Answer:
The final velocity of the block after the bullet passes through is 0.66 meters per second.
Explanation:
The interaction between the bullet and the block of woods is a clear example of a perfectly inelastic collision, which can be modelled after the Principle of Momentum Conservation. There are no external forces exerted on the bullet-block system. The equation describing the collision is described below:
[tex]m_{B}\cdot v_{B,o} + m_{W}\cdot v_{W,o} = m_{B}\cdot v_{B,f} + m_{W}\cdot v_{W,f}[/tex]
Where:
[tex]m_{B}[/tex], [tex]m_{W}[/tex]- Masses of the bullet and the block of wood, measured in kilograms.
[tex]v_{B,o}[/tex], [tex]v_{W,o}[/tex] - Initial speeds of the bullet and the block of wood, measured in meters per second.
[tex]v_{B,f}[/tex], [tex]v_{W,f}[/tex]- Final speeds of the bullet and the block of wood, measured in meters per second.
The final speed of the block is cleared:
[tex]v_{W,f} = \frac{m_{B}\cdot (v_{B,o}-v_{B,f})+m_{W}\cdot v_{W,o}}{m_{W}}[/tex]
[tex]v_{W,f} = v_{W,o} + \frac{m_{B}}{m_{W}} \cdot (v_{B,o}-v_{B,f})[/tex]
If [tex]v_{W,o} = 0\,\frac{m}{s}[/tex], [tex]m_{B} = 0.022\,kg[/tex], [tex]m_{W} = 2\,kg[/tex], [tex]v_{B,o} = 210\,\frac{m}{s}[/tex] and [tex]v_{B,f} = 150\,\frac{m}{s}[/tex], then the final velocity of the block after the bullet passes through is:
[tex]v_{W,f} = 0\,\frac{m}{s}+\left(\frac{0.022\,kg}{2\,kg}\right)\cdot \left(210\,\frac{m}{s}-150\,\frac{m}{s} \right)[/tex]
[tex]v_{W,f} = 0.66\,\frac{m}{s}[/tex]
The final velocity of the block after the bullet passes through is 0.66 meters per second.
Solve 3* +5-220t = 0
Answer:
t = 27.5
Explanation:
[tex]3 + 5 -220t = 0[/tex]
Well to solve for t we need to combine like terms and seperate t.
So 3+5= 8
8 - 220t = 0
We do +220 to both sides
8 = 220t
And now we divide 220 by 8 which is 27.5
Hence, t = 27.5
g a conductor consists of an infinite number of adjacent wires, each infinitely long. If there are n wires per unit length, what is the magnitude of B~
Answer:
B=uonI/2
Explanation:
See attached file
4. The Richter scale describes how much energy an earthquake releases. With every increase of 1.0 on the scale, 32 times more energy is released. How many times more energy would be released by a quake measuring 2.0 more units on the Richter scale?
Answer:
64 times
Explanation:
if increase of 1 gives you 32
then increase of 2 will give you its double
64
If you increase one, you get 32 then multiplying by 2 will give you 64, which is its double.
What is Earthquake?An earthquake is a sudden energy released in the Earth's lithosphere that causes shock wave, which cause the Earth's surface to shake. Earthquakes can range in strength from ones that are so small that no one can feel them to quakes that are so powerful that they uproot entire cities, launch individuals and objects into the air, and harm vital infrastructure.
The frequency, kind, and intensity of earthquakes observed over a specific time period are considered to be the seismic activity of an area.
The average rate of earthquake energy output per unit volume determines the basicity of a certain area of the Earth. The non-earthquake seismic rumbling is also alluded to as a tremor.
To know more about Earthquake:
https://brainly.com/question/1296104
#SPJ5
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 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
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
An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the new acceleration would be _____ m/s/s.
Hahahahaha. Okay.
So basically , force is equal to mass into acceleration.
F=ma
so when F=ma , we get acceleration=6m/s/s
Force is doubled.
Mass is 1/3 times original.
2F=1/3ma
Now , we rearrange , and we get 6F=ma
So , now for 6 times the original force , we get 6 times the initial acceleration.
So new acceleration = 6*6= 36m/s/s
A solenoid used to produce magnetic fields for research purposes is 2.2 mm long, with an inner radius of 30 cmcm and 1200 turns of wire. When running, the solenoid produced a field of 1.4 TT in the center. Given this, how large a current does it carry?
Answer:
The current is [tex]I = 2042\ A[/tex]
Explanation:
From the question we are told that
The length of the solenoid is [tex]l = 2.2 \ m[/tex]
The radius is [tex]r_i = 30 \ cm = 0.30 \ m[/tex]
The number of turn is [tex]N = 1200 \ turns[/tex]
The magnetic field is [tex]B = 1.4 \ T[/tex]
The magnetic field produced is mathematically represented as
[tex]B = \frac{\mu_o * N * I }{l }[/tex]
making [tex]I[/tex] the subject
[tex]I = \frac{B * l}{\mu_o * N }[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with values [tex]\mu_o = 4\pi *10^{-7} N/A^2[/tex]
substituting values
[tex]I = \frac{1.4 * 2.2 }{4\pi *10^{-7} * 1200 }[/tex]
[tex]I = 2042\ A[/tex]
As a wheel turns, the angle through which it has turned varies with time as β(t)=Ct + Bt3 where C=0.400rad/s and B=0.0120rad/s3. Calculate the angular velocity w(t) as a function of time.
Answer:
ω(t) = 0.4 + 0.036 t²
Explanation:
The angular displacement of the disk is given as the function of time:
β(t) = Ct + B t³
where,
C = 0.4 rad/s
B = 0.012 rad/s³
Therefore,
β(t) = 0.4 t + 0.012 t³
Now, for angular velocity ω(t), we must take derivative of angular displacement with respect to t:
ω(t) = dβ/dt = (d/dt)(0.4 t + 0.012 t³)
ω(t) = 0.4 + 0.036 t²
What is the wavelength λλlambda of the wave described in the problem introduction? Express the wavelength in terms of the other given variables and constants
Complete Question
The complete question is shown on the first uploaded image
Answer:
The wavelength is [tex]\lambda= \frac{2 \pi }{k}[/tex]
Explanation:
From the question we are told that
The electric field is [tex]\= E = E_o sin (kx - wt )\r j[/tex]
The magnetic field is [tex]\= B = B_0 sin (kx -wt) \r k[/tex]
From the above equation
and k is the wave number which is mathematically represented as
[tex]k = \frac{2 \pi }{\lambda }[/tex]
=> [tex]\lambda= \frac{2 \pi }{k}[/tex]
Where [tex]\lambda[/tex] is the wavelength
1.5 kg of air within a piston-cylinder assembly executes a Carnot power cycle with maximum and minimum temperatures of 800 K and 300 K, respectively. The heat transfer from the air during the isothermal compression is 80 kJ. At the end of the isothermal compression, the volume is 0.2 m3. Determine the volume at the beginning of the isothermal compression, in m3. Assume the ideal gas model for air and neglect kinetic and potential energy effects.
Answer:
Explanation:
Carton cycle consists of four thermodynamic processes . The first is isothermal expansion at higher temperature , then adiabatic expansion which lowers the temperature of gas . The third process is isothermal compression at lower temperature and the last process is adiabatic compression which increases the temperature of the gas to its original temperature .
So the given process of isothermal compression must have been done at the temperature of 300K , keeping the temperature constant .
Work done on gas at isothermal compression is equal to heat transfer .
work done on gas = 80 x 10³ J
work done on gas = n RT ln v₁ / v₂
n is number of moles v₁ and v₂ are initial and final volume
molecular weight of gas = 28.97 g
1.5 kg = 1500 / 28.97 moles
= 51.77 moles
work done on gas = n RT ln v₁ / v₂
Putting the values in the equation above
80 x 10³ = 51.78 x 8.31 x 300 x ln v₁ / .2
ln v₁ / .2 = .62
v₁ / .2 = 1.8589
v₁ = 0.37 m³
A typical electric oven has two separate heating elements: one on top and one on the bottom. The bottom element is used for baking while the top element is used to broil foods. When only the bottom element is active and glowing red hot, what heat transfer mechanisms carry most of the heat to the food in the oven?
Answer:
Convection and Radiation mechanisms carry most of the heat
Explanation:
This is because Convection proceeds strongy as heated air rises from the hot element while Radiation is also strong, although the material of the cooking pots will how effective it is.
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
Four 50-g point masses are at the corners of a square with 20-cm sides. What is the moment of inertia of this system about an axis perpendicular to the plane of the square and passing through its center
Answer:
moment of inertia I ≈ 4.0 x 10⁻³ kg.m²
Explanation:
given
point masses = 50g = 0.050kg
note: m₁=m₂=m₃=m₄=50g = 0.050kg
distance, r, from masses to eachother = 20cm = 0.20m
the distance, d, of each mass point from the centre of the mass, using pythagoras theorem is given by
= (20√2)/ 2 = 10√2 cm =14.12 x 10⁻² m
moment of inertia is a proportion of the opposition of a body to angular acceleration about a given pivot that is equivalent to the entirety of the products of every component of mass in the body and the square of the component's distance from the center
mathematically,
I = ∑m×d²
remember, a square will have 4 equal points
I = ∑m×d² = 4(m×d²)
I = 4 × 0.050 × (14.12 x 10⁻² m)²
I = 0.20 × 1.96 × 10⁻²
I = 3.92 x 10⁻³ kg.m²
I ≈ 4.0 x 10⁻³ kg.m²
attached is the diagram of the equation
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.
An elastic circular bar is fixed at one end and attached to a rubber grommet at the other end. The grommet functions as a torsional spring with spring constant k. If a concentrated torque of magnitude Ta is applied in the center of the bar, what is the rotation at the end of the bar, φ(L)? Assume a constant shear modulus G and polar moment of inertia J.
Answer:
2.1 rad(anticlockwise).
Explanation:
So, we are given the following data or parameters or information in the question above:
=> "The torsional stiffness of the spring support is k = 50 N m/rad. "
=> "If a concentrated torque of mag- nitude Ta = 500 Nm is applied in the center of the bar"
=> "L = 300 mm Assume a shear modu- lus G = 10 kN/mm2 and polar monnent of inertia J = 2000 mln"
Hence;
G × J = 10 kN/mm2 × 2000 mln = 20 Nm^2.
Also, L/2 = 300 mm /2 = 0.15 m (converted to metre).
==> 0.15/20 (V - w) + θ = 0.
==> 0.15/20 (V - w ) = -θ.
Where V = k = 50 N m/rad
w = 183.3 θ.
Therefore, w + Vθ = 500 Nm.
==> 183.3 + 50 θ = 500 Nm.
= 6.3
Anticlockwise,
θ = 2.1 rad.
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
Question 8
A spring is attached to the ceiling and pulled 8 cm down from equilibrium and released. The
damping factor for the spring is determined to be 0.4 and the spring oscillates 12 times each
second. Find an equation for the displacement, D(t), of the spring from equilibrium in terms of
seconds, t.
D(t) =
Can someone please help me ASAP?!!!!
Answer: D(t) = [tex]8.e^{-0.4t}.cos(\frac{\pi }{6}.t )[/tex]
Explanation: A harmonic motion of a spring can be modeled by a sinusoidal function, which, in general, is of the form:
y = [tex]a.sin(\omega.t)[/tex] or y = [tex]a.cos(\omega.t)[/tex]
where:
|a| is initil displacement
[tex]\frac{2.\pi}{\omega}[/tex] is period
For a Damped Harmonic Motion, i.e., when the spring doesn't bounce up and down forever, equations for displacement is:
[tex]y=a.e^{-ct}.cos(\omega.t)[/tex] or [tex]y=a.e^{-ct}.sin(\omega.t)[/tex]
For this question in particular, initial displacement is maximum at 8cm, so it is used the cosine function:
[tex]y=a.e^{-ct}.cos(\omega.t)[/tex]
period = [tex]\frac{2.\pi}{\omega}[/tex]
12 = [tex]\frac{2.\pi}{\omega}[/tex]
ω = [tex]\frac{\pi}{6}[/tex]
Replacing values:
[tex]D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)[/tex]
The equation of displacement, D(t), of a spring with damping factor is [tex]D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)[/tex].
Do the math: How many seconds would it take an echo sounder’s ping to make the trip from a ship to the Challenger Deep (10,994 meters) and back? Recall that depth in meters = ½ (1500 m/sec × Echo travel time in seconds). Round your answer to two decimal places.
Answer:
14.66secsExplanation:
Given the formula for calculating the depth in metres expressed as
depth in meters = ½ (1500 m/sec × Echo travel time in seconds)
Given depth of the challenger = 10, 994 meters, we will substitute this given value into the formula given to calculate the time take for the echo to travel.
10, 994 = depth in meters = ½ * 1500 m/sec × Echo travel time in seconds
10,994 = 750 * Echo travel time in seconds
Dividing both sides by 750;
Echo travel time in seconds = 10,994 /750
Echo travel time in seconds ≈ 14.66secs (to two decimal places)
Therefore, it would take an echo sounder’s ping 14.66secs to make the trip from a ship to the Challenger Deep and back
Determine the slit spacing d. Explain which measurement you made, show your calculation and your result for the slit spacing. There are several measurements you can make.
Answer:
The quantities to measure are:
* the distance to the screen
* The distance from the central maximum to each interference
* in order of interference
* wavelength
Explanation:
To determine the gap spacing we must use the constructive interference equation
d sin θ = m λ
as the angles are small
tan θ = sin θ / cos θ
tan θ = sin θ
and the definition of tangent is
tan θ = y / L
Thus
sin θ = y / L
when replacing
d y / L = m λ
d = m λ L / y
with this equation we can know what parameter should be measured.
The quantities to measure are:
* the distance to the screen
* The distance from the central maximum to each interference
* in order of interference
* wavelength
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 %
Huygens claimed that near the surface of the Earth the velocity downwards of an object released from rest, vy, was directly proportional to the square root of the distance it had fallen, . This is true if c is equal to
Answer:
the expression is False
Explanation:
From the kinematics equations we can find the speed of a body in a clean fall
v = v₀ - g t
v² = V₀² - 2 g y
If the body starts from rest, the initial speed is zero (vo = 0)
v= √ (2g y)
In the first equation it gives us the relationship between speed and time.
With the second equation we can find the speed in which the distance works, this is the expression, see that speed is promotional at the height of a delicate body.
Therefore the expression is False
Which of the following explains why metallic bonding only occurs between
metallic atoms?
A. Metallic atoms are less likely to give their electrons to nonmetallic
atoms
B. Electrical conductivity is higher in metallic atoms, which means
they are more likely to attract free electrons.
C. Metallic atoms are highly reactive and do not tend to form bonds
with other atoms.
D. Metallic atoms have valence shells that are mostly empty, which
means these atoms are more likely to give up electrons and allow
them to move freely.
Answer:
D. Metallic atoms have valence shells that are mostly empty, which
means these atoms are more likely to give up electrons and allow
them to move freely.
Explanation:
Metals usually contain very few electrons in their valence shells hence they easily give up these few valence electrons to yield metal cations.
In the metallic bond, metal cations are held together by electrostatic attraction between the metal ions and a sea of mobile electrons.
Since metals give up their electrons easily, it is very easy for them to participate in metallic bonding. They give up their electrons easily because their valence shells are mostly empty, metal valence shells usually contain only a few electrons.
A student builds a rocket-propelled cart for a science project. Its acceleration is not quite high enough to win a prize, so he uses a larger rocket engine that provides 39% more thrust, although doing so increases the mass of the cart by 13%. By what percentage does the cart's acceleration increase?
Answer:
Explanation:
a = F / m
where a is acceleration , F is thrust and m is mass
taking log and differentiating
da / a = dF / F - dm / m
(da / a)x 100 = (dF / F)x100 - (dm / m) x100
percentage increase in a = percentage increase in F - percentage increase in m
= percentage increase in acceleration a = 39 - 13 = 26 %
required increase = 26 %.
A typical home uses approximately 1600 kWh of energy per month. If the energy came from a nuclear reaction, what mass would have to be converted to energy per year to meet the energy needs of the home
Answer:
7.68×10^25kg
Explanation:
The formula for energy used per year is calculated as
Energy used per year =12 x Energy used per month
By substituting Energy used per month in the above formula, we get
Energy used per year =12 x 1600kWh
= 19200kWh
Conversion:
From kWh to J:
1 kWh=3.6 x 10^6 J
Therefore, it is converted to J as
19200 kWh =19200 x 3.6 x 10^6 J
= 6.912×10^10 J
Hence, energy used per year is 6.912×10^10 J
To find the mass that is converted to energy per year.
E = MC^2 ............1
E is the energy used per year
C is the speed of light = 3.0× 10^8m/s
Where E= 6.912×10^10 J
Substituting the values into equation 1
6.912×10^10 J = M × 3.0× 10^8m/s
M = 6.912×10^10 J / (3.0× 10^8m/s)^2
M = 6.912×10^10 J/9×10^16
M = 7.68×10^25kg
Hence the mass to be converted is
7.68×10^25kg
A uniform stick 1.5 m long with a total mass of 250 g is pivoted at its center. A 3.3-g bullet is shot through the stick midway between the pivot and one end The bullet approaches at 250 m/s and leaves at 140 m/s
With what angular speed is the stick spinning after the collision?
Answer:
63.44 rad/s
Explanation:
mass of bullet = 3.3 g = 0.0033 kg
initial velocity of bullet [tex]v_{1}[/tex] = 250 m/s
final velocity of bullet [tex]v_{2}[/tex] = 140 m/s
loss of kinetic energy of the bullet = [tex]\frac{1}{2}m(v^{2} _{1} - v^{2} _{2})[/tex]
==> [tex]\frac{1}{2}*0.0033*(250^{2} - 140^{2} )[/tex] = 70.785 J
this energy is given to the stick
The stick has mass = 250 g =0.25 kg
its kinetic energy = 70.785 J
from
KE = [tex]\frac{1}{2} mv^{2}[/tex]
70.785 = [tex]\frac{1}{2}*0.25*v^{2}[/tex]
566.28 = [tex]v^{2}[/tex]
[tex]v= \sqrt{566.28}[/tex] = 23.79 m/s
the stick is 1.5 m long
this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m
The angular speed will be
Ω = v/r = 23.79/0.375 = 63.44 rad/s
(a) Find the speed of waves on a violin string of mass 717 mg and length 24.3 cm if the fundamental frequency is 980 Hz. (b) What is the tension in the string? For the fundamental, what is the wavelength of (c) the waves on the string and (d) the sound waves emitted by the string? (Take the speed of sound in air to be 343 m/s.)
Answer:
a)v = 476.28 m / s , b) T = 6.69 10⁵ N , c) λ = 0.486 m , d) λ = 0.35 m
Explanation:
a) The speed of a wave on a string is
v = √T /μ
also all the waves fulfill the relationship
v = λ f
they indicate that the fundamental frequency is f = 980 Hz.
The wavelength that is fixed at its ends and has a maximum in the center
L = λ / 2
λ = 2L
we substitute
v = 2 L f
let's calculate
v = 2 0.243 980
v = 476.28 m / s
b) The tension of the rope
T = v² μ
the density of the string is
μ = m / L
T = v² m / L
T = 476.28² 0.717 / 0.243
T = 6.69 10⁵ N
c) λ = 2L
λ = 2 0.243
λ = 0.486 m
d) The violin has a resonance process with the air therefore the frequency of the wave in the air is the same as the wave in the string. Let's find the wavelength in the air
v = λ f
λ= v / f
λ = 343/980
λ = 0.35 m
At what temperature will water begin to boil and turn to steam?
212 degrees Celsius
100 degrees Fahrenheit
212 kelvins
100 degrees Celsius
Answer:
100 degrees Celsius
Explanation:
Water starts to boil at 100 degrees celcius or 212 degrees fahrenheit.
At 100 degrees Celsius water begin to boil and turns to steam.
What are the boiling point and melting point of water?The melting point for water is 0 degrees C (32 degrees F). The boiling point of water varies with atmospheric pressure. At lower pressure or higher altitudes, the boiling point is lower. At sea level, pure water boils at 212 °F (100°C).
Is boiling water always 212?If the temperature is much above 212°F, the water will boil. That means that it won't just evaporate from the surface but will form vapor bubbles, which then grow, inside the liquid itself. If the water has very few dust flecks etc.
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An inductor is connected to a 18 kHz oscillator. The peak current is 70 mA when the rms voltage is 5.4 V What is the value of the inductance L
Answer:
The value of the inductance is 0.955 mH
Explanation:
Given;
frequency of the oscillator, f = 18 kHz = 18,000 Hz
the peak current, I₀ = 70 mA = 0.07 A
the root mean square voltage, [tex]V_{rms}[/tex] = 5.4 V
The root mean square current is given as;
[tex]I_{rms}= \frac{I_o}{\sqrt{2} }[/tex]
[tex]I_{rms} = \frac{0.07}{\sqrt{2} } \\\\I_{rms} = 0.05 \ A[/tex]
Inductive reactance is given by;
[tex]X_L =\frac{V_{rms}}{I_{rms}} \\\\X_L = \frac{5.4}{0.05} \\\\X_L = 108 \ ohms[/tex]
Inductance is given by;
[tex]L = \frac{X_L}{2\pi f} \\\\L = \frac{108}{2\pi *18,000} \\\\L = 9.55 *10^{-4} \ H[/tex]
L = 0.955 mH
Therefore, the value of the inductance is 0.955 mH
The value of the inductance (L) for this oscillating circuit is equal to [tex]9.55 \times 10^{-4}[/tex] Henry.
Given the following data:
Oscillator frequency = 18 kHzPeak current = 70 mARms Voltage = 5.4 VTo determine the value of the inductance (L):
First of all, we would find the root mean square (rms) current by using the formula:
[tex]I_{rms} = \frac{I_o}{\sqrt{2} }\\\\I_{rms} = \frac{70 \times 10^{-3}}{1.4142} \\\\I_{rms} = 0.050 \;A[/tex]
Next, we would calculate the inductive reactance of the oscillator by using the formula:
[tex]X_L = \frac{V_{rms}}{I_{rms}} \\\\X_L = \frac{5.4}{0.050} \\\\X_L = 108 \; Ohms[/tex]
Now, we can solve for the value of the inductance (L):
[tex]L = \frac{X_L}{2\pi f}[/tex]
Where:
L is the inductance.f is the frequency.[tex]X_L[/tex] is the inductive reactance.Substituting the parameters into the formula, we have;
[tex]L = \frac{108}{2 \times 3.142 \times 18 \times 10^3} \\\\L = \frac{108}{113112}[/tex]
L = [tex]9.55 \times 10^{-4}[/tex] Henry.
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"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.