [tex]\frac{C_6}{X_6}[/tex]
Force exerted by one atom upon another
[tex]F_x = \frac{\partial U}{\partial X} = \frac{\partial}{\partial X} (-\frac{C_6}{X^6})[/tex]
or
[tex]F_x = \frac{\partial}{\partial X} (\frac{C_6}{X^6})[/tex]
or
[tex]F_x = -\frac{6 C_6}{2^7}[/tex]
As we can see that the [tex]C_6[/tex] comes in positive and constant which represents that the force is negative that means the force is attractive in nature
A parallel-plate capacitor has square plates that are 7.20 cm on each side and 3.40 mm apart. The space between the plates is completely filled with two square slabs of dielectric, each 7.20 cm on a side and 1.70 mm thick. One slab is Pyrex glass and the other slab is polystyrene. If the potential difference between the plates is 96.0 V, find how much electrical energy (in nJ) can be stored in this capacitor.
Answer:
U = 218 nJ
Explanation:
We are given;
Spacing between the plates; d = 3.4 mm = 3.4 × 10^(-3) m
Voltage across the capacitor; V = 96 V
Dimension of the square plates is 7.2cm x 7.2cm.
So, Area = 7.2 × 7.2 = 51.84 cm² = 51.84 × 10^(-4) m²
Permittivity of free space; ε_o = 8.85 × 10^(-12) C²/N.m²
From relative permeability table;
Dielectric constant of Pyrex; k1 = 5.6
Dielectric constant of polystyrene; k2 = 2.56
Now, formula for capacitance of a capacitor with Dielectric is;
C = kC_o
Where, C_o = ε_o(A/d)
Since there are 2 capacitors, d will now be d/2 = (3.4 × 10^(-3))/2 m = 1.7 × 10^(-3)
Since we have 2 capacitor, thus ;
C1 = k1*ε_o*(A/d)
C1 = (5.6 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))
C1 = 1.51 × 10^(-10) F
Similarly;
C2 = (2.56 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))
C2 = 0.691 × 10^(-10) F
For capacitors in series, formula for total capacitance(Cs) is;
1/Cs = (1/C1) + (1/C2)
Simplifying this, we have;
Cs = (C1*C2)/(C1 + C2)
Plugging in the relevant values ;
Cs = (1.51 × 10^(-10)*0.691 × 10^(-10))/((1.51 × 10^(-10)) + (0.691 × 10^(-10)))
Cs = 0.474 × 10^(-10) F
The formula for energy stored in a capacitor with 2 Dielectrics is given as;
U = ½Cs*V²
So,
U = ½ × 0.474 × 10^(-10) × 96²
U = 2.18 × 10^(-7) J = 218 × 10^(-9) = 218 nJ
A projectile is fired from ground level with an initial speed of 55.6 m/s at an angle of 41.2° above the horizontal. (a) Determine the time necessary for the projectile to reach its maximum height. (b) Determine the maximum height reached by the projectile. (c) Determine the horizontal and vertical components of the velocity vector at the maximum height. (d) Determine the horizontal and vertical components of the acceleration vector at the maximum heigh
Answer:
(a) t = 3.74 s
(b) H = 136.86 m
(c) Vₓ = 41.83 m/s, Vy = 0 m/s
(d) ax = 0 m/s², ay = 9.8 m/s²
Explanation:
(a)
Time to reach maximum height by the projectile is given as:
t = V₀ Sinθ/g
where,
V₀ = Launching Speed = 55.6 m/s
Angle with Horizontal = θ = 41.2°
g = 9.8 m/s²
Therefore,
t = (55.6 m/s)(Sin 41.2°)/(9.8 m/s²)
t = 3.74 s
(b)
Maximum height reached by projectile is:
H = V₀² Sin²θ/g
H = (55.6 m/s)² (Sin²41.2°)/(9.8 m/s²)
H = 136.86 m
(c)
Neglecting the air resistance, the horizontal component of velocity remains constant. This component can be evaluated by the formula:
Vₓ = V₀ₓ = V₀ Cos θ
Vₓ = (55.6 m/s)(Cos 41.2°)
Vₓ = 41.83 m/s
Since, the projectile stops momentarily in vertical direction at the highest point. Therefore, the vertical component of velocity will be zero at the highest point.
Vy = 0 m/s
(d)
Since, the horizontal component of velocity is uniform. Thus there is no acceleration in horizontal direction.
ax = 0 m/s²
The vertical component of acceleration is always equal to the acceleration due to gravity during projectile motion:
ay = 9.8 m/s²
The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+vit+12at2 where xf is the final position, xi is the initial position, vi is the initial velocity, a is the acceleration, and t is the time. Let's say a car starts with an initial speed of 15 m/s, and moves between the 1000 m and 5000 m marks on a roadway in a time of 60 s. What is its acceleration?
Answer:
a = 1.72 m/s²
Explanation:
The given kinematic equation is the 2nd equation of motion. The equation is as follows:
xf = xi + (Vi)(t) + (1/2)(a)t²
where,
xf = the final position = 5000 m
xi = the initial position = 1000 m
Vi = the initial velocity = 15 m/s
t = the time taken = 60 s
a = acceleration = ?
Therefore,
5000 m = 1000 m + (15 m/s)(60 s) + (1/2)(a)(60 s)²
5000 m = 1000 m + 900 m + a(1800 s²)
5000 m = 1900 m + a(1800 s²)
5000 m - 1900 m = a(1800 s²)
a(1800 s²) = 3100 m
a = 3100 m/1800 s²
a = 1.72 m/s²
What is the answer for this question
The mass of a particular eagle is twice that of a hunted pigeon. Suppose the pigeon is flying north at ,2=17.1 m/s when the eagle swoops down, grabs the pigeon, and flies off. At the instant right before the attack, the eagle is flying toward the pigeon at an angle =52.7 ° below the horizontal and a speed of ,1=41.5 m/s.
Answer:
31.4 m/s
44.4°
Explanation:
Momentum is conserved in the horizontal direction:
pₓᵢ = pₓ
m vᵢ₂ + 2m vᵢ₁ cos θ = (m + 2m) vₓ
vᵢ₂ + 2 vᵢ₁ cos θ = 3 vₓ
17.1 m/s + 2 (41.5 m/s) (cos -52.7°) = 3 vₓ
vₓ = 22.5 m/s
Momentum is conserved in the vertical direction:
pᵧᵢ = pᵧ
2m vᵢ₁ sin θ = (m + 2m) vᵧ
2 vᵢ₁ sin θ = 3 vᵧ
2 (41.5 m/s) (sin -52.7°) = 3 vᵧ
vᵧ = -22.0 m/s
The speed is:
v = √(vₓ² + vᵧ²)
v = √((22.5 m/s)² + (-22.0 m/s)²)
v = 31.4 m/s
The direction is:
θ = atan(vᵧ / vₓ)
θ = atan(-22.0 m/s / 22.5 m/s)
θ = -44.4°
The speed of the eagle at that instant is 31.4 m/s while it moves off in the direction of 44.4°.
Since momentum is conserved horizontally;
17.1 m/s + 2 (41.5 m/s) (cos -52.7°) = 3 vx
vx = 17.1 m/s + 2 (41.5 m/s) (cos -52.7°)/3
vx = 22.5 m/s
Also, momentum is conserved vertically hence;
2 (41.5 m/s) (sin -52.7°) = 3 vy
vy = 2 (41.5 m/s) (sin -52.7°) /3
vy = -22.0 m/s
The effective speed therefore, is;
v = √((22.5 m/s)² + (-22.0 m/s)²)
v = 31.4 m/s
The direction of this effective speed is;
θ = tan-1(22.0 m/s / 22.5 m/s)
θ = 44.4°
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Potential difference is measured in which units?
volts
amps
currents
watts
Answer:
Potential difference is measured in volts
Explanation:
The standard metric unit on electric potential difference is the volt, abbreviated V and named in honor of Alessandro Volta. One Volt is equivalent to one Joule per Coulomb.
Answer:
Your answer is A.) volts
Explanation:
A student writes down several steps of scientific method. Put the steps in the best order
Answer:
Make a hypothesis, conduct an experiment, Analyze the experimental data..
Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 165 cmcm , but its circumference is decreasing at a constant rate of 14.0 cm/scm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.800 TT , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.
(a) Find the emf induced in the loop at the instant when 9.0 s have passed.
(b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.
Answer:
(a) emf = 1.18 mV
(b) counter-clockwise sense
Explanation:
(a) The induced emf is given by the following formula:
[tex]emf=-\frac{d\Phi_B}{dt}[/tex] (1)
where:
ФB: magnetic flux = AB = (area of the loop)*(magnitude of the magnetic field)
A = πr^2
B = 0.800 T
You replace the expression for the magnetic flux in the equation (1):
[tex]emf=-B\frac{\Delta A}{\Delta t}=-B\frac{A_2-A_1}{t_2-t_1}[/tex]
A1: initial area
A2: final area
t2-t1: time interval = 9.0s
Then you have to calculate the change in the area of the loop, by using the information about the circumference of the loop. First you calculate the radius of the loop for a circumference of 165 cm = 1.65m
[tex]s=1.65m=2\pi r\\\\r=\frac{1.65m}{2\pi}=0.262m[/tex]
You calculate the initial area A1:
[tex]A_1=\pi (0.262m)^2=0.215m^2[/tex]
After 9.0 second the circumference will be:
[tex]s'=1.65m-0.14\frac{m}{s}(9.0s)=0.39m[/tex]
the new radius and the final area is:
[tex]r=\frac{0.39m}{2\pi}=0.062m[/tex]
[tex]A_2=\pi(0.062m)^2=0.012m^2[/tex]
Finally, you replace in the equation (1):
[tex]emf=-(0.800T)\frac{0.012m^2-0.215m^2}{9.0s}=1.8*10^{-3}V=1.8mV[/tex]
The induced emf in the circular loop is 1.18mV
(b) The induced emf generates an electric current, which produces a magnetic field that is opposite to the direction of the constant magnetic field of 0.800T. Due to this magnetic field point into the loop. The current has to have a direction in a counter-clockwise sense.
At a time when mining asteroids has become feasible, astronauts have connected a line between their 3220-kg space tug and a 6240-kg asteroid. They pull on the asteroid with a force of 362 N. Initially the tug and the asteroid are at rest, 311 m apart. How much time does it take for the ship and the asteroid to meet
-- F = m a ... ==> a = F/m
-- The tension in the rope is 362 N. That same force acts on the asteroid and on the tug, pulling them together.
-- The asteroid's acceleration is 362N / 6240 kg = 0.058 m/s², headed for a point on the rope somewhere between the asteroid and the tug.
-- The tug's acceleration is 362 N / 3220 kg = 0.112 m/s², also headed for a point on the rope somewhere between the tug and the asteroid.
-- So now we have a gap between them, initially 311 m long, closing with a speed that starts at zero and accelerates at 0.170 m/s² .
-- D = (1/2) a T²
311 m = (1/2) (0.170 m/s²) (T²)
T² = 311 m / 0.085 m/s²
T = √(311/0.085) seconds
T = 60.41 seconds
The answer I get is so durn near 60 seconds (1 minute) that it suggests two things to me: ==> That's where the weird numbers of 362N and 311m came from, and ==> there's a good chance that my answer is correct.
Note: It's important to me that you know that 5 points for this one is really cheap and chintzy, and the reason I decided to try it was only to see whether I could.
Leah is moving in a spaceship at a constant velocity away from a group of stars. Which one of the following statements indicates a method by which she can determine her absolute velocity through space?
A) She can measure her increases in mass.
B) She can measure the contraction of her ship.
C) She can measure the vibration frequency of a quartz crystal.
D) She can measure the changes in total energy of her ship.
E) She can perform no measurement to determine this quantity.
Answer:
E) She can perform no measurement to determine this quantity.
Explanation:
A spacecraft is a machine used to fly in outer space.
According to Isaac Newton's third law of motion, every action produces an equal and opposite reaction. When fuel is shoot out of one end of the rocket, the rocket moves forward for which no air is required.
As Leah is moving in a spaceship at a constant velocity away from a group of stars, she cannot measure to determine this quantity.
What is a substance?
At an accident scene on a level road, investigators measure a car’s skid mark (mass of car is M) to be of length d. It was a rainy day and the coefficient of friction was estimated to be μk.
A) Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes.B) Why does the car's mass not matter?1) Since both the change in kinetic energy and the work done by friction are proportional to the mass. The mass cancels out of the equation.2) Since the work done by friction does not depend on mass.3) Since the change in kinetic energy and the work done by friction do not depend on mass.
Answer:
1) Since both the change in kinetic energy and the work done by friction are proportional to the mass. The mass cancels out of the equation
Explanation:
The kinetic friction works against the kinetic energy of the car and the car stops when these two equalises .
friction force = μk x R , μk is coefficient of kinetic friction and R is reaction from the ground.
= μk x mg
work done by friction
= force x displacement
= μk x mg x d
kinetic energy of car at the time of accident = 1/2 m v²
kinetic energy = work done by friction
1/2 m v² = μk x mg x d
d = v² / (2 μk x g)
v² = 2dμk g
v = √(2dμk g)
Since both the change in kinetic energy and the work done by friction are proportional to the mass. The mass cancels out of the equation
PLEASE HELP !
Complete the following sentence. Choose the right answer from the given ones. The internal energy of the body can be changed A / B / C. A. only when the body is warmed or cooled B. when work is done on the body or heat flow C. only when the body does work
B
HOPE IT HELPS LET ME KNOW IF U NEED EXPLANATION
When an electromagnetic wave falls on a white, perfectly reflectingsurface, it exerts a force F on that surface. If the surfaceis now painted a perfectly absorbing black, the force that the samewave would exert on the surface is:___________.
A) F
B) F/2
C) F/4
D) 2F
E) 4F
Answer:
B. F/2
Explanation:
The radiation force per unit area (radiation pressure Prad) exerted by an electromagnetic wave on a perfectly absorbing body has been found by experiment to be equal to the energy density of the wave
i.e Prad = u
For a reflecting body, this force exerted per unit area has been found to be twice the energy density of the wave.
i.e Prad = 2u.
Therefore, if the force exerted on a perfectly reflective body is F, then the force exerted on a perfectly absorbing body will be F/2
If radio waves are used to communicate with an alien spaceship approaching Earth at 10% of the speed of light c, the aliens would receive our signals at speed of:_______.
a. 0.99c
b. 1.10c
c. 1.00c
d. 0.90c
e. 0.10c
Answer:
3×10^7 m/s or 0.10c (e)
Explanation: If the actual value of the speed of light were to be put into consideration.
Given that the speed of light is c = 3.0×10^8m/s
The alien spaceship is approaching at the rate of 10% of the speed of light.
10% of 3.0×10^8m/s
10/100 × 3.0×10^8m/s
0.1 ×3.0×10^8m/s
3×10^7 m/s. Which is the same thing as 0.1 of c = 0.1×c
Answer: 1.00c
Explanation: I got it correct on the homework
An aluminium pot whose thermal conductivity is 237 W/m.K has a flat, circular bottom
with diameter 15 cm and thickness 0.4 cm. Heat is transferred steadily to boiling water in
the pot through its bottom at a rate of 1400 W. If the inner surface of the bottom of the pot
is at 105 °C, determine the temperature at the outer surface of the bottom of the pot
Answer:
T₁ = 378.33 k = 105.33°C
Explanation:
From Fourier's Law of heat conduction, we know that:
Q = - KAΔT/t
where,
Q = Heat Transfer Rate = 1400 W
K = Thermal Conductivity of Material (Aluminum) = 237 W/m.k
A =Surface Area through which heat transfer is taking place=circular bottom
A = π(radius)² = π(0.15 m)² = 0.0707 m²
ΔT = Difference in Temperature of both sides of surface = T₂ - T₁
T₁ = Temperature of outer surface = ?
T₂ = Temperature of inner surface = 105°C + 273 = 378 k
ΔT = 388 k - T₁
t = thickness of the surface (Bottom of Pot) = 0.4 cm = 0.004 m
Therefore,
1400 W = - (237 W/m.k)(0.0707 m²)(378 k - T₁)/0.004 m
(1400 W)/(4188.14 W/k) = - (378 k - T₁)
T₁ = 0.33 k + 378 k
T₁ = 378.33 k = 105.33°C
EASY! WILL GIVE BRAINLIEST!
Find the conductivity of a conduit with a cross-sectional area of 0.50 cm2 and a length of 15 cm if its conductance G is 0.050 ohm-1.
σ = _____ ohm-1cm-l
3
75
1.5
0.0017
the answer is 1.5 hope this helps
Answer:
1.5
Explanation:
0.5=σ/(15/0.5)
σ=3/2 or 1.5
Someone please helppppppp!!!!!
a steel ball is dropped from a diving platform use the approximate value of g as 10 m/s^2 to solve the following problem what is the velocity of the ball 0.9 seconds after its released
Answer:
The final speed of the ball is 9 m/s.
Explanation:
We have,
A steel ball is dropped from a diving platform. It is required to find the velocity of the ball 0.9 seconds after its released. It will move under the action of gravity. Using equation of motion to find it as :
[tex]v=u+at[/tex]
u = 0 (at rest), a = g
[tex]v=gt\\\\v=10\times 0.9\\\\v=9\ m/s[/tex]
So, the final speed of the ball is 9 m/s.
Your electric drill rotates initially at 5.35 rad/s. You slide the speed control and cause the drill to undergo constant angular acceleration of 0.331 rad/s2 for 4.81 s. What is the drill's angular displacement during that time interval?
Answer:
The angular displacement is [tex]\theta = 29.6 \ rad[/tex]
Explanation:
From the question we are told that
The initial angular speed is [tex]w = 5.35 \ rad/s[/tex]
The angular acceleration is [tex]\alpha = 0.331 rad /s^2[/tex]
The time take is [tex]t = 4.81 \ s[/tex]
Generally the angular displacement is mathematically represented as
[tex]\theta = w * t + \frac{1}{2} \alpha * t^2[/tex]
substituting values
[tex]\theta = 5.35 * 4.81 + \frac{1}{2} * 0.331 * (4.81)^2[/tex]
[tex]\theta = 29.6 \ rad[/tex]
A wire of length L is made up of two sections of two different materials connected in series. The first section of length L1 = 17.7 m is made of steel and the second section of length L2 = 28.5 m is made of iron. Both wires have the same radius of 5.30 ✕ 10−4 m. If the compound wire is subjected to a tension of 148 N, determine the time taken for a transverse pulse to move from one end of the wire to the other. The density of steel is 7.75 ✕ 103 kg/m3 and the density of iron is 7.86 ✕ 103 kg/m3.
Answer:
Explanation:
velocity of wave in a tense wire is given by the expression
[tex]v= \sqrt{\frac{T}{m} }[/tex]
v is velocity . T is tension and m is mass per unit length .
for steel wire
m = π r² ρ where r is radius and ρ is density
= 3.14 x (5.3 x 10⁻⁴)²x7.75 x 10³
= 683.57 x 10⁻⁵ kg/m
v = [tex]\sqrt{\frac{148}{683.57\times 10^{-5}} }[/tex]
= 1.47 x 10² m /s
= 147 m /s
for iron wire
m = π r² ρ where r is radius and ρ is density
= 3.14 x (5.3 x 10⁻⁴)²x7.86 x 10³
= 693.27 x 10⁻⁵ kg/m
[tex]v = \sqrt{\frac{148}{693.27\times 10^{-5}} }[/tex]
= 146 m /s
Time taken to move from one end to another
= 17.7 / 147 + 28.5 / 146
= .12 + .195
= .315 s .
