The resultant of the concurrent forces is 300 lb pointing up along the Y axis. Compute the values of F and the angle required to
give the resultant.
2 answers:
The values of F and the angle required to give the resultant are F = 823.529 lbs and theta = 30.667 degrees
<h3>Further explanation </h3>The resultant of the concurrent forces is 300 lb pointing up along the Y axis. Compute the values of F and the angle required to give the resultant.
The first thing is to get the x and y components of each forces, then isolate the given figure:
The sum of y components is
The sum of x components is
Substitute the value of θ to either of the two working equations, we have
- Learn more about the values of F brainly.com/question/12251772
- Learn more about The resultant brainly.com/question/11784431
- Learn more about the angle brainly.com/question/10217662
Grade: 9
Subject: physics
Chapter: forces
Keywords: the values of F, the angle, The resultant, the concurrent forces, angle
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Answer:
82.76m
Explanation:
In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.
You use the formula for the circumference of the steel ring:
(1)
C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)
you solve for r in the equation (1):
Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:
(2)
L: final length of the tube = ?
Lo: initial length of the tube = 4*10^7m
ΔT = change in the temperature of the steel tube = 1°C
α: thermal coefficient expansion of steel = 13*10^-6 /°C
You replace the values of the parameters in the equation (2):
With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:
Finally, you compare both r and r' radius:
r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m
Hence, the distance to the ring from the ground is 82.76m
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
