Which nucleus completes the following equation?
Answer:
Option D. ²³⁹₉₃Np
Explanation:
From the question given above, the following data were:
²³⁹₉₂U —> ⁰₋₁e + __
Let ⁿₘX represent the unknown. Thus, the equation above becomes
²³⁹₉₂U —> ⁰₋₁e + ⁿₘX
Next, we shall determine n, m and X. This can be obtained as follow:
239 = 0 + n
239 = n
n = 239
92 = –1 + m
Collect like terms
92 + 1 = m
93 = m
m = 93
ⁿₘX => ²³⁹₉₃X => ²³⁹₉₃Np
Thus, the balanced equation becomes:
²³⁹₉₂U —> ⁰₋₁e + ⁿₘX
²³⁹₉₂U —> ⁰₋₁e + ²³⁹₉₃Np
Option D gives the correct answer to the question.
Answer:
D
Explanation:
239 93 Np
Each of the two grinding wheels has a diameter of 6 in., a thickness of 3/4 in., and a specific weight of 425 lb/ft3. When switched on, the machine accelerates from rest to its operating speed of 3450 rev/min in 5 sec. When switched off, it comes to rest in 35 sec. Determine the motor torque and frictional moment, assuming that each is constant. Neglect the effects of the inertia of the rotating motor armature.
Answer:
[tex]0.842\ \text{lb ft}[/tex]
[tex]0.1052\ \text{lb ft}[/tex]
Explanation:
d = Diameter of wheel = 6 in
r = Radius = 3 in = [tex]\dfrac{3}{12}=0.25\ \text{ft}[/tex]
t = Thickness = [tex]\dfrac{3}{4}=0.75\ \text{in}=\dfrac{0.75}{12}\ \text{ft}[/tex]
w = Specific weight = [tex]425\ \text{lb/ft}^3[/tex]
[tex]t_2[/tex] = Time taken to slow down = 35 s
[tex]t_1[/tex] = Time taken to reach operating speed = 5 s
[tex]\omega[/tex] = Angular velocity = [tex]3450\times \dfrac{2\pi}{60}\ \text{rad/s}[/tex]
Weight is given by
[tex]W=2\pi r^2tw\\\Rightarrow W=2\pi\times 0.25^2\times \dfrac{0.75}{12}\times 425\\\Rightarrow W=10.43\ \text{lbs}[/tex]
Mass is given by
[tex]m=\dfrac{W}{g}\\\Rightarrow m=\dfrac{10.43}{32}\\\Rightarrow m=0.326\ \text{lb}[/tex]
Moment of inertia is given by
[tex]I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{0.326\times 0.25^2}{2}\\\Rightarrow I=0.01019\ \text{lb ft}^2[/tex]
Angular acceleration while slowing down is given by
[tex]\alpha_f=\dfrac{\omega}{t_2}\\\Rightarrow \alpha_f=\dfrac{3450\times \dfrac{2\pi}{60}}{35}\\\Rightarrow \alpha_f=10.32\ \text{rad/s}^2[/tex]
Frictional moment is given
[tex]\tau_f=I\alpha_f\\\Rightarrow \tau_f=0.01019\times 10.32\\\Rightarrow \tau_f=0.1052\ \text{lb ft}[/tex]
Frictional moment is [tex]0.1052\ \text{lb ft}[/tex]
Angular acceleration while speeding up is given by
[tex]\alpha=\dfrac{\omega}{t_1}\\\Rightarrow \alpha=\dfrac{3450\times \dfrac{2\pi}{60}}{5}\\\Rightarrow \alpha=72.26\ \text{rad/s}^2[/tex]
Motor torque is given by
[tex]\tau_m=\tau_f+I\alpha\\\Rightarrow \tau_m=0.1052+0.01019\times 72.26\\\Rightarrow \tau_m=0.842\ \text{lb ft}[/tex]
Motor torque is [tex]0.842\ \text{lb ft}[/tex].
You are at a train yard observing trains (because why not). You see a train car (let's call it car 1) moving to the right ( x direction) towards a stationary train car (let's call this one car 2). Car 1 has an initial velocity of 15.0 m/s. A helpful train employee tells you that Car 1 also has a mass of 1,825 kg and Car 2 has a mass of 2,645 kg. Car 1 gently collides with Car 2, allowing them to connect. After the collision the two train cars stay connected. You can assume that there is no friction in the system. If you have never see train cars connect, you can watch the first 25ish seconds of this video to see two train cars couple. However, these cars have friction, so they stop - unlike our problem. What is the Final Velocity of the system consisting of Car 1 and Car 2
Answer:
6.12 m/s
Explanation:
Using the law of conservation of momentum
momentum before collision = momentum after collision
m₁v₁ + m₂v₂ = (m₁ + m₂)V (since the train cars become attached to each other) where m₁ = mass of car 1 = 1,825 kg, m₂ = mass of car 2 = 2,645 kg, v₁ = initial velocity of car 1 = + 15.0 m/s (positive since it is moving in the positive x direction), v₂ = initial velocity of car 2 = 0 m/s (since it is initially stationary) and V = velocity of both cars after collision,
So, m₁v₁ + m₂v₂ = (m₁ + m₂)V
m₁v₁ + m₂(0 m/s) = (m₁ + m₂)V
m₁v₁ + 0 = (m₁ + m₂)V
V = m₁v₁/(m₁ + m₂)
substituting the values of the other variables into the equation, we have
V = 1,825 kg × 15.0 m/s/(1,825 kg + 2,645 kg)
V = 27375 kgm/s/ 4470kg
V = 6.124 m/s
V ≅ 6.12 m/s
Which of the following actions will increase the current induced in a wire by a
magnetic field?
Answer:
The induced current can be increased in the coil in the following ways: By increasing the strength of the magnet. By increasing the speed of the magnet through the coil.
Explanation:
The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 41 mm , while nonathletes' stretch only 33 mm .
Hello. Your question is incomplete. However, I managed to find it completely on the internet and I realized that you forgot to mention that the question asks you for the maximum energy difference between velovistas and non-athletes, considering that the spring constant for the tendon of the two groups is equal to 33n/mm.
To make this calculation you will need to use Hooke's law, using the formula: ¹/2*K*x², where "K" will be the value of the spring constant for the tendon and "X" will be the value of the sprinter and non-athlete terms.
So for the sprinter we will have the calculation:
¹/2*33*41² -------> 0,5*33*1681 = 27736. 5 Nmm
(To facilitate the calculation, first solve the division of ¹/2 and then multiply 41 by 41, lastly, just multiply all the results.)
For the non-athlete we will have the calculation:
¹/2*33*33² -------> 0,5*33*1089 = 17968. 5 Nmm
(To facilitate the calculation, first solve the division of ¹/2 and then multiply 41 by 41, lastly, just multiply all the results.)
Now, to reach the final result, you only need to subtract the two values presented by the sprinter and the non-athlete.
27736.5 - 17968.5 = 9768 Nmm
The datasheet for a 5 V, three-terminal voltage regulator indicates that the output voltage changes by 3 mV when the input voltage is varied from 7 V to 25 V, and the output voltage varies by 5 mV when the load is varied from 0.01 A to the full load of 1 A. The line- and load regulation are respectively which of the following?
a. 6.7mΩ
b. 0.01Ω
c. 0.1Ω
d. 20mΩ
Answer:
0.01 Ω
Explanation:
Given that the output voltage varies by 5 mV when loaded from 0.01 A to 1 A
Therefore, the regulators output resistance is given by :
[tex]$I_L=\frac{V_L}{R_L}$[/tex]
[tex]$(1.00 - 0.01)A= \frac{5 \ mV}{R_L}$[/tex]
[tex]$0.99 \ A= \frac{5 \ mV}{R_L}$[/tex]
∴ The line and load regulation is 0.01 Ω
Convert (a) 50 oF, (b) 80 oF, (c) 95 oF to Celsius
The distance from the sun to Earth would be
Which phrase best completes the sentence?
any number of light years
more than one light year
exactly one light year
less than one light year
4
Answer:
less than one lightyear=d
Explanation:
I took the test.:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D::):):):):):):):):):):):):):):):):):):):):):)
A spiral staircase winds up to the top of a tower in an old castle. To measure the height of the tower, a rope is attached to the top of the tower and hung down the center of the staircase. However, nothing is available with which to measure the length of the rope. Therefore, at the bottom of the rope a small object is attached so as to form a simple pendulum that just clears the floor. The period of the pendulum is measured to be 6.82 s. What is the
Answer:
The answer is "[tex]11.55780\ m[/tex]"
Explanation:
Using formula:
[tex]= 2 \pi f= \frac{2\pi}{T} =\sqrt{\frac{g}{L}}[/tex]
L = length of pendulum.
[tex]= T =2 \pi \sqrt{\frac{L}{g}}[/tex]
Calculate the value for L:
[tex]L= g (\frac{T}{2 \pi})^2 \\\\[/tex]
[tex]= (9.80 \ \frac{m}{s^2}) (\frac{6.82 \ s}{2 \pi})^2\\\\= (9.80 \ \frac{m}{s^2}) (\frac{46.5124 \ s^2}{4 \times \pi^2})\\\\= (9.80 \ \frac{m}{s^2}) (\frac{46.5124\ s^2}{4 \times 9.8596 })\\\\= (9.80 \ \frac{m}{s^2}) (\frac{46.5124 \ s^2}{ 39.4384 })\\\\= \frac{455.82152}{39.4384} \ m\\\\=11.55780\ m[/tex]
The height of the tower = 11.55780 m
a train has an initial velocity of 30 m/s. If the train accelerates uniformly at a rate of 6.3 m/s ^ for 2.8 seconds what is the trains final velocity?
T
Answer:
the velocity is a second final to initial velocity of 39
A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If both objects have the same kinetic energy, what is the ratio of the speed of the cylinder to the speed of the spherical shell
Answer:
vcyl / vsph = 1.05
Explanation:
The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.The traslational part can be written as follows:[tex]K_{trans} = \frac{1}{2}* M* v_{cm} ^{2} (1)[/tex]
The rotational part can be expressed as follows:[tex]K_{rot} = \frac{1}{2}* I* \omega ^{2} (2)[/tex]
where I = moment of Inertia regarding the axis of rotation.ω = angular speed of the rotating object.If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:[tex]v = \omega * R (3)[/tex]
For a solid cylinder, I = M*R²/2 (4)Replacing (3) and (4) in (2), we get:[tex]K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2} (5)[/tex]
Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:[tex]K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2} +\frac{1}{4}* M* v_{cmc}^{2} = \frac{3}{4}* M* v_{cmc} ^{2} (6)[/tex]
Repeating the same steps for the spherical shell:[tex]I_{sph} = \frac{2}{3} * M* R^{2} (7)[/tex]
[tex]K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2} (8)[/tex]
[tex]K_{sph} = \frac{1}{2}* M* v_{cms} ^{2} +\frac{1}{3}* M* v_{cms}^{2} = \frac{5}{6}* M* v_{cms} ^{2} (9)[/tex]
Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.Rearranging, and taking square roots on both sides, we get:[tex]\frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)[/tex]
This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.Suppose a diode consists of a cylindrical cathode with a radius of 6.200×10−2 cm , mounted coaxially within a cylindrical anode with a radius of 0.5580 cm . The potential difference between the anode and cathode is 400 V . An electron leaves the surface of the cathode with zero initial speed (vinitial=0). Find its speed vfinal when it strikes the anode.
Answer:
The final speed will be "[tex]1.185\times 10^7 \ m/sec[/tex]".
Explanation:
The given values are:
Potential difference,
Δv = 400 v
Radius,
r = 0.5580 cm
As we know,
⇒ [tex]W=e \Delta v[/tex]
and,
⇒ [tex]\frac{1}{2}mv^2=e \Delta v[/tex]
then,
⇒ [tex]v=\sqrt{\frac{2e \Delta v}{m} }[/tex]
On substituting the values, we get
⇒ [tex]=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 400}{9.11\times 10^{-31}} }[/tex]
⇒ [tex]=\sqrt{\frac{1.6\times 10^{-19}\times 800}{9.11\times 10^{-31}}}[/tex]
⇒ [tex]=1.185\times 10^7 \ m/sec[/tex]
Which runner finished the 100 m race in the least amount of time?
Ming
Which runner stopped running for a few seconds during the race?
At what distance did Anastasia overtake Chloe in the race?
1: Ming
2: Chloe
3: 40m
May you please help?
Choice-A is the main reason that people use the thing in the picture.
Please solve for 15 points. Please don’t input a link.
Answer:
a). Single replacement.
Explanation:
Because one element replaces another element in a compound
Example of the center of the gravity
Answer:
The example of the center of the gravity is the middle of a seesaw
Explanation:
I hope this will help you and plz mark me brainlist
Look at the diagram showing the different wavelengths in sunlight.
A diagram showing the human eye and visible light. Visible light is broken down by color with wavelength in nanometers. Red is 700, orange is 600, yellow is 580, green is 550, blue is 475, indigo is 450, violet is 400.
Which has a wavelength of 350 nanometers?
red light
violet light
infrared light
ultraviolet light
Answer:
ultraviolet light
plz mark me as brainliest.
Answer:
Ultra violet
Explanation:
What fuel does a main-sequence star use for nuclear fusion?
oxygen (0)
petroleum
helium (He)
hydrogen (H)
Answer:
A main sequence star is powered by fusion of hydrogen into helium in its core
Explanation:
Easy question I’ll give extra points. Help
Answer:
the answer is A
I wish you a good day!
Answer:
THe answer is graph 2 as it represents the puck going in a linear motion
Explanation:
light of wavelength 485 nm passes through a single slit of width 8.32 *10^-6m. what is the single between the first (m=1) and second (m=2) interference minima?
Answer:
3.35
Explanation:
Got it on Acellus
The light of wavelength 485 nm passes through a single slit. The single between the first (m=1) and second (m=2) interference minima is 3.36°.
What is diffraction?Diffraction is the phenomenon of bending of waves through obstacles.
Given is the wavelength λ= 485 nm, silt width d = 8.32 *10⁻⁶ m, then the angle θ will be
d sinθ =mλ
for m=1, sin θ₁ = λ/d
for m=2, sin θ₂ = 2λ/d
Substitute the values into both expressions to find the angles,
sin θ₁ = 485 x 10⁻⁹ / 8.32 *10⁻⁶
θ₁ = 3.34°
and sin θ₂ = (2 x 485 x 10⁻⁹ )/ 8.32 *10⁻⁶
θ₂ = 6.7°
The angle between m =1 and m=2 will be
θ₂ -θ₁ = 6.7° - 3.34° =3.36°
Thus, angle between the first (m=1) and second (m=2) interference minima is 3.36°.
Learn more about diffraction.
https://brainly.com/question/12290582
#SPJ2
Acellus
What are harmful substances in the air,
soil, and water called?
A. pollutants
B. toxins
C. carcinogens
D. mutants
The eight plants of the Solar System orbit the Sun in a chaotic random way.
True
False
Answer:
The Solar System has plants? I assume you meant planets. If so, that is false
Explanation:
An 80.0-kg skydiver jumps out of a balloon at an altitude of 1,000 m and opens his parachute at an altitude of 200 m. A. Assuming the total friction (resistive) force on the skydiver is constant at 50.0 N with the parachute closed and constant at 3,600 N with the parachute open, find the speed of the skydiver when he lands on the ground. B. At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground is 5.00 m/s
Answer:
[tex]24.9\ \text{m/s}[/tex]
[tex]206.7\ \text{m}[/tex]
Explanation:
m = Mass of skydiver = 80 kg
[tex]x_1[/tex] = Height for which the parachute is closed = 1000-200 = 800 m
[tex]x_2[/tex] = Height for which the parachute is open = 200 m
[tex]f_1[/tex] = Resistive force when parachute is closed = 50 N
[tex]f_2[/tex] = Resistive force when parachute is open = 3600 N
v = Velocity of skydiver on the ground
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
h = Height from which the skydiver jumps = 1000 m
The energy balance of the system will be
[tex]mgh-f_1x_1-f_2x_2=\dfrac{1}{2}mv^2\\\Rightarrow 80\times 9.81\times 1000-50\times 800-3600\times 200=\dfrac{1}{2}\times 80\times v^2\\\Rightarrow v=\sqrt{\dfrac{2(80\times 9.81\times 1000-50\times 800-3600\times 200)}{80}}\\\Rightarrow v=24.9\ \text{m/s}[/tex]
The velocity fo the skydiver when he lands will be [tex]24.9\ \text{m/s}[/tex]
x = Height where the person opens the parachute
v = 5 m/s
[tex]mgh-f_1x_1-f_2x_2=\dfrac{1}{2}mv^2\\\Rightarrow 80\times 9.81\times 1000-50\times (1000-x)-3600\times x=\dfrac{1}{2}\times 80\times 5^2\\\Rightarrow 80\times 9.81\times 1000-50000+50x-3600x=\dfrac{1}{2}\times 80\times 5^2\\\Rightarrow x=\dfrac{80\times 9.81\times 1000-50000-\dfrac{1}{2}\times 80\times 5^2}{3550}\\\Rightarrow x=206.7\ \text{m}[/tex]
The height at which the parachute is to be opened is [tex]206.7\ \text{m}[/tex]
a disk of a radius 50 cm rotates at a constant rate of 100 rpm. what distance in meters will a point on the outside rim travel during 30 seconds of rotation?
Answer:
Wait lang po sandali po wait lang
Explain why your image never disappears and never flips over as you bring the convex mirror
close to your eye.
Explanation:
When you get closer to the mirror than the focal point a virtual image is formed behind the mirror and this image is not inverted. That's why the image flips as you get closer. ... With a virtual image the light rays never come to a focus so there is no place you can put a piece of paper to see the image.
A 5kg cart moving to the right with a velocity of 16 m/s collides with a concrete wall and
rebounds with a velocity of 22 m/s. Is the change in momentum of the cart
Explanation:
mass, m = 5kg
initial velocity, u = 16m/s
final velocuty, v = -22m/s
change in momentum, ∆p = ?
∆p = m (v-u)
5(-22-16)
5(38)
∆p = 190kgm/s
check the calculations!
Two identical loudspeakers are driven in phase by the same amplifier. The speakers are positioned a distance of 3.2 m apart. A person stands 5.0 m away from one speaker and 6.2 m away from the other. Calculate the second lowest frequency that results in destructive interference at the point where the person is standing. Assume the speed of sound to be 330 m/ s. A) 183 Hz B) 275 Hz C) 413 Hz D) 137 Hz E) 550 Hz
Answer:
C) 413 Hz
Explanation:
For destructive interference, the path difference ΔL = (n + 1/2)λ where ΔL = L₂ - L₁ where L₁ = person's distance from one speaker (the closer one) = 5.0m and L₂ = person's distance from other speaker (the farther one) = 6.2 m and λ = wavelength = v/f where v = speed of sound = 330 m/s and f = frequency
So, ΔL = (n + 1/2)λ
L₂ - L₁ = (n + 1/2)v/f
f = (n + 1/2)v/(L₂ - L₁)
At the second lowest frequency that results in destructive interference at the point where the person is standing, n = 1.
So,
f = (1 + 1/2)v/(L₂ - L₁)
f = 3v/2(L₂ - L₁)
Substituting the values of the variables into the equation, we have
f = 3v/2(L₂ - L₁)
f = 3(330 m/s)/2(6.2 m - 5.0 m)
f = 3(330 m/s)/2(1.2 m)
f = 990 m/s ÷ 2.4 m)
f = 412.5 Hz
f ≅ 413 Hz
if matter cannot be created nor destroyed where did the matter in the univerce com from ???????
BIG BRAIN
i have no idea
I don't even know how we are here
If a 75 W lightbulb is 15% efficient, how many joules of light energy does the bulb produce every minute?
Answer:
1 W = 1 J / sec Definition of watt is 1 joule / sec
So if a bulb uses 75 J / sec it must use
75 J/s * 60 sec / min = 4500 J/min energy used by bulb
If bulb is 15% efficient then the light delivered is
P = 4500 J / min * .15 = 675 J / min
A solenoid of 2100 turns, area 10 cm2, and length 30 cm carries a current of 4.0 A. (a) Calculate the magnetic energy stored in the solenoid from 1/2 LI 2. J [2 points] 0 attempt(s) made (maximum allowed for credit
Answer:
E = 0.1472 J
Explanation:
Given that,
The number of turns in the solenoid, N = 2100
Area of the solenoid, A = 10 cm² = 0.001 m²
The length of the solenoid, l = 30 cm = 0.3 m
Current in the solenoid, I = 4 A
We need to find the magnetic energy stored in the solenoid. The expression for the stored energy is :
[tex]E=\dfrac{1}{2}LI^2[/tex]
Where
L is self inductance of the solenoid,
[tex]L=\dfrac{\mu_oN^2A}{l}\\\\L=\dfrac{4\pi \times 10^{-7}\times 2100^2\times 0.001}{0.3}\\\\L=0.0184\ H[/tex]
So,
[tex]E=\dfrac{1}{2}\times 0.0184\times 4^2\\\\E=0.1472\ J[/tex]
So, 0.1472 J of energy is stored in the solenoid.