Answer: A. The total number of protons and the total number of neutrons both remain the same.
Explanation:
Nuclear fission is the process in which a large nucleus splits into two smaller nuclei with the release of energy. In other words, fission is the process in which a nucleus is divided into two or more fragments, and neutrons and energy are released.
In fission processes, the total number of protons and neutrons both remain the same because nucleons are neither created nor destroyed, they just rearrange into new nuclei.
Explanation:In fission processes, it is true that the overall number of protons and the total number of neutrons remains the same after the event. This is because nucleons, which include both protons and neutrons, are neither created nor destroyed during fission; they simply rearrange themselves into new nuclei. The mass number, which is the sum of protons and neutrons, is preserved, and though there can be a conversion between protons and neutrons (e.g., via beta decay), the total count of nucleons stays constant. However, the total mass of the product nuclei is less than the mass of the reactants due to the release of nuclear energy as a result of the conversion of mass to energy according to Einstein's equation E=mc2.
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g A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.462 of the escape speed from Earth and (b) its initial kinetic energy is 0.462 of the kinetic energy required to escape Earth
Answer:
a.)r=4RE/4
b.)r=2RE
c.)ZERO
Explanation:
a.) given v= 0.462 which is V= 0.466Ve
Since the projectile is shot directly away from earth surface the speed it escape with;
v=√2GM/RE......................eqn(1)
M is the Earth's Mass
RE is the Radius
From the Law of conservation of Energy
K+U=0...............................eqn(2)
K₁+U₁=K₂+U₂....................eqn(3)
where K₁ and U₁ is initial kinetic and potential energy
K₂ and U₂ are final kinetic and potential energy
Kinetic Energy (K.E) decrease with time as the projectile moves up and there is decrease in Potential Energy (P.E) , it will let to a point where K.E will turn to zero i.e K₂=0
U₂=K₁U₂ ..........................eqn(3)
From Gravitational Law
U₁= -GMm/RE ..................(5)
U₂= -GMm/r .....................(6)
Where "r" is the distance
v= 0.462√2GM/RE
v= √GM/2RE
GM/4RE - GM/RE = -GM/r
r= 4RE/3
b.) "r" is calculated by this equation;
K₁=0.466
K₁= 1/4MVe².............................eqn(9)
substitute eqn(1) into eqn (9) then
1/4m2GM/RE=0.466GM/2Re
GM/2RE - GM/RE =-GM/r
r=2RE
c.)The potential energy and kinetic energy is the same in terms of their size both in different directions, while the potential energy face outward, the kinetic energy face inward therefore the least initial mechanical energy
required at launch if the projectle is to escape is ZERO
Derivation of Eq. (3): o Basic physics principles: Justify equations (1) and (2) in your own words. . Doing the algebra: From equations (1) and (2), show that equation (3) holds. mu? =eu B This would be nice if we knew the velocity. Fortunately, we know the voltage through which the electrons are accelerated. Setting the change in electrical potential energy equal to the final kinetic energy of the electrons, we find: eV = From this point, some algebra lets us eliminate the velocity variable entirely, giving us: e 2V m (B_T)
Answer:
About eq (1)
[tex]mv^2/r = eVB[/tex]
when a charged particle (electron) enters into the magnetic field which is perpendicular to direction of motion than there will be magnetic force on particle and particle will travel in circular path in with constant speed.
So using force balance on charged particle:
[tex]F_{net} = Fc - Fm[/tex]
Since particle is traveling at constant speed, So acceleration is zero, and
[tex]F_{net} = 0[/tex]
[tex]Fc - Fm=0[/tex]
[tex]Fc = Fm[/tex]
Fc = centripetal force on particle [tex]= m*v^2/r[/tex]
Fm = magnetic force on electron = [tex]q*VxB = q*V*B*sin \theta[/tex]
q = charge on electron = e
since magnetic field is perpendicular to the velocity of particle, So theta = 90 deg
sin 90 deg = 1
So,
[tex]m*v^2/r = e*v*B[/tex]
About equation 2:
When this charged particle is released from rest in a potential difference V, and then it enters into above magnetic field, then using energy conservation on charge particle
[tex]KEi + PEi = KEf + PEf[/tex]
KEi = 0, since charged particle started from rest
[tex]PEi - PEf = q*dV[/tex]
[tex]PEi - PEf = eV[/tex]
KEf = final kinetic energy of particle when it leaves [tex]= (1/2)*m*v^2[/tex]
So,
[tex]0 + eV = (1/2)*m*v^2[/tex]
[tex]eV = (1/2)*m*v^2[/tex]
From above equation (1) and (2)
[tex]m*v^2/r = evB[/tex]
[tex]e/m = v/(r*B)[/tex]
Now
[tex]eV = (1/2)*m*v^2[/tex]
[tex]v = \sqrt{(2*e*V/m)}[/tex]
[tex]e/m = \sqrt{ (2*e*V/m)/(r*B)}[/tex]
[tex]\frac{e^2}{m^2} = \frac{2*e*V}{(m*r^2*B^2)}[/tex]
[tex]e/m = 2*V/(r^2*B^2)[/tex]
[tex]e/m = 2V/(Br)^2[/tex]
A step-up transformer is connected to a generator that is delivering 141 V and 110 A. The ratio of the turns on the secondary to the turns on the primary is 1110 to 5. What voltage is across the secondary
Final answer:
To find the voltage across the secondary winding of a step-up transformer with a primary voltage of 141 V and a turns ratio of 1110 to 5, the secondary voltage is calculated to be 31302 V.
Explanation:
Calculating the Voltage Across the Secondary in a Step-Up Transformer
To calculate the voltage across the secondary winding of a step-up transformer, we use the transformer equation that relates the primary and secondary voltages with the number of turns on the primary (Np) and secondary (Ns) windings:
Vs / Vp = Ns / Np
Given that the primary voltage (Vp) is 141 V, and the ratio of the number of turns is 1110 to 5 (Ns / Np), we can rearrange the equation to solve for the secondary voltage (Vs) as follows:
Vs = Vp × (Ns / Np)
Vs = 141 V × (1110 / 5)
Vs = 141 V × 222
Vs = 31302 V
Thus, the voltage across the secondary winding is 31302 V.
If the vertical displacement from crest to trough is 0.50 cm, what is the amplitude?
cm
0.50
0.25
1.0
Answer:
0.25
Explanation:
Since there is a vertical displacement of 0.5cm from the crest to the trough, there is half of that displacement to the midline, which is also known as the amplitude. Therefore, the amplitude is 0.5/2=0.25cm. Hope this helps!
One end of a steel rod of radius R = 10 mm and length L = 80cm is held in a vise. A force of magnitude F = 60kN is then applied perpendicularly to the end face (uniformly across the area) at the other end, pulling directly away from the vise. What is the stress on the rod?
Answer: 1.91*10^8 N/m²
Explanation:
Given
Radius of the steel, R = 10 mm = 0.01 m
Length of the steel, L = 80 cm = 0.8 m
Force applied on the steel, F = 60 kN
Stress on the rod, = ?
Area of the rod, A = πr²
A = 3.142 * 0.01²
A = 0.0003142
Stress = Force applied on the steel/Area of the steel
Stress = F/A
Stress = 60*10^3 / 0.0003142
Stress = 1.91*10^8 N/m²
From the calculations above, we can therefore say, the stress on the rod is 1.91*10^8 N/m²
A copper rod is sliding on two conducting rails that make an angle of 19o with respect to each other, as in the drawing. The rod is moving to the right with a constant speed of 0.60 m/s. A 0.63-T uniform magnetic field is perpendicular to the plane of the paper. Determine the magnitude of the average emf induced in the triangle ABC during the 7.5-s period after the rod has passed point A.
Answer:
0.2923 V
Explanation:
Given that
Angle between the rails, θ = 19°
Speed of the rod, v = 0.6 m/s
Magnetic field present, B = 0.63 T
Time used, t = 7.5 s
E = -ΔΦ/Δt
where, Φ = BA, so
E = -BΔA / Δt
To get the area, if we assume the rails are joined in a triangular fashion(see attachment)
E = -B(1/2 * AC * BC) / Δt
E = -B(vΔt * vΔt tanθ) / 2Δt
E = -(B * v² * Δt² * tanθ) / 2Δt
E = -Bv²Δt.tanθ/2
E = -(0.63 * 0.6² * 7.5 * tan 19) / 2
E = -0.5857 / 2
E = -0.2923
Thus, the magnitude of average emf induced if 0.2923 V
A closed loop conductor that forms a circle with a radius of is located in a uniform but changing magnetic field. If the maximum emf induced in the loop is what is the maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies
Final answer:
The maximum rate at which the magnetic field strength is changing in a closed loop conductor can be determined using Faraday's law. The rate is equal to the maximum induced emf divided by the area of the loop.
Explanation:
The maximum rate at which the magnetic field strength is changing can be determined using Faraday's law of electromagnetic induction. According to Faraday's law, the induced emf (voltage) in a closed loop conductor is equal to the negative rate of change of magnetic flux through the loop. In this case, the magnetic field is perpendicular to the plane of the loop, so the magnetic flux is given by the product of the magnetic field strength and the area of the loop. Therefore, the maximum rate of change of magnetic field strength is equal to the maximum induced emf divided by the area of the loop.
So, the maximum rate at which the magnetic field strength is changing is given by: (d(B)/dt) = (Emax / A)
where (d(B)/dt) is the rate of change of magnetic field strength, Emax is the maximum induced emf, and A is the area of the loop.
The rotor of an electric motor has rotational inertia Im = 4.36 x 10-3 kg·m2 about its central axis. The motor is used to change the orientation of the space probe in which it is mounted. The motor axis is mounted along the central axis of the probe; the probe has rotational inertia Ip = 6.07 kg·m2 about this axis. Calculate the number of revolutions of the rotor required to turn the probe through 30.6° about its central axis.
Answer:
The number of revolutions of the rotor required to turn the probe is 118 revolutions
Explanation:
Given;
rotational inertia of the electric motor, Im = 4.36 x 10⁻³ kg·m²
rotational inertia of the probe, Ip = 6.07 kg·m²
the angular position of the probe, θ = 30.6°
From the principle of conservation of angular momentum;
[tex]I_m \omega _m = I_p \omega _p \\\\Also;\\\\I_m \theta _m = I_p \theta _p[/tex]
where;
[tex]\omega _m[/tex] is the angular velocity of the electric motor
[tex]\omega _p[/tex] is the angular velocity of the probe
[tex]\theta _m[/tex] is the angular position of the electric motor
[tex]\theta _p[/tex] is the angular position of the probe
[tex]\theta _m = \frac{I_p \theta_p}{I_m} \\\\\theta _m = \frac{6.07* 30.6^o}{4.36*10^{-3}} = 42601.4^o[/tex]
360° = One revolution
42601.4° = ?
Divide 42601.4° by 360°
= 118 revolutions
Therefore, the number of revolutions of the rotor required to turn the probe is 118 revolutions
Why are the peaks opposite in direction?
a. The peaks are opposite in direction because the change in magnetic field at one end of the coil is opposite to the change in magnetic field at the other end. Faraday's law predicts that the direction of the induced voltage is dependent on the nature of the change in magnetic field.
b. The peaks are opposite in direction because the change in magnetic field at one end of the coil has the same direction to the change in magnetic field at the other end. Faraday's law predicts that the direction of the induced voltage is dependent on the nature of the change in magnetic field.
c. The peaks are in the same direction because the change in magnetic field at one end of the coil has the same direction to the change in magnetic field at the other end. Faraday's law predicts that the direction of the induced voltage is dependent on the nature of the change in magnetic field
Answer:
A
Explanation:
Peaks are in opposite direction because change in magnetic field at one end of the coil is opposite to the change in magnetic field at other end of the coil. Faraday's law predict that the direction of induced voltage is dependent on the nature of change in magnetic field
In astronomy, distances are often expressed in light-years. One light-year is the distance traveled by light in one year. If the distance to a star is 3.6 light-years, what is this distance in meters? (There are 365.25 days in one year.)
Answer:
The distance of the star is [tex]3.40x10^{16}[/tex] meters
Explanation:
It is known that the speed of light has a value of [tex]3x10^{8}m/s[/tex] in vacuum. That is, it travels [tex]3x10^{8]m[/tex] in one second, according with the following equation:
[tex]v = \frac{x}{t}[/tex]
Where v is the speed, x is the distance and t is the time.
[tex]x = v\cdot t[/tex] (1)
Equation 1 can be used to determine the distance that the light travels in 1 year:
It is necessary to find how many seconds are in 1 year (365.25 days).
[tex]365.25 days \cdot \frac{86400s}{1 day}[/tex] ⇒ [tex]31557600s[/tex]
[tex]x = (3x10^{8}m/s)(31557600s)[/tex]
[tex]x = 9.46x10^{15}m[/tex]
Therefore, in 1 year, light travels [tex]9.46x10^{15}[/tex] meters.
If the distance to a star is 3.6 light-years, what is this distance in meters?
A simple conversion between units can be used to get the distance in meters
[tex]x_{star} = 3.6ly \cdot \frac{9.46x10^{15}m}{1ly}[/tex] ⇒ [tex]3.40x10^{16}m[/tex]
Hence, the distance of the star is [tex]3.40x10^{16}[/tex] meters.
A light-year, a distance unit used in astronomy, is approximately 9.46 x [tex]10^{15[/tex] meters. Hence, a distance of 3.6 light years would convert to approximately 3.4 x [tex]10^{16[/tex] meters.
Explanation:A light-year is a unit of distance used in astronomy, which represents the distance light travels in one year. Light travels at a speed of 186,000 miles per second. Over the course of a year, this distance adds up significantly.
To calculate a light-year in meters, we would carry out the following calculation - light travels at 299,792 kilometers per second. This converts to exactly 299,792,000 meters per second. If we multiply this by the number of seconds in a year (60 seconds/minute, 60 minutes/hour, 24 hours/day, 365.25 days/year) we find that one light-year is approximately 9.46 x [tex]10^{15[/tex] meters.
Therefore, if the distance to a star is 3.6 light-years, then this would convert to about 3.4 x [tex]10^{16[/tex] meters.
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Newton's law of gravity and Coulomb's law are both inverse-square laws. Consequently, there should be a "Gauss's law for gravity." The electric field was defined as E⃗ =F⃗ onq/q , and we used this to find the electric field of a point charge. Using analogous reasoning, what is the gravitational field g⃗ of a point mass? Write your answer using the unit vector r^ , but be careful with signs; the gravitational force between two "like masses" is attractive, not repulsive. Express your answer in terms of the variables G , M , r , and r^ .
The correct answer for the gravitational field of a point mass is [tex]g = -\dfrac{GM_1M_1}{r^2} * \dfrac{r}{m}[/tex]
The gravitational force between two masses is given by Newton's law of gravitation:
[tex]F = \dfrac{GM_1M_2}{r^2}[/tex]
[tex]F[/tex] is the gravitational force between [tex]M_1[/tex] and [tex]M_2[/tex]
[tex]G[/tex] is the gravitational constant,
[tex]r^2[/tex] is the square of the distance between two masses.
The gravitational force is a vector quantity, and it is directed along the line connecting the two masses.
[tex]g= \dfrac{F}{m}[/tex]
[tex]g[/tex] is the gravitational field.
[tex]F[/tex] is the gravitational force.
Let [tex]m[/tex] be the mass test object.
Substitute the value of gravitational force from Newton's law of gravity:
[tex]g = \dfrac{GM_1M_1}{r^2} * \dfrac{r}{m}[/tex]
[tex]r[/tex]is the unit vector pointing from the mass M to the test mass, which represents the direction of the gravitational field.
[tex]g = -\dfrac{GM_1M_1}{r^2} * \dfrac{r}{m}[/tex]
The negative sign indicates the direction of the gravitational force is attractive.
The gravitational field is [tex]g = -\dfrac{GM_1M_1}{r^2} * \dfrac{r}{m}[/tex]
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The gravitational field of a point mass is given by -GM/(r^2)*r^, where the negative sign indicates the attractive nature (opposite direction to r^) of gravity.
Explanation:The gravitational field g of a point mass using the same reasoning as for the electric field would be analogous but with some slight modifications to account for the attractive nature of gravity, unlike the repulsive nature of electric charges of the same sign. The formula is given by -GM/(r^2)*r^, where G is the gravitational constant, M is the mass of the object, and r is the distance to the object. The negative sign indicates that the force is attractive and acts in the direction opposite to r (the vector pointing directly away from the mass).
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A 7.7 kg sphere makes a perfectly inelastic collision with a second sphere initially at rest. The composite system moves with a speed equal to one third the original speed of the 7.7 kg sphere. What is the mass of the second sphere
Answer:
15.4 kg.
Explanation:
From the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
mu+m'u' = V(m+m').................... Equation 1
Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, V = common velocity of both sphere.
Given: m = 7.7 kg, u' = 0 m/s (at rest)
Let: u = x m/s, and V = 1/3x m/s
Substitute into equation 1
7.7(x)+m'(0) = 1/3x(7.7+m')
7.7x = 1/3x(7.7+m')
7.7 = 1/3(7.7+m')
23.1 = 7.7+m'
m' = 23.1-7.7
m' = 15.4 kg.
Hence the mass of the second sphere = 15.4 kg
Answer:
The mass of the second sphere is 15.4 kg
Explanation:
Given;
mass of the first sphere, m₁ = 7.7 kg
initial velocity of the second sphere, u₂ = 0
let mass of the second sphere = m₂
let the initial velocity of the first sphere = u₁
final velocity of the composite system, v = ¹/₃ x u₁ = [tex]\frac{u_1}{3}[/tex]
From the principle of conservation of linear momentum;
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = v(m₁ + m₂)
Substitute the given values;
[tex]7.7u_1 + 0=\frac{u_1}{3} (7.7+m_2)[/tex]
Divide through by u₁
7.7 = ¹/₃(7.7 + m₂)
multiply both sides by 3
23.1 = 7.7 + m₂
m₂ = 23.1 - 7.7
m₂ = 15.4 kg
Therefore, the mass of the second sphere is 15.4 kg
A man pushes a shopping cart at a constant speed. What happens if the man increases the force on the cart?
Answer:
The cart would speed up.
Explanation:
According to Newton's 1st law, object subjected to no force, or net force 0, would have a constant speed. In our case the cart is initially at constant speed, meaning the man exerts a force that is equal to friction force. If he increases the force on the cart, the net force would no longer be 0. The cart would gain an acceleration and increases its speed.
Answer:
A. The cart accelerates
Explanation:
Interactive LearningWare 22.2 reviews the fundamental approach in problems such as this. A constant magnetic field passes through a single rectangular loop whose dimensions are 0.46 m x 0.68 m. The magnetic field has a magnitude of 3.0 T and is inclined at an angle of 67o with respect to the normal to the plane of the loop. (a) If the magnetic field decreases to zero in a time of 0.49 s, what is the magnitude of the average emf induced in the loop
Explanation:
The dimension of a single rectangular loop is 0.46 m x 0.68 m.
Magnetic field, B = 3 T
The loop is inclined at an angle of 67 degrees with respect to the normal to the plane of the loop.
It is required to find the magnitude of the average emf induced in the loop if the magnetic field decreases to zero in a time of 0.49 s.
Te induced emf in the loop is given by :
[tex]\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(BA)}{dt}\\\\\epsilon=A\dfrac{dB}{dt}\cos\theta\\\\\epsilon=0.46\times 0.68\times \dfrac{3}{0.49}\times \cos(67)\\\\\epsilon=0.74\ V[/tex]
So, the magnitude of the average emf induced in the loop is 0.74 V.
Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of north. The resultant vector A + B is given by:
(a) 9.8 m at an angle of 26° north of east,
(b) 3.3 m at an angle of 26° north of east,
(c) 7.2 m at an angle of 26° east of north,
(d) 3.3 m at an angle of 64° east of north or
(e) 9.8 m at an angle of 64° east of north
Answer:
The resultant vector A + B is given by 7.2 m at an angle of 26° east of north,
C
Explanation:
Resolving the vectors to vertical and horizontal component;
Vertical;
Vector A = 6sin30
Vector B = 4sin60
Resultant vertical = 6sin30 + 4sin60 = 6.464m
Horizontal;
Vector A = 6cos30
Vector B = -4cos60
Resultant horizontal = 6cos30 - 4cos60 = 3.196m
Resultant R = √(6.464^2 + 3.196^2) = 7.2m
Tanθ = 6.464/3.196
θ = taninverse (6.464/3.196) BN
θ = 64° north of East.
Or
26° east of north
The resultant vector A + B is given by 7.2 m at an angle of 26° east of north,
The resultant vector A + B is given by 7.2 m at an angle of 26° east of north. Hence, option (c) is correct.
Given data:
The magnitude of vector A is, A = 6.0 m.
The direction of vector A is, 30° north of east.
The magnitude of vector B is, B = 4.0 m.
The direction of vector B is, 30° west of north.
The quantity having both the magnitude as well as the magnitude are known as vector quantities.
Resolving the vectors to vertical and horizontal component;
Along the Vertical direction;
Vector A = 6sin30
Vector B = 4sin60
Resultant vertical vector = 6sin30 + 4sin60 = 6.464 m
Along the Horizontal direction;
Vector A = 6cos30
Vector B = -4cos60
Resultant horizontal vector = 6cos30 - 4cos60 = 3.196 m
Now, the resultant vector is calculated as,
[tex]R=\sqrt{6.464^{2}+3.196^{2}}\\\\R = 7.2 \;\rm m[/tex]
And the resultant direction of vectors is,
[tex]tan \theta = 6.464/3.196\\\\\theta = tan^{-1} (6.464/3.196) \\\theta =64 ^{\circ}[/tex]( North of East)
θ = 64° north of East.
Or
26° east of north
Thus, we can conclude that the resultant vector A + B is given by 7.2 m at an angle of 26° east of north. Hence, option (c) is correct.
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For safety reasons, a worker’s eye travel time in a certain operation must be separated from the manual elements that follow. The distance the worker’s eyes must travel is 20 in. The perpendicular distance from her eyes to the line of travel is 24 in. No refocus is required. What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element?
Answer:
The answer is 12.67 TMU
Explanation:
Recall that,
worker’s eyes travel distance must be = 20 in.
The perpendicular distance from her eyes to the line of travel is =24 in
What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element = ?
Now,
We solve for the given problem.
Eye travel is = 15.2 * T/D
=15.2 * 20 in/24 in
so,
= 12.67 TMU
Therefore, the MTM -1 of normal time that should be allowed for the eye travel element is = 12.67 TMU
A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 254 kcal of heat is added to the gas, the volume is observed to increase slowly from 12.0 m3 to 16.2 m3 . Part A Calculate the work done by the gas. Express your answer with the appropriate units. W = nothing nothing Request Answer Part B Calculate the change in internal energy of the gas. Express your answer with the appropriate units. ΔU = nothing nothing
The work done by the gas when the volume increases from 12.0 m³ to 16.2 m³ at atmospheric pressure is 425.565 kJ. The change in internal energy of the gas when 254 kcal of heat is added is 636.931 kJ.
Explanation:To calculate the work done by the gas during a quasi-static expansion, we can use the formula W = P ΔV, where W is work, P is pressure, and ΔV is the change in volume. Given the atmospheric pressure is 1 atm or 101,325 Pa, and the volume change is from 12.0 m³ to 16.2 m³, the work done by the gas can be calculated as:
W = P ΔV = 101,325 Pa × (16.2 m³ - 12.0 m³)
W = 101,325 Pa × 4.2 m³
W = 425,565 J or 425.565 kJ
To calculate the change in internal energy (ΔU) of the gas, we can use the first law of thermodynamics, which states that ΔU = Q - W, where Q is the heat added to the system. The heat is given as 254 kcal, which needs to be converted to joules (1 kcal = 4.184 kJ).
ΔU = Q - W = (254 kcal × 4.184 kJ/kcal) - 425.565 kJ
ΔU = 1062.496 kJ - 425.565 kJ
ΔU = 636.931 kJ
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An open-delta three-phase transformer system has one transformer center-tapped to provide a neutral for single-phase voltages. If the voltage from line to center tap is 277 V, what is the high-leg voltage?
Answer:
The high-leg voltage is [tex]V_h= 497.76V[/tex]
Explanation:
A diagram showing the arrangement of this three phase transformer is shown on the first uploaded image
The high - leg terminal is in the diagram is from [tex]Z_1 - n[/tex]
Generally the high-leg terminal is 1.732 times of the the voltage from line to center tap which is given as 277 V
The high -voltage can be computed as follows
[tex]V_h = 1.732 *277[/tex]
[tex]V_h= 497.76V[/tex]
In a Millikan oil-drop experiment, the condenser plates are spaced 2.00 cm apart, the potential across the plates is 4000 V, the rise or fall distance is 4.00 mm, the density of the oil droplets is 0.800 g/cm3 , and the viscosity of the air is 1.81 105 kg m1 s 1 . The average time of fall in the absence of an electric field is 15.9 s. The following different rise times in seconds are observed when the field is turned on: 36.0, 17.3, 24.0, 11.4, 7.54. (a) Find the radius and mass of the drop used in this experiment. (b) Calculate the charge on each drop, and show that charge is quantized by considering both the size of each charge and the amount of charge gained (lost) when the rise time changes. (c) Determine the electronic charge from these data. You may assume that e lies between 1.5 and 2.0 1019 C. 7.
Answer:
Answer is in the following attachment
Explanation:
By considering the magnetic force in the second region, develop a mathematical expression that relates the mass of the particle to the other variables. Do not include the velocity in your expression. You can use the condition that the particle passed through the region of electric and magnetic fields undeflected to eliminate v from your expression. Your expression will also contain the radius of the circular path. i. Your expression for m should depend on B, E, r, and q
Answer:
M = [tex]\frac{qrB^{2} }{E}[/tex]
Explanation:
considering the magnetic force in the second region derive a mathematical expression that equates the mass of the particle to other variables
In a magnetic field
q = charge, M = mass of particle, E = electric field,B= magnetic field
qvb = [tex]\frac{mv^{2} }{r}[/tex] = therefore m = [tex]\frac{qrb}{v}[/tex] (equation 1)
note : the particle passes through the region undeflected
therefore : qvb = qE therefore (E = VB)
hence v = [tex]\frac{E}{B}[/tex] ( equation 2 )
insert equation 2 into equation 1
m = [tex]\frac{qrB^{2} }{E}[/tex]
As a dilligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.11-m-long rod as you jog at 3.07 m/s, holding the rod perpendicular to your direction of motion. What is the strength of the magnetic field that is perpendicular to both the rod and your direction of motion and that induces an EMF of 0.287 mV across the rod
Answer:
The strength of the magnetic field is 0.0842 mT
Explanation:
Given:
Velocity of rod [tex]v = 3.07[/tex] [tex]\frac{m}{s}[/tex]
Length of rod [tex]l = 1.11[/tex] m
Induced emf across the rod [tex]\epsilon = 0.287 \times 10^{-3}[/tex] V
According to the faraday's law
We have a special case for moving rod in magnetic field.
Induced emf in moving rod is given by,
[tex]\epsilon = Blv[/tex]
Where [tex]B =[/tex] strength of magnetic field
[tex]B = \frac{\epsilon}{lv}[/tex]
[tex]B = \frac{0.287 \times 10^{-3} }{3.07 \times 1.11}[/tex]
[tex]B = 0.0842 \times 10^{-3}[/tex] T
[tex]B = 0.0842[/tex] mT
Therefore, the strength of the magnetic field is 0.0842 mT
if two objects each have a mass of 10 kg , then the force of gravity between them...
A) is constant
B) is 100 kg
C)is greater when they are closer together
D) depends only on their masses
if two objects each have a mass of 10 kg, then the force of gravity between them C) is greater when they are closer together.
We have two objects, each with a mass of 10 kg. We can calculate the force of gravity (F) between them using Newton's law of universal gravitation.
[tex]F = G \frac{m_1m_2}{r^{2} }[/tex]
where,
G: gravitational constantm₁, m₂: masses of both objectsr: distance between the objectsAs we can see from this expression. the gravitational force between the objects depends on their masses and the distance between them. The closer they are, the stronger the gravity force.
if two objects each have a mass of 10 kg, then the force of gravity between them...
A) is constant. No. It varies with the distance.
B) is 100 kg. No. kg is not a unit of force.
C) is greater when they are closer together. Yes.
D) depends only on their masses. No. It also depends on the distance between them.
if two objects each have a mass of 10 kg, then the force of gravity between them C) is greater when they are closer together.
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An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner's 1.80 T field with his fingers pointing in the direction of the field. His wedding ring has a diameter of 2.23 cm, and it takes 0.320 s to move it into the field.
1. What average current is induced in the ring in A if its resistance is 0.0100 Ω?
2. What average power is dissipated in mW?
3. What magnetic field is induced at the center of the ring in T?
Given that,
Magnetic field strength is
B = 1.8T
The wedding ring has a diameter of
d = 2.23 cm = 0.023m
Time take t = 0.32 secs
A. Current induced
From ohms law
V= iR, given that R = 0.01Ω
So, we need to get the induced emf
Using
ε = -NdΦ / dt
Where Φ = BA
ε = -A ∆B / ∆t
ε = -¼πd²(B2-B1) / (t2-t1)
ε = -¼ × π × 0.023² × -1.8 / 0.32
ε = 0.0023371 V
Then
I = ε / R
I = 0.002337 / 0.01
I =0.2337 A
B. Power discippated?
Power is given as
P = iV
P = 0.2337 × 0.002337
P = 0.0005462 W
P = 0.56 mW
C. The magnetic field at the centre of the ring.
The electric field at the centre of the ring is zero because each part of the ring will cause a symmetrical opposite magnitude at every point,
Then, B = 0 T
The eyes of amphibians such as frogs have a much flatter cornea but a more strongly curved (almost spherical) lens than do the eyes of air-dwelling mammals. In mammalian eyes, the shape (and therefore the focal length) of the lens changes to enable the eye to focus at different distances. In amphibian eyes, the shape of the lens doesn't change. Amphibians focus on objects at different distances by using specialized muscles to move the lens closer to or farther from the retina, like the focusing mechanism of a camera. In air, most frogs are nearsighted; correcting the distance vision of a typical frog in air would require contact lenses with a power of about −6.0D .A frog can see an insect clearly at a distance of 10cm . At that point the effective distance from the lens to the retina is 8mm .If the insect moves 5cm farther from the frog, by how much and in which direction does the lens of the frog's eye have to move to keep the insect in focus?0.02cm, toward the retina.0.02cm, away from the retina.0.06cm, toward the retina.0.06cm, away from the retina.
Answer:
0.2cm towards the retina.
Explanation:
the focal length of the frog eye is
(1/f) = (1/10) + (1/0.8)
f = 0.74cm
Since the distance of the object is 15cm Hence
(1/0.74) = (1/15) + (1/V)
V = 0.78cm
Therefore the distance the retina is to move is
0.78cm - 0.8cm = 0.02cm towards the retina.
Suppose that the resistive force of the air on a skydiver can be approximated by f = −bv2. If the terminal velocity of an 82.0 kg skydiver is 33.4 m/s, what is the value of b (in kg/m)?
Answer:
The value of b 0.7351 kg/m
Explanation:
Given that;
Mass of sky diver = 82 kg
Velocity = 33.4 m/s
f = −bv²
It is a resistance force, therefore the negative sign is ignored.
since; f = mg
∴ mg = bv²
b = mg / v ² ........ (1)
At terminal velocity a = 0
Put parameters in (1)
b = (82 × 10) / (33.4)²
b = 820 / 1,115.56
b = 0.7351
Suppose we take a 1 m long uniform bar and support it at the 21 cm mark. Hanging a 0.40 kg mass on the short end of the beam results in the system being in balance. Find the mass of the beam.
Answer:
The mass of the beam is = 29 kg.
Explanation:
A beam with mass 40 kg is shown in figure. Point S is the support point. Point B is the middle point on the beam where mass of the beam acts.
Taking moment about Point S
40 × 21 = [tex]M_{beam}[/tex] × 29
[tex]M_{beam}[/tex] = 29 kg
Therefore the mass of the beam is = 29 kg.
A bothersome feature of many physical measurements is the presence of a background signal (commonly called "noise"). In Part 2.2.4 of the experiment, some light that reflects off the apparatus or from neighboring stations strikes the photometer even when the direct beam is blocked. In addition, due to electronic drifts, the photometer does not generally read 0.0 mV even in a dark room. It is necessary, therefore, to subtract off this background level from the data to obtain a valid measurement. Suppose the measured background level is 5.1 mV. A signal of 20.7 mV is measured at a distance of 29 mm and 15.8 mV is measured at 32.5 mm. Correct the data for background and normalize the data to the maximum value. What is the normalized corrected value at 32.5 mm?
Answer:
0.685
Explanation:
the background-corrected light measurement at 29mm: 20.7mv- 5.1mv⇒15.6mv
at 32.5mm: 15.8mv- 5.1mv⇒ 10.7mv
So, Normalize the data to the maximum value probably means to set the peak value to unity and scale the remaining data by the same factors.
Therefore, the normalized corrected value at 32.5mm is ⇒ 10.7mv/15.6mv ⇒0.685
Rank from largest to smallest the values of the magnetic field at the following distances from the axis of the conducting cylinder: Ra = 7.75 cm , Rb = 4.95 cm , r = 5.40 cm , and r>Ra.
The ranking from largest to smallest the values of the magnetic field should be like
Rb = 4.95 cm
r= 5.40 cm
Ra =7.75 cm
r> Ra
What is the magnetic field?It is a vector field that explained the magnetic impact on moving electric charges, electric currents, and magnetic materials.
In this, the force should be perpendicular to the velocity and the magnetic field. Also, it should be inversely proportional with respect to the distance with the axis of the cylinder.
So,
The largest magnetic field =Rb =4.95 cm
And,
Smallest magnetic field = r>Ra
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The magnetic field inside a conducting cylinder is maximized at certain points before dropping off outside. The ranked values from largest to smallest are: r = 5.40 cm, Rb = 4.95 cm, Ra = 7.75 cm, and r > Ra.
To determine the magnetic field at different distances from the axis of a conducting cylinder, we need to understand how the magnetic field varies inside and outside a cylinder carrying a current.
Given the distances Ra = 7.75 cm, Rb = 4.95 cm, r = 5.40 cm, and r > Ra, we will rank the magnetic field intensities:
1.) r = 5.40 cm:
Since r is within the cylinder where current is distributed uniformly, the magnetic field at this point can be calculated using Ampère's Law.
2.) Rb = 4.95 cm:
Still within the cylinder but closer to the center, thus the field here will not yet be maximized.
3.) Ra = 7.75 cm:
Outside the cylinder where the current configuration influences how the field drops off with increasing distance.
4.) r > Ra:
This point is furthest from the axis, hence the magnetic field will be the smallest.
The magnetic field reaches a maximum inside the conducting cylinder before decreasing outside of it.
In an interference pattern, the intensity is Group of answer choices smaller in regions of constructive interference than in regions of destructive interference. unchanged in regions of destructive interference but smaller in regions of constructive interference. the same in both the regions of constructive interference and the regions of destructive interference. unchanged in regions of destructive interference but greater in regions of constructive interference. greater in regions of constructive interference than in regions of destructive interference.
Answer:
The same in both the regions of constructive interference and the regions of destructive interference.
Explanation:
Interference is a phenomenon which occurs when two waves meet while moving along the same medium . The amplitude formed as a result of the interference could be greater, lower, or the same amplitude.
Constructive and destructive interference result from the interaction of waves that are correlated or coherent with each other. This is because arose from the same source or they have the same or nearly the same frequency.
The waves being coherent, arising from the same source and having the same frequency explains why it’s the same in both the regions of constructive interference and the regions of destructive interference.
Answer:
Option E (greater in regions of constructive interference than in regions of destructive interference)Explanation:
Constructive interference occurs when the maxima of two waves add together (the two waves are in phase), so that the amplitude of the resulting wave is equal to the sum of the individual amplitudes. Whereas in destructive interference opposite is true.
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A sphere of radius 5.00 cm c m carries charge 3.00 nC n C . Calculate the electric-field magnitude at a distance 4.00 cm c m from the center of the sphere and at a distance 6.00 cm c m from the center of the sphere if the sphere is a solid insulator with the charge spread uniformly throughout its volume.
To calculate the electric-field magnitude at points inside and outside a uniformly charged insulating sphere, we apply Gauss's Law, considering the proportion of enclosed charge within a Gaussian surface for points inside, and treating the sphere as a point charge for points outside.
Explanation:We apply Gauss's Law to understand the electric-field magnitude around a uniformly charged insulating sphere, which states that the electric flux through a closed surface is proportional to the enclosed charge. We'll calculate the electric field (E) at two distances from the centre of the sphere: 4.00 cm, which is inside the sphere, and 6.00 cm, which is outside the sphere.
For the inside point (r = 4.00 cm), we only consider the charge enclosed by a Gaussian surface of radius 4.00 cm. Using the fact that charge is distributed uniformly throughout the volume of the sphere, we calculate the enclosed charge (q_enclosed) by using the ratio of volumes: q_enclosed = (Q * (4/3 * π * r^3)) / (4/3 * π * R^3), where Q is the total charge of the sphere. R is the radius of the sphere.
The electric field inside the sphere at 4.00 cm is then E = (1 / (4 * π * ε_0)) * (q_enclosed / r^2). We substitute the values using the given radius and charge and solve for E.
For the outside point (r = 6.00 cm), we use the formula for the electric field outside a spherically symmetric charge distribution, which is E = (1 / (4 * π * ε_0)) * (Q / r^2), and solve for E using the sphere's total charge and the distance of 6.00 cm.
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