If an object is projected upward from ground level with an initial velocity of 144144 ft per sec, then its height in feet after
1 answer:
Answer:
4.5 s, 324 ft
Explanation:
The object is projected upward with an initial velocity of
The equation that describes its height at time t is
(1)
where t, the time, is measured in seconds.
In order to find the time it takes for the object to reach the maximum height, we must find an expression for its velocity at time t, which can be found by calculating the derivative of the position, s(t):
(2)
At the maximum heigth, the vertical velocity is zero:
v(t) = 0
Substituting into the equation above, we find the corresponding time at which the object reaches the maximum height:
And by substituting this value into eq.(1), we also find the maximum height:
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Answer:
2.84403 seconds
2.91483 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
It takes 2.84403 seconds to reach the highest point
The ball will travel 39.67431+2 = 41.67431 m while going down to the ground
The ball takes 2.91483 seconds to hit the ground after it reaches its highest point.
Answer:
Incomplete questions check attachment for circuit diagram.
Explanation:
We are going to use superposition
So, we will first open circuit the current source and find the voltage Voc.
So, check attachment for open circuit diagram.
From the diagram
We notice that R3 is in series with R4, so its equivalent is given below
Req(3-4) = R3 + R4
R(34) = 20+40 = 60 kΩ
Notice that R2 is parallel to the equivalent of R3 and R4, then, the equivalent of all this three resistor is
Req(2-3-4) = R2•R(34)/(R2+R(34))
R(234) = (100×60)/(100+60)
R(234) = 37.5 kΩ
We notice that R1 and R(234) are in series, then, we can apply voltage divider rule to find voltage in R(234)
Therefore
V(234) = R(234) / [R1 + R(234)] × V
V(234) = 37.5/(25+37.5) × 100
V(234) = 37.5/62.5 × 100
V(234) = 60V.
Note, this is the voltage in resistor R2, R3 and R4.
Note that, R2 is parallel to R3 and R4. Parallel resistor have the same voltage, then voltage across R2 equals voltage across R34
V(34) = 60V.
Now, we also know that R3 and R4 are in series,
So we can know the voltage across R4 which is the Voc we are looking for.
Using voltage divider
V4 = Voc = R4/(R4 + R(34)) × V(34)
Voc = 40/(40+60) × 60
Voc = 24V
This is the open circuit Voltage
Now, finding the short circuit voltage when we short circuit the voltage source
Check attachment for circuit diagram.
From the circuit we notice that R1 and R2 are in parallel, so it's equivalent becomes
Req(1-2) = R1•R2/(R1+R2)
R(12) = 25×100/(25+100)
R(12) = 20 kΩ
We also notice that the equivalent of Resistor R1 and R2 is in series to R3. Then, the equivalent resistance of the three resistor is
Req(1-2-3) = R(12) + R(3)
R(123) = 20 + 20
R(123) = 40 kΩ
We notice that, the equivalent resistance of the resistor R1, R2, and R3 is in series to resistor R4.
So using current divider rule to find the current in resistor R4.
I(4) = R(123) / [R4+R(123)] × I
I(4) = 40/(40+40) × 8
I(4) = 4mA
Then, using ohms law, we can find the voltage across the resistor 4 and the voltage is the required Voc
V = IR
V4 = Voc = I4 × R4
Voc = 4×10^-3 × 40×10^3
Voc = 160V
Then, the sum of the short circuit voltage and the open circuit voltage will give the required Voc
Voc = Voc(open circuit) + Voc(short circuit)
Voc = 24 + 160
Voc = 184V.
Answer:
- k = 167.33 N/m
Explanation:
- The radio waves have a fixed relationship between the propagation speed (the speed of light in vacuum), the frequency and the wavelength, as follows:
- v = c = λ*f
where c= speed of light in vacuum = 3*10⁸ m/s, λ = wavelength =
4.92*10⁷ m.
Solving for f, we get the frequency of the radio waves:
f = 6.1 Hz
- Now, from the Hooke's law, we know that the mass attached at the end of the spring oscillates with an angular frequency defined by a fixed relationship between the spring constant k and the mass m, as follows:
- Now, we know that there exists a fixed relationship between the angular frequency and the frequency, as follows:
- We also know that f in (2) is the same that we got for the radio waves, so replacing (2) in (1), and rearranging terms, we can solve for k, as follows: