Answer:
4.981 MeV
Explanation:
The quantity of energy Q can be calculated using the formula
Q = (mass before - mass after) × c²
Atomic Mass of thorium = 232.038054 u, atomic of Radium = 228.0301069 u and mass of Helium = 4.00260. The difference of atomic number and atomic mass between the thorium and radium ( 232 - 228) and ( 90 - 88) show α particle was emitted.
1 u = 931.494 Mev/c²
Q = (mass before - mass after) × c²
Q = ( mass of thorium - ( mass of Radium + mass of Helium ) )× c²
Q = 232.038054 u - ( 228.0301069 + 4.00260) × c²
Q = 0.0053471 u × c²
replace 1 u = 931.494 MeV/ c²
Q = 0.0053471 × c² × (931.494 MeV / c²)
cancel c² from the equation
Q = 0.0053471 × 931.494 MeV = 4.981 MeV
A rock of mass m is thrown straight up into the air with initial speed |v0 | and initial position y = 0 and it rises up to a maximum height of y = h. A second rock with mass 2m (twice the mass of the original) is thrown straight up with an initial speed of 2|v0 |. What maximum height does the second rock reach?
Answer:
Explanation:
Case 1:
mass = m
initial velocity = vo
final velocity = 0
height = y
Use third equation of motion
v² = u² - 2as
0 = vo² - 2 g y
y = vo² / 2g ... (1)
Case 2:
mass = 2m
initial velocity = 2vo
final velocity = 0
height = y '
Use third equation of motion
v² = u² - 2as
0 = 4vo² - 2 g y'
y ' = 4vo² / 2g
y' = 4 y
Thus, the second rock reaches the 4 times the distance traveled by the first rock.
The maximum height the second rock reach is :
-4 times the distance traveled by the first rock.
"Mass"Case 1:
mass = m
initial velocity = vo
final velocity = 0
height = y
using Third equation of motion
v² = u² - 2as
0 = vo² - 2 g y
y = vo² / 2g ... (1)
Case 2:
mass = 2m
initial velocity = 2vo
final velocity = 0
height = y '
Use third equation of motion
v² = u² - 2as
0 = 4vo² - 2 g y'
y ' = 4vo² / 2g
y' = 4 y
Therefore, the second rock reaches the 4 times the distance traveled by the first rock.
Learn more about "Mass":
https://brainly.com/question/15959704?referrer=searchResults
Calculate the rotational inertia of a meter stick, with mass 0.71 kg, about an axis perpendicular to the stick and located at the 18 cm mark. (Treat the stick as a thin rod.)
To solve this problem we will use the parallel axis theorem for which the inertia of a point of an object can be found through the mathematical relation:
[tex]I = I_{cm} +mx^2[/tex]
Where
[tex]I_{cm}[/tex] = Inertia at center of mass
m = mass
x = Displacement of axis.
Our mass is given as 0.71kg,
m = 0.71kg
Para a Stick with length (L) the Moment of Inertia of the stick about and axis passing through the center and perpendicular to stick is
[tex]I_{cm} = \frac{1}{12} mL^2[/tex]
[tex]I_{cm} = \frac{1}{12} (0.71)(1)^2[/tex]
[tex]I_{cm} = 0.05916Kg\cdot m^2[/tex]
The distance between center of mass to the specific location is
[tex]x = 50cm - 18cm[/tex]
[tex]x = 38cm = 0.38m[/tex]
So, from parallel axis theorem ,
[tex]I = I_{cm} + mx^2[/tex]
[tex]I =0.05916Kg\cdot m^2+ (0.71kg)(0.38m)^2[/tex]
[tex]I = 0.161684Kg\cdot m^2[/tex]
Therefore the rotational inertia is [tex]0.161684Kg\cdot m^2[/tex]
8–4. The tank of the air compressor is subjected to an internal pressure of 90 psi. If the internal diameter of the tank is 22 in., and the wall thickness is 0.25 in., determine the stress components acting at point A. Draw a volume element of the material at this point, and show the results on the element.
Answer:
The stress S = 1935 [Psi]
Explanation:
This kind of problem belongs to the mechanical of materials field in the branch of the mechanical engineering.
The initial data:
P = internal pressure [Psi] = 90 [Psi]
Di= internal diameter [in] = 22 [in]
t = wall thickness [in] = 0.25 [in]
S = stress = [Psi]
Therefore
ri = internal radius = (Di)/2 - t = (22/2) - 0.25 = 10.75 [in]
And using the expression to find the stress:
[tex]S=\frac{P*D_{i} }{2*t} \\replacing:\\S=\frac{90*10.75 }{2*0.25} \\S=1935[Psi][/tex]
In the attached image we can see the stress σ1 & σ2 = S acting over the point A.
A forward-biased silicon diode is connected to a 12.0-V battery through a resistor. If the current is 12 mA and the diode potential difference is 0.70 V, what is the resistance?
To solve this problem we will use the concepts related to Ohm's law for which voltage, intensity and resistance are related.
Mathematically this relationship is given as
[tex]V = IR \rightarrow R= \frac{V}{I}[/tex]
Where,
V= Voltage
I = Current
R = Resistance
The value of the given voltage is 12V, while the current is 12mA, therefore the resistance would be
[tex]R = \frac{12}{12*10^{-3}}[/tex]
[tex]R = 1000 \Omega[/tex]
Therefore the resistance is [tex]1000\Omega[/tex]
There are usually _________ collisions in a motor vehicle crash.
Answer:
3
Explanation:
During a crash 3 types of collisions can occur.
There are usually three types of collisions involved in a motor vehicle crash.
The first type of collision is the vehicle collision, which involves the physical impact between two or more vehicles. When vehicles collide, their structures deform, and the forces involved can cause severe damage to the involved vehicles. The severity of this collision depends on factors like the speed, mass, and angle of impact.
The second type of collision is the human collision, which occurs inside the vehicle. During an accident, passengers inside the vehicle can collide with each other or with interior components, such as the dashboard, steering wheel, or windows. These collisions can result in injuries like whiplash, head injuries, or broken bones.
The third type of collision is the internal collision, which involves the organs and tissues within the human body. When a collision occurs, the human body is subjected to rapid deceleration, causing organs to collide with each other or with the skeletal structure. These internal collisions can lead to internal bleeding, organ damage, and other life-threatening injuries.
To know more about collision here
https://brainly.com/question/30636941
#SPJ6
The inductance in the drawing has a value of L = 9.4 mH. What is the resonant frequency f0 of this circuit?
Answer:
The resonant frequency of this circuit is 1190.91 Hz.
Explanation:
Given that,
Inductance, [tex]L=9.4\ mH=9.4\times 10^{-3}\ H[/tex]
Resistance, R = 150 ohms
Capacitance, [tex]C=1.9\ \mu F=1.9\times 10^{-6}\ C[/tex]
At resonance, the capacitive reactance is equal to the inductive reactance such that,
[tex]X_C=X_L[/tex]
[tex]2\pi f_o L=\dfrac{1}{2\pi f_oC}[/tex]
f is the resonant frequency of this circuit
[tex]f_o=\dfrac{1}{2\pi \sqrt{LC}}[/tex]
[tex]f_o=\dfrac{1}{2\pi \sqrt{9.4\times 10^{-3}\times 1.9\times 10^{-6}}}[/tex]
[tex]f_o=1190.91\ Hz[/tex]
So, the resonant frequency of this circuit is 1190.91 Hz. Hence, this is the required solution.
A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 4.10 L ? (The temperature was held constant.)
Answer:
0.358g
Explanation:
Density of Helium = 0.179g/L
ρ=m/v
m=ρv
when the volume was 2L
m1= 0.179*2
m1=0.358g
when the volume increased to 4L
m2= 0.179*4
m2=0.716g
gram of helium added = 0.716g-0.358g
=0.358g
A block of mass m = 0.775 kg is fastened to an unstrained horizontal spring whose spring constant is k = 83.6 N/m. The block is given a displacement of +0.113 m, where the + sign indicates that the displacement is along the +x axis, and then released from rest. What is the force (magnitude) that the spring exerts on the block just before the block is released?
Answer:
F= 9.45 N
Explanation:
If the mass is fastened to an unstrained horizontal spring, this means that at this position, the spring doesn't exert any force, because it keeps his equilibrium length.
If then the block is given a displacement of +0.113m, this means that the spring has been stretched in the same length.
According to Hooke's Law, the spring exerts a restoring force (trying to return to his equilibrium state) that opposes to the displacement, and which is proportional (in magnitude) to it, being the proportionality constant, a quantity called spring constant, which depends on the type of spring.
We can write the Hooke's Law as follows:
F = - k * Δx
Just before the block is released, we can get the value of F as follows:
⇒ F = 83.6 N/m* 0.113 m = 9.45 N (in magnitude)
Scientists are working on a new technique to kill cancer cells by zapping them with ultrahigh-energy (in the range of 1012 W) pulses of electromagnetic waves that last for an extremely short time (a few nanoseconds). These short pulses scramble the interior of a cell without causing it to explode, as long pulses would do. We can model a typical such cell as a disk 4.6 μm in diameter, with the pulse lasting for 3.4 ns with an average power of 2.46×1012 W . We shall assume that the energy is spread uniformly over the faces of 100 cells for each pulse.
Part A
How much energy is given to the cell during the pulse?
Express your answer to two significant figures.
The energy given to each cell during the pulse can be calculated by multiplying the power of the pulse by its duration, and then dividing by the number of cells.
Explanation:The energy supplied to the cell during the pulse is determined by the power multiplied by the duration of the pulse. In this scenario, the power is 2.46×1012 W and the duration is 3.4 ns (which is 3.4x10-9 s when converted to seconds for mathematical calculations).
We use the formula:
E = P * t
Where,
E is the Energy
P is the Power
t is the time (duration of the pulse)
Substituting the given values into the formula:
E = 2.46x1012 W * 3.4x10-9 s
This gives the total energy supplied. We know the energy is spread uniformly over the faces of 100 cells, so each cell will get 1/100 of the total energy. Using these calculations, we can determine the amount of energy given to each cell during the pulse.
Learn more about Energy Calculation here:https://brainly.com/question/31994371
#SPJ12
(a) A proton is confined to the nucleus of an atom. Assume the nucleus has diameter 5.5 x 10-15 m and that this distance is the uncertainty in the proton's position. What is the minimum uncertainty in the momentum of the proton? Dpmin = kg-m/s
(b) An electron is confined in an atom. Assume the atom has diameter 1 x 10-10 m and that this distance is the uncertainty in the electron's position. What is the minimum uncertainty in the momentum of the electron?
Answer:
[tex]1.91738\times 10^{-20}\ kgm/s[/tex]
[tex]1.05456\times 10^{-24}\ kgm/s[/tex]
Explanation:
h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]
[tex]\Delta x[/tex] = Uncertainty in the position
[tex]\Delta p[/tex] = Uncertainty in the momentum
From the uncertainty principle
[tex]\Delta x\Delta p=\dfrac{h}{2\pi}\\\Rightarrow \Delta p=\dfrac{h}{2\pi \Delta x}\\\Rightarrow \Delta p=\dfrac{h}{2\pi \Delta x}\\\Rightarrow \Delta p=\dfrac{6.626\times 10^{-34}}{2\pi\times 5.5\times 10^{-15}}\\\Rightarrow \Delta p=1.91738\times 10^{-20}\ kgm/s[/tex]
The minimum uncertainty in the momentum of the proton is [tex]1.91738\times 10^{-20}\ kgm/s[/tex]
[tex]\Delta x\Delta p=\dfrac{h}{2\pi}\\\Rightarrow \Delta p=\dfrac{h}{2\pi \Delta x}\\\Rightarrow \Delta p=\dfrac{h}{2\pi \Delta x}\\\Rightarrow \Delta p=\dfrac{6.626\times 10^{-34}}{2\pi\times 1\times 10^{-10}}\\\Rightarrow \Delta p=1.05456\times 10^{-24}\ kgm/s[/tex]
The minimum uncertainty in the momentum of the electron is [tex]1.05456\times 10^{-24}\ kgm/s[/tex]
Final answer:
The minimum uncertainty in the momentum of a proton confined to the nucleus of an atom is 9.6 × 10^-12 kg m/s. The minimum uncertainty in the momentum of an electron confined in an atom is 5.3 × 10^-24 kg m/s.
Explanation:
The Heisenberg Uncertainty Principle states that there is a limit to the precision with which we can know both the position and momentum of a particle. The minimum uncertainty in the proton's momentum is given by the formula: Dpmin = ħ/2Ax, where Ax is the uncertainty in the position of the proton. In this case, Ax is given as the diameter of the nucleus, so we have:
Dpmin = (1.055 × 10^(-34) kg m^2/s) / (2(5.5 × 10^(-15) m)) = 9.6 × 10^(-12) kg m/s.
Similarly, for the electron:
Dpmin = (1.055 × 10^(-34) kg m^2/s) / (2(1 × 10^(-10) m)) = 5.3 × 10^(-24) kg m/s.
I am standing next to the edge of a cliff. I throw a ball upwards and notice that 4 seconds later it is traveling downwards at 10 m/s. Where is the ball located at this time? (Pick the answer closest to the true value.)A. 120 meters above me B. 30 meters below meC. 30 meters above meD. 120 meters below meE. At the same height that it started
Answer:
Explanation:
Given
Velocity after t=4 sec is v=10 m/s downward
assuming u is the initial upward velocity
[tex]v=u+at[/tex]
[tex]-10=u-gt[/tex]
[tex]u=9.8\times 4-10=29.2 m/s[/tex]
[tex]v^2-u^2=2 as[/tex]
[tex](-10)^2-(29.2)^2=2\times (-9.8)\cdot s[/tex]
[tex]s=\frac{29.2^2-10^2}{2\times 9.8}[/tex]
[tex]s=38.4 m[/tex]
i.e. 38.4 m above the initial thrown Position
The Earth’s radius is 6378.1 kilometers. If you were standing at the equator, you are essentially undergoing uniform circular motion with the radius of your circular motion being equal to the radius of the Earth. You are an evil mad scientist and have come up with the simultaneously awesome and terrifying plan to increase the speed of the Earth’s rotation until people at the Earth’s equator experience a centripetal (radial) acceleration with a magnitude equal to g, (9.81 m/s2 ), effectively making them experience weightlessness. If you succeed in your dastardly plan, what would be the new period of the Earth’s rotation?
a. 2.7 minutes b. 84 minutes c. 48 minutes d. 76 minutes
Answer:
b. 84 minutes
Explanation:
[tex]a_c=g[/tex] = Centripetal acceleration = 9.81 m/s²
r = Radius of Earth = 6378.1 km
v = Velocity
Centripetal acceleration is given by
[tex]a_c=\dfrac{v^2}{r}\\\Rightarrow v=\sqrt{a_cr}\\\Rightarrow v=\sqrt{9.81\times 6378100}\\\Rightarrow v=7910.06706\ m/s[/tex]
Time period is given by
[tex]T=\dfrac{2\pi r}{v60}\\\Rightarrow T=\dfrac{2\pi 6378.1\times 10^3}{7910.06706\times 60}\\\Rightarrow T=84.43835\ minutes[/tex]
The time period of Earth’s rotation would be 84.43835 minutes
The new period of the Earth’s rotation is mathematically given as
T=84.43835 min
What would be the new period of the Earth’s rotation?Question Parameter(s):
The Earth’s radius is 6378.1 kilometers.
g= (9.81 m/s2 ),
Generally, the equation for the is mathematically given as
[tex]a_c=\dfrac{v^2}{r}[/tex]
Therefore
[tex]v=\sqrt{a_cr}\\\\v=\sqrt{9.81*6378100}[/tex]
v=7910.06706 m/s
In conclusion
[tex]T=\dfrac{2\pi r}{v60}[/tex]
Hence
[tex]T=\dfrac{2\pi 6378.1*10^3}{7910.06706*60}[/tex]
T=84.43835 min
Read more about Time
https://brainly.com/question/4931057
Why is it impossible for an astronaut inside an orbiting space station to go from one end to the other by walking normally?A. In an orbiting station, the gravitational force is too large and the astronaut can't take his feet off the floor.B. It is impossible to walk inside an orbiting space station because its rotation is too fast.C. In an orbiting station, after one foot pushes off there isn't a friction force to move forward. The astronaut "jumps" on the same place.D. In an orbiting station, after one foot pushes off there isn't a force to bring the astronaut back to the "floor" for the next step.
Final answer:
An astronaut cannot walk normally in a space station because there's no frictional force to move forward in the near-weightless environment. To move, astronauts use handholds and walls, pushing against them to create a reaction force.
Explanation:
It is impossible for an astronaut inside an orbiting space station to go from one end to the other by walking normally because C. In an orbiting station, after one foot pushes off there isn't a friction force to move forward. The astronaut would indeed "jump" in place due to the lack of friction between their feet and the floor of the space station, which is a result of the near-weightlessness they experience. In space, normal walking is ineffective because walking relies on gravity to pull the body back down to the floor after each step, which isn't present in the same way on a space station in orbit.
In order to move in such an environment, an astronaut must push against a solid object, creating a reaction force in the opposite direction according to Newton's third law of motion. This principle allows the astronaut to propel and steer themselves around the space station using handholds and walls. The environment inside the ISS is similar to that inside a freely falling box where gravity still exists, but occupants do not feel its effects because they are in free fall around Earth, which creates the sensation of weightlessness.
Final answer:
Astronauts cannot walk normally in an orbiting space station due to the lack of gravity and friction. They are in a state of free fall, creating a sensation of weightlessness. Movement can be achieved by utilizing the conservation of momentum and Newton's third law of motion. Therefore option C is the correct answer.
Explanation:
The reason it is impossible for an astronaut inside an orbiting space station to walk from one end to the other by walking normally is C. In an orbiting station, after one foot pushes off there isn't a friction force to move forward. The astronaut cannot walk from one end to the other by walking normally because, in the microgravity environment of an orbiting spacecraft, traditional walking, which relies on the force of gravity and friction between the feet and the ground, does not work. Instead, astronauts move about by pushing off surfaces or floating through the air.
In orbit, the International Space Station (ISS) and everything inside it, including the astronauts, are in a state of free fall. They are falling around Earth at the same rate as the space station, creating a sensation of weightlessness. This is akin to the sensation of temporary weightlessness one experiences at the topmost point of a roller coaster ride or when an elevator suddenly descends.
Achieving locomotion for an astronaut stranded in the center of the station without contact with any solid surface would necessitate a method that does not rely on gravity or friction. The astronaut would have to utilize the principle of conservation of momentum. For instance, by throwing an object in one direction, the astronaut would move in the opposite direction, as described by Newton's third law of motion: for every action, there is an equal and opposite reaction.
To understand the formula representing a traveling electromagnetic wave.Light, radiant heat (infrared radiation), X rays, and radio waves are all examples of traveling electromagnetic waves. Electromagnetic waves comprise combinations of electric and magnetic fields that are mutually compatible in the sense that the changes in one generate the other.The simplest form of a traveling electromagnetic wave is a plane wave. For a wave traveling in the x direction whose electric field is in the y direction, the electric and magnetic fields are given byE? =E0sin(kx??t)j^,B? =B0sin(kx??t)k^.This wave is linearly polarized in the y direction.1.a. In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables E0 and B0are the __________ of the electric and magnetic fields.Choose the best answer to fill in the blank.1. maxima2. amplitudes3. wavelengths4. velocitiesb. The variable ? is called the __________ of the wave.Choose the best answer to fill in the blank.1. velocity2. angular frequency3. wavelengthc. The variable k is called the __________ of the wave.1. wavenumber
2. wavelength
3. velocity
4. frequency
Answer:
1) Eo and Bo. They are maximum amplitudes. Answer 1 and 2
2) .w is angular frequency. Answer 2
3) k is wave number. Answer 1
Explanation:
The electromagnetic wave is given by
[tex]E_{y}[/tex] = E₀ sin (kx –wt)
This is the equation of a traveling wave on the x axis with the elective field oscillating on the y axis
The terms represent E₀ the maximum amplitude of the electric field,
The wave vector
k = 2π /λ
Angular velocity
w = 2π f
To answer the questions let's use the previous definitions
1) Eo and Bo. They are maximum amplitudes. Answer 1 and 2
2) .w is angular frequency. Answer 2
3) k is wave number. Answer 1
The headlights of a car are 1.6 m apart and produce light of wavelength 575 nm in vacuum. The pupil of the eye of the observer has a diameter of 4.0 mm and a refractive index of 1.4. What is the maximum distance from the observer that the two headlights can be distinguished?
To solve this problem it is necessary to apply the concepts related to angular resolution, for which it is necessary that the angle is
[tex]\theta = 1.22\frac{\lambda}{nd}[/tex]
Where
d = Diameter of the eye
n = Index of refraction
D = Distance between head lights
[tex]\lambda[/tex]= Wavelength
Replacing with our values we have that
[tex]\theta = 1.22 \frac{(1.22)(575*10{-9})}{1.4(4*10^{-3})}[/tex]
[tex]\theta = 1.252*10^{-4}rad[/tex]
Using the proportion of the arc length we have to
[tex]L = \frac{D}{\theta}[/tex]
Where L is the maximum distance, therefore
[tex]L = \frac{1.6}{1.252*10^{-4}}[/tex]
[tex]L = 12.77km[/tex]
Therefore the maximum distance from the observer that the two headlights can be distinguished is 12.77km
Arace car accelerates uniformly at 11.3 m/s2. If the race car starts from rest how fast will it
be going after 6.7 seconds.
Answer:
75.71 m/s
Explanation:
From equation of motion, acceleration is given by
[tex]a=\frac {v-u}{t}[/tex]where v is the final velocity, u is the initial velocity and t is time taken.
Making v the subject of the above formula
v=at+u
Substituting 6.7 s for time, t and 11.3 for a and taking u as zero since it starts from rest
v=11.3*6.7=75.71 m/s
A metal bar is used to conduct heat. When the temperature at one end is 100°C and at the other is 20°C, heat is transferred at a rate of 16 J/s.
If the temperature of the hotter end is reduced to 80°C, what will be the rate of heat transfer?
a. 4 J/sb. 8 J/sc. 9 J/sd. 12 J/s
Answer:
a. 4 J/s
Explanation:
Fourier's law states for the case in which there is stationary heat flow in only one direction, that is, linearly, the heat transmitted per unit of time is proportional to the temperature difference:
[tex]\frac{Q}{t}\propto \Delta T[/tex]
When the temperature at one end is 100°C and at the other is 20°C, we have:
[tex]\Delta T_1=100^\circ C-20^\circ C\\\Delta T_1=80^\circ C[/tex]
If the temperature of the hotter end is [tex]80^\circ C[/tex], we have:
[tex]\Delta T_2=100^\circ C-80^\circ C\\\Delta T_2=20^\circ C[/tex]
So:
[tex]\Delta T_1=4\Delta T_2\\\Delta T_2=\frac{\Delta T_1}{4}[/tex]
Finally, we calculate the rate of heat transfer:
[tex]\frac{Q_2}{t_2}=\frac{\frac{Q_1}{t_1}}{4}\\\\\frac{Q_2}{t_2}=\frac{16\frac{J}{s}}{4}\\\frac{Q_2}{t_2}=4\frac{J}{s}[/tex]
Did you think about this over Christmas? I did (-: Before Christmas a 65kg student consumes 2500 Cal each day and stays at the same weight. For three days in a row while visiting her parents she eats 3500 Cal and, wanting to keep from gaining weight decides to "work off" the excess by jumping up and down at the Christmas tree. With each jump she accelerates to a speed of 3.2 m/s before leaving the ground. a) How high will she jump each jump? b) How many jumps must she do to keep her weight? Assume that the efficiency of the body in using energy is 25%. c) Do you suggest that is a reasonable way for the student not to gain weight over Christmas? d) Possible enhancement: What other way/ways would you suggest for the student to keep her weight?
Answer:
a) Em = 332.8 J , b) # jump = 13, c) It is reasonable since there are not too many jumps , d) lower the calories consumed
Explanation:
a) Let's use energy conservation
Initial. On the floor
Em₀ = K = ½ m v²
Final. The highest point
Emf = U = m g h
Energy is conserved
Em₀ = Emf
½ m v² = m g h
h = ½ v² / g
h = ½ 3.2² /9.8
h = 0.52 m
b) When he was at home he maintained his weight with 2500 cal / day. In his parents' house he consumes 3500 cal / day, the excess of calories is
Q = 3500 -2500 = 1000cal / day
Let's reduce this value to the SI system
Q = 1000 cal (4,184 J / 1 cal) = 4186 J / day
Now the energy in each jump is
Em = K = ½ m v²
Em = ½ 65 3.2²
Em = 332.8 J
They indicate that the body can only use 25% of this energy
Em effec = 0.25 332.8 J
Em effec = 83.2 J
This is the energy that burns the body
Let's use a Proportion Rule (rule of three), if a jump spends 83.2J how much jump it needs to spend 1046 J
# jump = 1046 J (1 jump / 83.2 J)
# jump = 12.6 jumps / day
# jump = 13
c) It is reasonable since there are not too many jumps
d) That some days consume more vegetables to lower the calories consumed
A continuous and aligned fiber-reinforced composite having a cross-sectional area of 1130 mm2 is subjected to an external tensile load. If the stresses sustained by the fiber and matrix phases are 156 MPa and 2.75 MPa, respectively, the force sustained by the fiber phase is 74,000 N and the total longitudinal strain is 1.25 x 10-3, what is the value of the modulus of elasticity of the composite material in the longitudinal direction?
Answer:
Ec=53.7×10⁹N/m² =53.7Gpa
Explanation:
To calculate the modulus of elasticity in the longitudinal direction. This is possible realizing Ec=σ/ε where σ=(Fm+Ff)/Ac
[tex]Ec=Sigma/E\\Ec=\frac{(Fm+Ff)/E}{Ac}\\ Ec=\frac{1802+74,000}{(1.25*10^{-3})(1130)(1/1000)^{2} }\\ Ec=53.7*10^{9}N/m^{2}\\or\\Ec=53.7GPa[/tex]
Final answer:
The modulus of elasticity of the composite material in the longitudinal direction is 124,800 MPa.
Explanation:
To find the modulus of elasticity of the composite material in the longitudinal direction, we can use the formula:
E = (stress sustained by the fiber phase)/(longitudinal strain)
Given that the stress sustained by the fiber phase is 156 MPa and the total longitudinal strain is 1.25 x 10^-3, we can plug in these values to calculate the modulus of elasticity:
E = 156 MPa / (1.25 x 10^-3) = 124,800 MPa
Therefore, the modulus of elasticity of the composite material in the longitudinal direction is 124,800 MPa.
A singly charged positive ion has a mass of 3.46 × 10−26 kg. After being accelerated through a potential difference of 215 V the ion enters a magnetic field of 0.522 T in a direction perpendicular to the field. The charge on the ion is 1.602 × 10−19 C. Find the radius of the ion’s path in the field. Answer in units of cm.
Answer:
1.8 cm
Explanation:
[tex]m[/tex] = mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg
[tex]q[/tex] = charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C
[tex]\Delta V[/tex] =Potential difference through which the ion is accelerated = 215 V
[tex]v[/tex] = Speed of the ion
Using conservation of energy
Kinetic energy gained by ion = Electric potential energy lost
[tex](0.5) m v^{2} = q \Delta V\\(0.5) (3.46\times10^{-26}) v^{2} = (1.6\times10^{-19}) (215)\\(1.73\times10^{-26}) v^{2} = 344\times10^{-19}\\v = 4.5\times10^{4} ms^{-1}[/tex]
[tex]r[/tex] = Radius of the path followed by ion
[tex]B[/tex] = Magnitude of magnetic field = 0.522 T
the magnetic force on the ion provides the necessary centripetal force, hence
[tex]qvB = \frac{mv^{2} }{r} \\qB = \frac{mv}{r}\\r =\frac{mv}{qB}\\r =\frac{(3.46\times10^{-26})(4.5\times10^{4})}{(1.6\times10^{-19})(0.522)}\\r = 0.018 m \\r = 1.8 cm[/tex]
A projectile of mass m is fired straight upward from the surface of an airless planet of radius R and mass M with an initial speed equal to the escape speed vesc (meaning the projectile will just barely escape the planet's gravity -- it will asymptotically approach infinite distance and zero speed.) What is the correct expression for the projectile's kinetic energy when it is a distance 9R from the planet's center (8R from the surface). Ignore the gravity of the Sun and other astronomical bodies. KE (at r = 9R) is:a. GMm/9Rb. GMm/8Rc. 1/2mvesc^2d. -GMm/8Re. None of these
Answer:
K = G Mm / 9R
Explanation:
Expression for escape velocity V_e = [tex]\sqrt{\frac{2GM}{R} }[/tex]
Kinetic energy at the surface = 1/2 m V_e ²
= 1/2 x m x 2GM/R
GMm/R
Potential energy at the surface
= - GMm/R
Total energy = 0
At height 9R ( 8R from the surface )
potential energy
= - G Mm / 9R
Kinetic energy = K
Total energy will be zero according to law of conservation of mechanical energy
so
K - G Mm / 9R = 0
K = G Mm / 9R
A proton initially traveling at 50,000 m/s is shot through a small hole in the negative plate of a parallal-plate capacitor. The electric field strength inside the capacitor is 1,500 V/m. How far does the proton travel above the negative plate before temporarily coming to rest and reversing course? Assume the proton reverses course before striking the positive plate.
Answer:
x = 8.699 10⁻³ m
Explanation:
The proton feels an electric charge that is the opposite direction of speed, let's look for acceleration using Newton's second law
F = m a
F = q E
a = q E / m
a = 1.6 10⁻¹⁹ 1500 / 1.67 10⁻²⁷
a = 1,437 10¹¹ m / s²
Now we can use kinematic relationships
v² = v₀² - 2 a x
When at rest the speed is zero (v = 0)
x = v₀² / 2 a
Let's calculate
x = 50,000² / (2 1,437 10¹¹)
x = 8.699 10⁻³ m
In Bob Shaw's short story, "The Light of Other Days," he describes something called slow glass. In the story, a married couple buys a 4-foot-wide window of slow glass that has been out on a beautiful hillside in Ireland, collecting light for 10 years. The idea is that the light takes 10 years to pass through the glass, so if you mount the window in your house it will give a view of the Irish landscape for the next 10 years, slowly unveiling everything that happened there. You can read the full short story via the link below, if you are interested.
Link to Bob Shaw's short story: The Light of Other Days.
(a) In the short story, the couple buys a window that is one-quarter-inch thick, and takes light 10 years to pass through. Let's say that you were able to locate a supplier of slow glass, and you bought some glass that was 5.00 mm thick, with the light taking 7.00 years to pass through. Taking one year to be 365.24 days, calculate the index of refraction of your piece of slow glass.
In 1999, Lene Hau, a physicist at Harvard University, received quite a bit of attention for getting light to travel at bicycle speed (later, she was able to temporarily stop light completely). The speed of a bicycle is a lot faster than light travels through the slow glass from the story, but it is still orders of magnitude less than the speed at which light travels through vacuum. If you're interested, you can follow this link to learn more about Lene Hau.
(b) Lene Hau used something called a Bose-Einstein condensate to slow down light. If the light is traveling at a speed of 40.0 km/hr through the Bose-Einstein condensate, what is the effective index of refraction of the condensate?
Answer:
Consider the following calculations
Explanation:
a ) velocity of the glass is v = distance / time
= 5 X 10-3 / 7 X 365.24 X 24 X 60 X 60
v = 2.263 X 10-11 m/sec
the speed of the light in vaccum is C
C = 3 X 108 m/sec
n = C / v
n = 3 X 108 / 2.263 X 10-11
n = 1.32567 X 1019
b ) given is 40 km/hr
= 40 X 103 / 60 X 60
= 11.11 m/sec
n = C / v
n = 3 X 108 / 11.11
n = 27002700.27
The index of refraction for the 5.00 mm thick slow glass taking light 7.00 years to pass through is approximately 1.33 x 10^19. For a Bose-Einstein condensate where light travels at 40.0 km/hr, the effective index of refraction is about 2.70 x 10^7.
In Bob Shaw's short story 'The Light of Other Days', a fictional material called slow glass is described, which delays the passage of light. To calculate the index of refraction of a 5.00 mm thick piece of slow glass where light takes 7.00 years to pass through, we can use the formula n = c/v, where c is the speed of light in a vacuum (3.00 x 108 m/s), and v is the speed of light through the material.
To find v, we can calculate the total distance light travels in 7.00 years and divide it by the time it takes to travel through the slow glass. Since the slow glass is 5.00 mm thick, which is equivalent to 5.00 x 10-3 m, and one year is 365.24 days, we calculate the speed as follows:
v = distance/time = 5.00 x 10-3 m / (7.00 years x 365.24 days/year x 24 hours/day x 3600 seconds/hour) = 5.00 x 10-3 m / 220,937,280 seconds ≈ 2.263 x 10-11 m/s.
Then, the index of refraction, n, can be calculated as n = c/v ≈ 3.00 x 108 m/s / 2.263 x 10-11 m/s ≈ 1.33 x 1019.
For Lene Hau's experiment using a Bose-Einstein condensate with a light speed of 40.0 km/hr, the index of refraction can also be calculated using n = c/v. Converting 40.0 km/hr to m/s:
v = 40.0 km/hr x (1000 m/km) / (3600 s/hr) = 11.1 m/s.
Using this value for v, we calculate n as n = c/v ≈ 3.00 x 108 m/s / 11.1 m/s ≈ 2.70 x 107.
If a nucleus decays by gamma decay to a daughter nucleus, which of the following statements about this decay are correct? (There may be more than one correct choice.)
a)The daughter nucleus has fewer protons than the original nucleus.
b)The daughter nucleus has the same number of nucleons as the original nucleus.
c)The daughter nucleus has more protons than the original nucleus.
d)The daughter nucleus has fewer neutrons than the original nucleus. The daughter nucleus has more neutrons than the original nucleus
Answer: Option (b) is the correct answer.
Explanation:
A gamma particle is basically a photon of electromagnetic radiation with a short wavelength.
Symbol of a gamma particle is [tex]^{0}_{0}\gamma[/tex]. Hence, charge on a gamma particle is also 0.
For example, [tex]^{234}_{91}Pa \rightarrow ^{234}_{91}Pa + ^{0}_{0}\gamma + Energy[/tex]
So, when a nucleus decays by gamma decay to a daughter nucleus then there will occur no change in the number of protons and neutrons of the parent atom but there will be loss of energy as a nuclear reaction has occurred.
Thus, we can conclude that the statement daughter nucleus has the same number of nucleons as the original nucleus., is correct about if a nucleus decays by gamma decay to a daughter nucleus.
Answer: Option (b) is the correct answer.
Explanation:
A gamma particle is basically a photon of electromagnetic radiation with a short wavelength.
A 100 g aluminum calorimeter contains 250 g of water. The two substances are in thermal equilibrium at 10°C. Two metallic blocks are placed in the water. One is a 50 g piece of copper at 75°C. The other sample has a mass of 66 g and is originally at a temperature of 100°C. The entire system stabilizes at a final temperature of 20°C. Determine the specific heat of the unknown second sample. (Pick the answer closest to the true value.)A. 1950 joules Co/kgB. 975 joules Co/kgC. 3950 joules Co/kgD. 250 joules Co/kgE. 8500 joules Co/kg
Answer:
A. 1,950 J/kgºC
Explanation:
Assuming that all materials involved, finally arrive to a final state of thermal equilibrium, and neglecting any heat exchange through the walls of the calorimeter, the heat gained by the system "water+calorimeter" must be equal to the one lost by the copper and the unknown metal.
The equation that states how much heat is needed to change the temperature of a body in contact with another one, is as follows:
Q = c * m* Δt
where m is the mass of the body, Δt is the change in temperature due to the external heat, and c is a proportionality constant, different for each material, called specific heat.
In our case, we can write the following equality:
(cAl * mal * Δtal) + (cH₂₀*mw* Δtw) = (ccu*mcu*Δtcu) + (cₓ*mₓ*Δtₓ)
Replacing by the givens , and taking ccu = 0.385 J/gºC and cAl = 0.9 J/gºC, we have:
Qg= 0.9 J/gºC*100g*10ºC + 4.186 J/gºC*250g*10ºC = 11,365 J(1)
Ql = 0.385 J/gºC*50g*55ºC + cₓ*66g*80ºC = 1,058.75 J + cx*66g*80ºC (2)
Based on all the previous assumptions, we have:
Qg = Ql
So, we can solve for cx, as follows:
cx = (11,365 J - 1,058.75 J) / 66g*80ºC = 1.95 J/gºC (3)
Expressing (3) in J/kgºC:
1.95 J/gºC * (1,000g/1 kg) = 1,950 J/kgºC
The specific heat of the unknown metal can be determined from the equilibrium of heat transfer in the system. The heat lost by the hot substances is equal to the heat gained by the cooler substances. Solving for the specific heat of the unknown substance involves calculating the heat gained and lost and equating their values.
Explanation:The specific heat of a substance is a measure of the amount of heat energy required to raise the temperature of a certain mass of the substance by a certain amount. In this case, we're solving for the specific heat (c) of an unknown substance. As the system is in thermal equilibrium, the heat lost by hot substances (copper and unknown metal) is equal to the heat gained by the cooler substances (water and the calorimeter).
The specific heat (c) of the unknown substance can therefore be determined by setting the heat gained (Q_gained = m*c*ΔT) by the cooler substances equal to the heat lost (Q_lost = m*c*ΔT) by the hot substances and solving for the specific heat (c) of the unknown substance. Given that ΔT is the change in temperature, m is the mass, and c is the specific heat, and using the specific heat values for water, aluminum, and copper.
Learn more about Specific Heat here:https://brainly.com/question/28852989
#SPJ3
A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e-0.2t T. What is the magnitude of the current induced in the coil at the time t = 2 s? A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e-0.2t T. What is the magnitude of the current induced in the coil at the time t = 2 s? 1.3 mA 7.5 mA 2.6 mA 4.2 mA 9.2 mA
Answer:
the question is incomplete, the complete question is
"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"
2.6mA
Explanation:
we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.
using the formula be low,
[tex]E=-\frac{d}{dt}(BACOS\alpha )\\[/tex]
where B is the magnitude of the field and A is the area of the circular coil.
First, let determine the area using [tex]\pi r^{2} \\[/tex] where r is the radius of 5cm or 0.05m
[tex]A=\pi *(0.05)^{2}\\ A=0.00785m^{2}\\[/tex]
since we no that the angle is at [tex]0^{0}[/tex]
we determine the magnitude of the magnetic filed
[tex]B=0.5e^{-t} \\t=2s[/tex]
[tex]E=-(0.5e^{-2} * 0.00785)[/tex]
[tex] E=-0.000532v\\[/tex]
the Magnitude of the voltage is 0.000532V
Next we determine the current using ohm's law
[tex]V=IR\\R=0.2\\I=\frac{0.000532}{0.2} \\I=0.0026A[/tex]
[tex]I=2.6mA[/tex]
The magnitude of the induced current in the coil at t = 2s in the given scenario is 2.4 mA. This is calculated using Faraday's law of electromagnetic induction and Ohm's law.
Explanation:To find the magnitude of the current induced in the coil, we need to consider Faraday's law of electromagnetic induction. This law states that the induced electromotive force (emf) in any closed circuit is equal to the rate of change of the magnetic flux through the circuit.
In this situation, we have: B = 0.5 e-0.2t T, and the time derivative of the magnetic field is dB/dt = -0.1 e-0.2t T/s. The area A of the coil is πr²= π(0.05)² m². The induced emf (ε) equals -A dB/dt. Thus, we have ε = -π(0.05)² × -0.1 e-0.2t = 0.0007875 e-0.2t V.
Now, according to Ohm's law, I = ε/R, where R is the resistance of the coil. Substituting the given values, we have I = 0.0007875 e-0.2t / 0.2 = 0.0039375 e-0.2t A. At t=2s, we can substitute into the equation to get I = 0.0039375 e-0.4 = 0.0024 A or 2.4 mA. Therefore, the magnitude of the induced current at t = 2s is 2.4 mA.
Learn more about Electromagnetic Induction here:https://brainly.com/question/32444953
#SPJ3
The magnetic field in a plane monochromatic electromagnetic wave with wavelength λ = 598 nm, propagating in a vacuum in the z-direction is described by B =(B1sin(kz−ωt))(i^+j^) where B1 = 8.7 X 10-6 T, and i-hat and j-hat are the unit vectors in the +x and +y directions, respectively. What is k, the wavenumber of this wave?
Answer:
For this given plane monochromatic electromagnetic wave with wavelength λ=598 nm, the wavenumber is [tex]k=0,0105\ x\ 10^{-9}\ m^{-1}[/tex] .
Explanation:
For a plane electromagnetic wave we have that the electrical and magnetic field are:
[tex]E(r,t)=E_{0}\ cos ( wt-kr)\\\ B(r,t)=B_{0}\ cos(wt-kr)[/tex]
In this case we have the data for the magnetic field. We are told that the magnetic field in a plane electromagnetic wave with wavelength λ=598 nm, propagating in a vacuum in the z direction ([tex]\hat k[/tex]) is described by
[tex]B=8.7\ x\ 10^{-6}\ T sin(kz-wt) (\hat i+\hat j)[/tex]
([tex]\hat i,\hat j, \hat k[/tex] are the unit vectors in the x,y,z directions respectively)
The wavenumber k is a measure of the spatial frequency of the wave, is defined as the number of radians per unit distance:
[tex]k=\frac{2\pi}{\lambda}[/tex]
where λ is the wavelength
So we get that
[tex]k=\frac{2\pi}{\lambda} \rightarrow k=\frac{2\pi}{598 nm} \rightarrow k=0,0105\ x\ 10^{9}\ m^{-1}[/tex]
The wavenumber is
[tex]k=0,0105\ x\ 10^{9}\ m^{-1}[/tex] .
Suppose that you lift four boxes individually, each at a constant velocity. The boxes have weights of 3.0 N, 4.0 N, 6.0 N, and 2.0 N, and you do 12 J of work on each. Match each box to the vertical distance through which it is lifted.
Answer:
The vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m
Explanation:
Worked : work can be defined as the product of force and distance.
The S.I unit of work is Joules (J).
Mathematically it can be represented as,
W = F×d.................. Equation 1
d = W/F.............................. Equation 2
where W = work, F = force, d = distance.
Given: W = 12 J
(i) for the 3.0 N weight,
using equation 2
d = 12/3
d= 4 m.
(ii) for the 4.0 N weight,
d = 12/4
d = 3 m.
(iii) for the 6.0 N weight,
d = 12/6
d = 2 m.
(iv) for the 2.0 N weight,
d = 12/2
d = 6 m
Therefore vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m
A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per unit length of 0.01 Ω/m. If a current of 100 A flows through the wire and the convection coefficient is 500W/m2K, what is the steady state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature within 1-degC of the steady state value? The density of the wire is 8,000kg/m3, its heat capacity is 500 J/kgK and its thermal condu
To determine the steady state temperature of the wire, one can use the power dissipation formula and the convection heat transfer equation. The time for the wire to reach within 1-degree Celsius of steady state involves transient heat transfer calculations using the given material properties.
The student has asked about the steady state temperature of a 1-meter-long wire with a 1mm diameter submerged in an oil bath at 25 degrees Celsius when a current of 100A flows through it. We also need to calculate how long it takes for the wire to reach within 1-degree Celsius of the steady state temperature. To find the steady state temperature, we use the formula P = I2R, where P is the power, I is the current, and R is the resistance. Given that R = 0.01
Ω/m and I = 100A, we find P = (100A)2 x 0.01
Ω/m = 100W/m. Then, using the convection heat transfer equation Q = hA(Ts - T
bath), where Q is the heat transfer rate, h is the convection coefficient, A is the surface area, Ts is the wire surface temperature, and Tbath is the oil bath temperature, we equate Q to P since the wire is in steady state, and solve for Ts. The time to reach within 1-degree Celsius of steady state temperature requires calculating the transient heat transfer, which involves solving the heat transfer equation with the given material properties such as density, heat capacity, and thermal conductivity.
The steady-state temperature of the wire is approximately [tex]\(343.471 {°C}\)[/tex], and it takes approximately [tex]\(1.539[/tex], for the wire to reach within 1°C of the steady-state value.
Steady-State Temperature Calculation:
- Calculate the radius [tex](\(r\))[/tex] of the wire:
[tex]\[ r = \frac{d}{2} = \frac{0.001 \, \text{m}}{2} = 0.0005 \, \text{m} \][/tex]
- Calculate the surface area [tex](\(A\))[/tex] of the wire:
[tex]\[ A = 2\pi r l = 2\pi \times 0.0005 \times 1 = 0.00314 \, \text{m}^2 \][/tex]
- Calculate the heat transfer rate [tex](\(q\))[/tex]:
[tex]\[ q = I^2 R = (100)^2 \times 0.01 = 1000 \, \text{W} \][/tex]
- Calculate the steady-state temperature [tex](\(T_{\text{wire}}\))[/tex]:
[tex]\[ T_{\text{wire}} = \frac{q}{hA} + T_{\text{fluid}} \][/tex]
[tex]\[ T_{\text{wire}} \approx \frac{1000}{500 \times 0.00314} + 298.15 \][/tex]
[tex]\[ T_{\text{wire}} \approx 343.471 \, \text{°C} \][/tex]
Time to Reach Within 1°C of Steady-State:
- Calculate the volume [tex](\(V\))[/tex] of the wire:
[tex]\[ V = \pi r^2 l = \pi \times (0.0005)^2 \times 1 = 7.854 \times 10^{-7} \, \text{m}^3 \][/tex]
- Calculate the thermal time constant [tex](\(\tau\))[/tex]:
[tex]\[ \tau = \frac{\rho V c}{hA} \][/tex]
[tex]\[ \tau \approx \frac{8000 \times 7.854 \times 10^{-7} \times 500}{500 \times 0.00314} \][/tex]
[tex]\[ \tau \approx 0.7854 \, \text{s} \][/tex]
- Calculate the time [tex](\(t\))[/tex] it takes for the wire to reach within 1°C of the steady-state value:
[tex]\[ t = \tau \ln\left(\frac{T_{\text{steady}} - T_{\text{initial}}}{T_{\text{steady}} - T_{\text{fluid}}}\right) \][/tex]
[tex]\[ t \approx 0.7854 \times \ln\left(\frac{343.471 - 25}{343.471 - 298.15}\right) \][/tex]
[tex]\[ t \approx 0.7854 \times \ln\left(\frac{318.471}{45.321}\right) \][/tex]
[tex]\[ t \approx 0.7854 \times \ln(7.032) \][/tex]
[tex]\[ t \approx 1.539 \, \text{s} \][/tex]
The girl makes a microscope with a 3.0 cm focal length objective and a 5.0 cm eyepiece. The microscope tube length is 10 cm. Use the simple formula to find the expected magnification of this microscope.
(A) 12
(B) 14
(C) 17
(D) 20
(E) 24
To solve this problem we will use the concepts related to Magnification. Magnification is the process of enlarging the apparent size, not physical size, of something. This enlargement is quantified by a calculated number also called "magnification".
The overall magnification of microscope is
[tex]M = \frac{Nl}{f_ef_0}[/tex]
Where
N = Near point
l = distance between the object lens and eye lens
[tex]f_0[/tex]= Focal length
[tex]f_e[/tex]= Focal of eyepiece
Given that the minimum distance at which the eye is able to focus is about 25cm we have that N = 25cm
Replacing,
[tex]M = \frac{25*10}{3*5}[/tex]
[tex]M = 16.67\approx 17\\[/tex]
Therefore the correct answer is C.