a skier is gliding along 3.0 m/s on horizontal, frictionless snow. he suddenly starts down a 10 degrees incline. his speed at the bottom is 15 m/s. what is the lenght of the incline? how long does it take him to reach the bottom?
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(way 1) determine acceleration using force
only force act on skier is mg vertically. spilt vector we get force along the incline = mgsin(10) and f= ma so
ma = mgsin(10) or a = gsin(10)
a (along the incline)= gsin(10) = 10sin(10) = 1.74
v^2 = u^2 + 2as
15^2 = 3^2 + 2(1.74)s
s = 62.06 m
(way 2) using conservation of energy
energy (KE+PE) on top = energy at bottom
0.5m3^2 + mgh = 0.5m15^2 +0
h (height of incline) = (112.5 - 4.5)/10 = 19.8 m
length of incline = h/sin(10) = 62.2 m ; trigonometry
... find time
s = (u+v)t/2
t = 2s/(u+v) = 2(62.2)/(3+15) = 6.91 s
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Answer:
Explanation:
Creating an even parity circuit in Multisim involves designing a digital logic circuit that checks if the number of '1's in a binary input is even. If the input has an even number of '1's, the circuit should output '1' (indicating even parity); otherwise, it should output '0' (indicating odd parity).
Here's how to create an even parity circuit in Multisim, along with the truth table:
**Creating the Even Parity Circuit:**
1. Open Multisim and create a new blank schematic.
2. Add the following components to your schematic:
- Input pins (for binary input bits)
- XOR gates
- An AND gate
- An inverter (NOT gate)
- Output display (LED or probe)
3. Connect the input pins to the XOR gates. Each input pin corresponds to one bit of the binary input.
4. Connect the outputs of the XOR gates to the inputs of the AND gate.
5. Connect the output of the AND gate to the input of the inverter (NOT gate).
6. Connect the output of the inverter to the output display.
7. Label your input pins for clarity (e.g., A0, A1, A2, ...).
**Designing the Even Parity Truth Table:**
To create the truth table for even parity, you'll need to list all possible input combinations (binary numbers) along with the corresponding output (even or odd).
Assuming you have a 3-bit input (A2, A1, A0), here's the truth table:
| A2 | A1 | A0 | Output (Even Parity) |
|----|----|----|-----------------------|
| 0 | 0 | 0 | 1 (Even) |
| 0 | 0 | 1 | 0 (Odd) |
| 0 | 1 | 0 | 0 (Odd) |
| 0 | 1 | 1 | 1 (Even) |
| 1 | 0 | 0 | 0 (Odd) |
| 1 | 0 | 1 | 1 (Even) |
| 1 | 1 | 0 | 1 (Even) |
| 1 | 1 | 1 | 0 (Odd) |
Each row in the truth table represents a unique combination of input bits (A2, A1, A0) and specifies whether the output is '1' (Even) or '0' (Odd).
Once you have created the circuit in Multisim and designed the truth table, you can simulate the circuit to verify its functionality. Ensure that the circuit produces the expected output (even parity) based on the input values.
Answers
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Answer:
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Explanation:
As a hole is punched in a full milk carton, and we have to calculate the initial velocity of outflow 10 cm below the top. We use the concept of conservation of energy.
Further Explanation:
Using the conservation of energy, below the top the potential energy converted in kinetic energy
P.E =K.E
As mass of milk does not change at the top and 10 cm below the top, therefore
Here, v is the initial velocity,
g is acceleration due to gravity
h is height.
Given: h = 10 cm and g = 9.8 m/s2
Substituting the given values, we get
Lear more:
Key word:
Conservation of energy, Kinetic energy, potential energy.
Milk has almost same density as water: (Rho)= 1,000 kg /m³
P 2 = 101,325 Pa + 1,000 kg/m³ · 9.81 m/s² · 0.1 m = 102,306 Pa
The hydrostatic equation:
P 1 + (Rho)v1² / 2 = P 2 + (Rho)·g·h2
101,325 + 1,000 v1²/2 = 102,306 + 1000 · 9.81 · 0.1
500 v 1² = 102,306 + 981 - 101,325
v 1² = 3.924
v 1 = √ 3.924
v 1 = 1.98 m/s
The initial velocity of outflow is 1.98 m/s.
Answers
Final answer:
The question is related to Physics and deals with kinematic equations. With the supplied information, one can calculate elements such as velocity or applied force in the jump.
Explanation:
The subject of the question pertains to the field of Physics, specifically the area of kinematic equations which deal with the motion of objects. The provided information in the question pertains to the rise of a person's body during a jump. Given the average height of 60cm that a person typically attains and the approximate rise of the body from the knees up being 50cm, these figures can be used in a Physics context to determine different factors of the jump such as velocity or force applied.
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Learn more about Kinematic Equations here:
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