The expression for the speed of waves in materials allows us to find the result for the speed of sound in aluminum is:
Sound speed is: v = 5050 m / s
The speed of a wave in a material is determined by the relationship between its volumetric modulus and its density, it is given by the expression.
v = [tex]\sqrt{ \frac{B}{\rho} }[/tex]
where v is the speed of the wave in the material (sound), B is the volume modulus and ρ the density.
They indicate that the volumetric or elastic modulus of aluminum is;
B = 6.89 10⁴ Mpa ( [tex]\frac{10^6 \ Pa}{1 \ MPa}[/tex] ) = 6.89 10¹⁰ Pa
The density of aluminum is tabulated ρ = 2.7 10³ kg / m³
We calculate.
v = [tex]\sqrt{ \frac{6.89 \ 10^{10} }{2.7 \ 10^3 }}[/tex]
v = 5.05 10³ m / s
In conclusion using the expression for the speed of waves in materials we can find the result for the speed of sound in aluminum is:
v = 5050 m / s
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The equation for a progressive wave is y=6 cos(20t-4x) What is the equation of another progressive wave which has twice the amplitude and frequency, and moving in the same direction?
The equation of the progressive wave is y = 12 cos(40t - 4x)
The general wave equation is given by:
y = A sin(ωt - kx)
Where A is the amplitude, ω is the angular frequency = 2πf, f is the frequency, k is the wave number and y, x is the displacement.
Given the equation for a progressive wave is y=6 cos(20t-4x). Hence:
The amplitude A = 6,
ω = 20 = 2πf
f = 20/2π = 3.183 Hz
Twice the amplitude = 2 * 6 = 12, twice the frequency = 2 * 3.183.
ω = 2π(3.183*2) = 40
Therefore the other progressive wave has an equation of:
y = 12 cos(40t - 4x)
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Suppose a ball is thrown vertically upward (positive direction) from an initial height LaTeX: h_0 with initial velocity LaTeX: v_0. Find the position function LaTeX: s(t) of the ball after LaTeX: t seconds assuming the gravitational acceleration LaTeX: g is a positive constant pointing downward (negative direction).
After time t, the position function of the ball is determined as [tex]y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2[/tex]
The given parameters;
initial velocity of the ball, = [tex]v_0[/tex]initial position of the ball, = [tex]h_0[/tex]acceleration due to gravity, = gThe position function of the ball after time t, is calculated as follows;
[tex]y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2[/tex]
The negative sign of acceleration of due to gravity is because the ball is moving upward against gravity.
Thus, after time t, the position function of the ball is determined as [tex]y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2[/tex]
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science thanks sa points
Answer: Are these free point?
Explanation:
Which is the main gas that makes up the Earth's atmosphere?
Answer:
78 percent nitrogenExplanation:
I hope it's helpful for you
Objects 1 and 2 attract each other with a gravitational force of 16 units. If the mass of object 1 is one-third the original value AND the mass of object 2 is doubled AND the distance separating objects 1 and 2 is doubled, then the new gravitational force will be _____ units.
Explanation:
Fgravity = G*(mass1*mass2)/D²
G is the gravitational constant, which has the same value throughout our universe.
D is the distance between both objects.
so, now some numbers change
Fgravitynew = G*((1/3)*mass1*2*mass2)/(2D)² =
= G*((2/3)*mass1*mass2)/(4D²) =
= (2/3)* (G*(mass1*mass2)/D²) / 4 =
= ((2/3)/4) * G*(mass1*mass2)/D² =
= (2/12) * Fgravity = Fgravity/6
the new gravitational force will be 16/6 = 8/3 units.
difference between speed and velocity
Answer:
Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object's movement.
Explanation:
calculate the surface area of a box whose mass is 200 kg and exerts a pressure of 100 Pascal on the floor.
Answer:
Explanation:
If 2×2 is 4 so 1 kg can be 1 gram if it belive on it self some people change
What happens to the iron in the coilgun if the electricity in the coil was turned on
A wheel has a radius of 0.40 m and is mounted on frictionless bearings. A block is suspended from a rope that is wound on the wheel and attached to it (see figure). The wheel is released from rest and the block descends 1.5 m in 2.00 s without any slipping of the rope. The tension in the rope during the descent of the block is 20 N. What is the moment of inertia of the wheel?
The moment of inertia of the wheel is 4.27 kg.m²
The kinematics equation explains the variables associated and related of motion.
From the information given, applying the kinematic equation of motion to determine the acceleration of the block, we have:
[tex]\mathbf{y = ut + \dfrac{1}{2}at^2}[/tex]
[tex]\mathbf{y = (0)t + \dfrac{1}{2}at^2}[/tex]
[tex]\mathbf{y = \dfrac{1}{2}at^2}[/tex]
Making acceleration (a) the subject, we have:
[tex]\mathbf{a = \dfrac{2y}{t^2}}[/tex]
where;
y = 1.5 mt = 2.0 s[tex]\mathbf{a = \dfrac{2\times 1.5 }{2.0^2}}[/tex]
a = 0.75 m/s²
The angular acceleration of the wheel can be estimated by the formula:
[tex]\mathbf{\alpha = \dfrac{a}{r}}[/tex]
[tex]\mathbf{\alpha = \dfrac{0.75 \ m/s^2}{0.40 \ m}}[/tex]
[tex]\mathbf{\alpha = 1.875 \ rad/s^2}[/tex]
Finally, the torque acting on the wheel is:
[tex]\mathbf{\tau = I \alpha}[/tex]
[tex]\mathbf{Tr = I \alpha}[/tex]
where;
T = tensionr = radiusI = moment of inertia∝ = angular acceleration∴
[tex]\mathbf{I =\dfrac{T\times r}{\alpha} }[/tex]
[tex]\mathbf{I =\dfrac{20 \ N\times 0.40 \ m}{1.875 \ rad/s^2} }[/tex]
I = 4.27 kg.m²
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A train slows its speed from 52 kilometers per hour to 46 kilometers per hour in 0.04 hour. What is the acceleration o the train during this time?
Answer: here you go i have to put 20 letters in so just ignore this and look at the link.
A 6.5 N ball is thrown with an initial velocity of 20 m/s at a 35° angle from a height of 1.5 m, what is the velocity if it is caught at 1.5 m?
Answer:
20 m/s at -35°
Explanation:
Ignoring air resistance, the initial vertical velocity will be reversed and the initial horizontal velocity will remain constant.
Describe the concept of energy quanta of EM radiation which was explained by Planck.
Answer:
Planck postulated that the energy of light is proportional to the frequency, and the constant that relates them is known as Planck's constant (h). His work led to Albert Einstein determining that light exists in discrete quanta of energy, or photons.
Explanation:
Answer:
Energy does not occur in continuous amounts but in discrete amounts described by:
E = N h ∨ where N is the number of quanta (energy units), ∨ the frequency of the energy, and h Planck's constant (6.63E-34 J-sec)
Some students conduct an experiment to prove conservation of momentum. They use two objects that collide Measurements
are taken before and after the collision.
Which two quantities will the students multiply together before and after the collision?
A. mass and velocity
B. distance and time
C. mass and acceleration
D. velocity and time
This question involves the concepts of the law of conservation of momentum, velocity, and mass.
The two quantities, the students should multiply before and after the collision are "A. mass and velocity".
According to the law of conservation of momentum, In an isolated system, the total momentum of the system before the collision is always equal to the total momentum of the system after the collision.
To prove the law of conservation of momentum, consider two balls of masses ‘m₁’ and ‘m₂’, moving with velocities ‘u₁’ and ‘u₂’, respectively, such that u₁ is greater than u₂. After some time, these balls collide with each other and their velocities become ‘v₁’ and ‘v₂’, respectively.
This situation is illustrated in the attached picture.
So, according to the law of conservation of momentum:
Total Momentum Before Collision = Total Momentum After Collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
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Sorry this is a year late, but here it is for those of you who are stuck on the same thing.
======================================Proving Conservation of Momentum Quick Check - 5/5NOTE: Please Check and Confirm That You Are On The Same Assignment with The Same Questions and Number of Questions. Thank You and Good Luck!
=======================================1. Mass & velocity
2. The total momentum after the collision is the same as the total momentum before the collision.
3. 0.54 kg⋅m/s
4. The system has external forces, such as friction and air resistance, acting on it.
5. 3.0 m/s
Accelerations are produced by
A. Masses
B.accelerations
C. Velocities
D.unbalanced, net forces
How are a series and parallel car different?
3) A 60. kg person is in an elevator. The elevator starts from rest and then accelerates upwards at 2.0 m/s^2 for 4.0 seconds. Calculate the work done by the normal force on the person. *
Answer:
WD = 960 J
Explanation:
WD = work done (J)
F = force (N)
s = displacement (m)
m = mass (kg) = 60
a = acceleration (m/s²) = 2
t = time (s) = 4
u = initial velocity (m/s) = 0
The formulas or equations that are relevant ate:
WD = F × s
F = m × a
s = u + at
We want to find WD, so we need to now the force and the displacement (or distance);
We calculate force, in Newtons, with the formula F = ma:
F = 60 × 2
F = 120 N
We also need displacement, which get with the formula s = u + at:
s = 0 + 2(4)
s = 8 m
Now we have F and s, we can calculate WD:
WD = 120 × 8
WD = 960 J
Methodology:
Starting with what you want to find, in this case WD, list the formula/s you could use;
Then, identify the information you need for the formula and whether or not you are given that information;
Next, list the formulas for the information you don't have and once again, identify whether the information you are given is sufficient to use those formulas;
Once you can calculate all necessary information, then proceed to calculate the values and finally, the answer;
I suggest also keeping a list of all the variables as I've done at the top of my working so it is clear for you to see and use.
What is the relationship between balancing equations and the law of conservation of matter
A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. What is the total energy of the system?
Hi there!
With the work-energy theorem for oscillating springs:
ME = KE + PE
[tex]ME = \frac{1}{2}mv^2 + \frac{1}{2}kx^2[/tex]
Where:
m = mass (kg)
v = velocity (m/s)
k = Spring Constant (N/m)
x = displacement from equilibrium (m)
If the object is at the equilibrium position, there is NO potential energy since:
[tex]\frac{1}{2}k(0^2) = 0 J[/tex]
Thus:
[tex]ME = \frac{1}{2}mv^2[/tex]
Plug in the given values:
[tex]ME = \frac{1}{2}(0.50)(1.5^2) = \boxed{0.5625 \text{ J}}[/tex]
At which type of boundary is new oceanic crust created?
A. a convergent plate boundary
B. a divergent plate boundary
C. a subduction plate boundary
D. a transform plate boundary
Answer:
c.
Explanation:
If the two plates that meet at a convergent plate boundary both are of oceanic crust, the older, denser plate will subduct beneath the less dense plate. The older plate subducts into a trench, resulting in earthquakes. Melting of mantle material creates volcanoes at the subduction zone.
When two oceanic plates converge, the denser plate will end up sinking below the less dense plate, leading to the formation of an oceanic subduction zone. Old, dense crust tends to be subducted back into the earth. An example of a subduction zone formed from a convergent boundary is the Chile-Peru trench….
Answer:
a divergent plate boundary
A 1300 watt hair blow dyer is designed to operate on 120 Volts. How much current does the dryer require
Answer:
10.83 Amperes
Explanation:
if A ⇒ current
W = VA
1300 = 120 x A
1300 / 120 = A
10.83 = A
The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.2 m above the ground, how steep a slope can the truck be parked on without tipping over
The slope of the road can be given as the ratio of the change in vertical
distance per unit change in horizontal distance.
The maximum steepness of the slope where the truck can be parked without tipping over is approximately 54.55 %.Reasons:
Width of the truck = 2.4 meters
Height of the truck = 4.0 meters
Height of the center of gravity = 2.2 meters
Required:
The allowable steepness of the slope the truck can be parked without tipping over.
Solution:
Let, C represent the Center of Gravity, CG
At the tipping point, the angle of elevation of the slope = θ
Where;
[tex]tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}[/tex]
The steepness of the slope is therefore;
[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100[/tex]
Where;
[tex]\overline{AM}[/tex] = Half the width of the truck = [tex]\dfrac{2.4 \, m}{2}[/tex] = 1.2 m
[tex]\overline{CM}[/tex] = The elevation of the center of gravity above the ground = 2.2 m
[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%[/tex]
[tex]tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}[/tex]
[tex]Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right) \approx 28.6 ^{\circ}[/tex]
The maximum steepness of the slope where the truck can be parked is 54.55 %.
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plz help me on this question thank you
Answer:
D
Explanation:
The volume of a toy car was calculated by displacing water. The water
rose by 20ml when the object was placed into the graduated cylinder. The balance showed the toy car had a
mass of 500grams. Calculate the density of the toy car
Answer:
D = 25g/cm³
Explanation:
1ml = 1cm³
D = m/V
D = 500g/20cm³
D = 25g/cm³
What is sixth state of matter?
A car was traveling at 25 m/s when it slammed on the brakes and came to a complete stop in 3 seconds. What is the cars INITIAL/FINAL VELOCITY?
Answer:
Explanation:
Initial velocity 25 m/s
final velocity 0 m/s
The ratio of the two is undefined as dividing by zero is wonky.
calculate the mass of a block of ice having volume 5m³. (density of ice≈920 kg/m³)
Answer:
4600kg
Explanation:
Density=mass÷volume
920=m/5
m=920×5=4600kg
How is the wavelength of a sound affected when (a) a sound source moves toward a stationary observer and (b) the observer moves away from a stationary sound source
Answer:
If the observer is stationary but the source moves toward the observer at a speed vs, the observer still intercepts more waves per second and the frequency goes up. This time it is the wavelength of the wave received by the observer that is effectively shifted by the motion, rather than the speed.
The oscillation of the 2.0-kg mass on a spring is described by x = 3.0 cos (4.0 t) where x is in centimeters and t is in seconds. What is the force constant k of the spring?
The force constant k of the spring, if The oscillation of the 2 kg mass of spring is described by x = 3.0 cos (4.0 t) is 32 N / m.
What is force?Force is the influence of either pull or pushes in the body. Basically, gravitation forces, nuclear forces, and friction forces are the types of forces. For e.g. when the wall is hit by a hand then a force is exerted by the hand on the wall as well as the wall also exerts a force on the hand. There are different laws given to Newton to understand force.
Newton is a unit of force used by physicists that is part of the International System (SI). The force required to move a body weighing one kilogram one meter per second is known as a newton.
Given:
The mass of the block, m = 2 kg,
The oscillation of spring, x = 3 cos 4t,
Calculate the omega by comparing the standard equation given below,
[tex]x = A cos \omega t[/tex]
ω = 4
Calculate the spring constant by the formula given below,
[tex]\omega = \sqrt{\frac{k}{m} }[/tex]
4² = k / 2
k = 32 N / m
Therefore, the force constant k of the spring, if The oscillation of the 2 kg mass of spring is described by x = 3.0 cos (4.0 t) is 32 N / m.
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identify the following prefixes:
1) Di-
2) Tetra-
3) Deca-
4) Hepta-
Explanation:
Di -. 2
Tetra. -3
deca. -. 10
Hepta. -- 7
Carbon tetrachloride (CCl4) is diffusing through benzene (C6H6), as the drawing illustrates. The concentration of CCl4 at the left end of the tube is maintained at 1.71 x 10-2 kg/m3, and the diffusion constant is 21.9 x 10-10 m2/s. The CCl4 enters the tube at a mass rate of 5.86 x 10-13 kg/s. Using these data and those shown in the drawing, find (a) the mass of CCl4 per second that passes point A and (b) the concentration of CCl4 at point A.
We have that for the Question "find (a) the mass of CCl4 per second that passes point A and (b) the concentration of CCl4 at point A."
Answers:
Mass of CCI_4 per second = [tex]5.86*10^{-13} kg/s[/tex] Concentration of CCI_4 = [tex]12.6*10^{-3}kg/m^3[/tex]
From the question we are told
The concentration of [tex]CCl_4[/tex] at the left end of the tube is maintained at 1.71 x 10-2 kg/m3, and the diffusion constant is 21.9 x 10-10 m2/s. The CCl4 enters the tube at a mass rate of 5.86 x 10-13 kg/s
A) the mass flow rate of CCI_4 as it passes point A is the same as the mass flow rate at which CCI_4 enters the left end of the tube
Therefore, the mass flow rate of CCI_4 at point A
= [tex]5.86*10^{-13} kg/s[/tex]
B) From Fick's law
[tex]\deltaC = \frac{mL}{DAt}\\\\ Assume L = 5*10^{-3}, A = 3*10^{-4}\\\\\deltaC = \frac{5.86*10^{-13} * 5*10^{-3}}{21.9*10^{-10} * 3*10^{-4}}\\\\\deltaC = 4.46*10^{-3}kg/m^3[/tex]
Then,
[tex]Concentration = 1.71*10^{-2} - 4.46*10^{-3}\\\\= 12.6*10^{-3}kg/m^3[/tex]
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