Answer:
Explanation:
The magnetic field due to straight wire is into the square coil.
As the current in straight wire decreases the magnetic flux in the coil decreases . The induced magnetic field is into the coil.The induced current is along +y direction
Answer:
Approximately .
Explanation:
The formula for the kinetic energy of an object is:
,
where
Important: Joule () is the standard unit for energy. The formula for
requires two inputs: mass and speed. The standard unit of mass is
while the standard unit for speed is
. If both inputs are in standard units, then the output (kinetic energy) will also be in the standard unit (that is: joules,
Convert the unit of the arrow's mass to standard unit:
.
Initial of this arrow:
.
That's the same as the energy output of this bow. Hence, the efficiency of energy transfer will be:
.
Answer:
we have to find out the critical resolved shear stress. As it it given in the question
Ф = 28.1°and the possible values for λ are 62.4°, 72.0° and 81.1°.
a) Slip will occur in the direction where cosФ cosλ are maximum. Cosine for all possible λ values are given as follows.
cos(62.4°) = 0.46
cos(72.0°) = 0.31
cos(81.1°) = 0.15
Thus, the slip direction is at the angle of 62.4° along the tensile axis.
b) now the critical resolved shear stress can be find out by the following equation.
τ = σ
( cosФ cosλ)
now by putting values,
= (1.95MPa)[ cos(28.1) cos(62.4)] = 0.80 MPa (114 Psi) 7.23
Answer:
Hello your question has some missing parts and the required diagram attached below is the missing part and the diagram
Digital circuits require actions to take place at precise times, so they are controlled by a clock that generates a steady sequence of rectangular voltage pulses. One of the most widely
used integrated circuits for creating clock pulses is called a 555 timer. shows how the timer’s output pulses, oscillating between 0 V and 5 V, are controlled with two resistors and a capacitor. The circuit manufacturer tells users that TH, the time the clock output spends in the high (5V) state, is TH =(R1 + R2)*C*ln(2). Similarly, the time spent in the low (0 V) state is TL = R2*C*ln(2). Design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a 500 pF capacitor. What values do you need to specify for R1 and R2?
ANSWER : R1 = 144.3Ω, R2 = 72.2 Ω
Explanation:
Frequency = 10 MHz
Time period = 1 / F = 0.1 <em>u </em>s
Duty cycle = 75% = 0.75
Duty cycle can be represented as : Ton / T
Also: Ton = Th = 0.75 * 0.1 <em>u </em>s = 75 <em>n</em> s
TL = T - Th = 100 <em>n</em>s - 75 <em>n</em> s = 25 <em>n</em> s
To find the value of R2 we use the equation for time spent in the low (0 V) state
TL = R2*C*ln(2)
hence R2 = TL / ( C * In 2 )
c = 500 pF
Hence R2 = 25 / ( 500 pF * 0.693 ) = 72.2 Ω
To find the value of R1 we use the equation for the time the clock output spends in the high (5V) state,
Th = (R1 + R2)*C*ln(2)
from the equation make R1 the subject of the formula
R1 = (Th - ( R2 * C * In2 )) / (C * In 2)
R1 = ( 75 ns - ( 72.2 * 500 pF * 0.693)) / ( 500 pF * 0.693 )
R1 = ( 75 ns - ( 25 ns ) / 500 pf * 0.693
= 144.3Ω
