The speed when all the fuel has been exhausted is 2415m/s
∆V = V(e) ln(m1/m2)
Hence;
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Answer:
Explanation:
given,
exhaust velocity of fuel(v_e) = 1500 m/s
initial speed of rocket,v₁ = 0 m/s
final speed of rocket, v₂ = ?
fuel weigh = 80 % of total weight
using Tsiolkovsky rocket equation
Δ v = v₂ - v₁
v_e is the exhaust speed
m₁ is the initial total mass.
m₂ is the is the final total mass without propellant.
m₂ = m₁ - 0.8 m₁
m₂ = 0.2 m₁
When all the fuel is exhausted speed of the fuel is equal to
Answer:
Yes, the paths of the two particles cross.
Location of path intersection = ( 1 , 2 , 3)
Explanation:
In order to find the point of intersection, we need to set both locations equal to one another. It should be noted however, that the time for each particle can vary as we are finding the point where the paths meet, not the point where the particles meet themselves.
So, we can name the time of the first particle , and the time of the second particle .
Setting the locations equal, we get the following equations to solve for and :
Equation 1
Equation 2
Equation 3
Solving these three equations simultaneously we get:
2 seconds
4 seconds
Since, we have an answer for when the trajectories cross, we know for a fact that they indeed do cross.
The point of crossing can be found by using the value of or in the location matrices. Doing this for the first particle we get:
Location of path intersection = ( -1 + 2 , 4 - 2 , -1 + 2(2) )
Location of path intersection = ( 1 , 2 , 3)
Complete Question
A loop of wire lies flat on the horizontal surface in an area with uniform magnetic field directed vertically up. The loop of wire suddenly contracts to half of its initial diameter. As viewed from above induced electric current in the loop is
a. counterclockwise
b. clockwise
c. there is no current in the loop because magnetic field is uniform
d. there is no current in the loop because magnetic field does not change
Answer:
Option A is the correct answer
Explanation:
According to the question the loop of wire contracts to half it initial diameter and will mean that less number of electric field line will pass through the loop and this change in magnetic flux will cause current to flow in the loop of wire and from Lenz's law this current will in the opposite direction of what produced it which is the change in magnetic flux so the current will flow in a counterclockwise direction
Answer:
75
Explanation:
just took it e2020
Answer:
60%
Explanation:
efficiency= useful/input x 100%,
Here, kinetic energy is useful for food processor (i.e. spinning blades)
600J/1000J=60%
b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity]
c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have?
d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?
Answer:
A. 2.083 MV/m from anode to cathode.
B. 93648278.15 m/s
C. 2.5x10^-5 C and there are about 1.56x10^14 electrons
D. 4x10^-15 Joules
Explanation:
Voltage V across plate is 25 kV = 25x10^3 V
Distance apart x = 1.2 cm = 1.2x10^-2 m
A. Electric field strength is the potential difference per unit distance
E = V/x = 25x10^3/1.2x10^-2 = 2083333.3 V/m
= 2.083 MV/m
B. Energy of electron is electron charge times the voltage across
i.e eV
Charge on electron = 1.6x10^-19 C
Energy of electron = 1.6x10^-19 x 25x10^3 = 4x10^-15 Joules
Mass of electron m is 9.12x10^-31 kg
Kinetic energy of electron = 0.5mv^2
Where v is the speed
4x10^-15 = 0.5 x 9.12x10^-31 x v^2
v^2 = 8.77x10^15
v = 93648278.15 m/s
C. From Q = CV
Q = charge
C = capacitance = 1 nF 1x10^-9 F
V = voltage = 25x10^3 V
Q = 1x10^-9 x 25x10^3 = 2.5x10^-5 C
Total number of electrons = Q/e
= 2.5x10^-5/1.6x10^-19 = 1.56x10^14 electrons
D. To push electron from cathode to anode, I'll have to do a work of about
4x10^-15 Joules
Answer:
The ratio of the model size is 1 : 2000
Explanation:
Given
Real Diameter = 0.012 um
Scale Diameter = 24 um
Required
Determine the scale ratio
The scale ratio is calculated as follows;
Substitute values for real and scale measurements
Divide the numerator and the denominator by 0012um
Represent as ratio
Hence, the ratio of the model size is 1 : 2000
The ratio of the model size to the actual size is 1 : 2000. This means the model represents the white blood cell's diameter 2000 times larger than its actual size.
The ratio of the model size to the actual size can be calculated using the given measurements:
Actual Diameter = 0.012 um
Model Diameter = 24 um
Ratio = Model Diameter / Actual Diameter
Ratio = 24 um / 0.012 um
Ratio = 2000
So, the ratio of the model size to the actual size is 1 : 2000. This means the model represents the white blood cell's diameter 2000 times larger than its actual size.
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(b) the exergy destroyed during this process.
A) The exergy of the refrigerant at the initial and final states are :
B) The exergy destroyed during this process is : - 1048.4397 kJ
Given data :
Mass ( M ) = 5 kg
P1 = 0.7 Mpa = P2
T1 = 60°C = 333 k
To = 24°C = 297 k
P2 = 100 kPa
A) Determine the exergy at initial and final states
At initial state :
U = 274.01 kJ/Kg , V = 0.034875 m³/kg , S = 1.0256 KJ/kg.k
exergy ( Ф ) at initial state = M ( U + P₂V - T₀S )
= 5 ( 274.01 + 100* 10³ * 0.034875 - 297 * 1.0256)
≈ - 135.5285 kJ
At final state :
U = 84.44 kJ / kg , V = 0.0008261 m³/kg, S = 0.31958 kJ/kg.k
exergy ( ( Ф ) at final state = M ( U + P₂V - T₀S )
= -51.96 kJ
B) Determine the exergy destroyed
exergy destroyed = To * M ( S2 - S1 )
= 297 * 5 ( 0.31958 - 1.0256 )
= - 1048.4397 KJ
Hence we can conclude that A) The exergy of the refrigerant at the initial and final states are : Initial state = - 135.5285 kJ, Final state = -51.96 kJ and The exergy destroyed during this process is : - 1048.4397 kJ
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Exergy of refrigerant-134a at initial and final states is obtained from property tables and by multiplying the mass of the refrigerant with its specific exergy at each state. The difference in exergy between the two states represents the exergy destroyed.
To solve the given question, we need the property values of
refrigerant-134a
at the initial and the final states.
At an initial state of 0.7 MPa and 60°C, the specific exergy for refrigerant-134a can be obtained from property tables which are standard in thermodynamics textbooks. Same for the final state at 0.7 MPa and 24°C, the specific exergy can be obtained from the same property tables.
The exergy of the refrigerant at the initial and the final states can be calculated by multiplying the mass of the refrigerant with its specific exergy at each state.
Exergy destruction during this process can be calculated using the relation between exergy change and exergy destruction. The exergy change of a system between initial and final states is equal to the difference of the exergy of the system at final and initial states.
Based on the second law of thermodynamics, the difference in exergy should be equal to the exergy destroyed during the process.
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