Answer:
I can give you a list of noncontact forces off of my head, if that is what you need.
Explanation:
First, gravity. It acts everywhere and pulls down every object that is in the air towards the ground, unless they have an equal and opposite force that counters it.
Second, wind. I don't think this will count because wind is a physical thing, but it can be used in the same way as we did with gravity. If the force counteracting it is larger, equal, and opposite of the force it is countering, then it will move forward, through the wind.
Answer:
Explanation:
Work done by torque is given as
Word one = torque × angular displacement
W = τ × θ
Given that,
τ = 2000Nm
Mass of motor = 300kg
Radius r = 55cm = 0.55m
Work done by wheel in first t= 23second.
Now we need to find the angular displacement
We know that,
τ = I•α
Moment of inertia of wheel
I = MR²
I = 300 × 0.55²
I = 90.75 kgm²
Then, τ = I•α
α = τ / I
α = 2000/90.75
α = 22.04rad/s²
Then, using circular motion,
∆θ = wit + ½αt²
wi = 0rad/s
∆θ = ½αt²
∆θ = ½ × 22.04 × 23²
∆θ = 5829.2 rad.
Then,
Work done?
W = τ × θ
W = 2000 × 5829.2
W = 1.17 × 10 ^7 J
I think when particles are in the gas form they are most likely to cause a chemical reaction considering the fact that gas particles aren't very controllable and it would make sense that they would react unexpectedly
Answer:
I₁ = 1.6 A (through 7 Ohm Resistor)
I₂ = 1.3 A (through 8 Ohm Resistor)
I₃ = I₁ - I₂ = 1.6 A - 1.3 A = 0.3 A (through 4 Ohm Resistor)
Explanation:
Here we consider two loops doe applying Kirchhoff's Voltage Law (KVL). The 1st loop is the left side one with a voltage source of 12 V and the 2nd Loop is the right side one with a voltage source of 9 V. We name the sources and resistor's as follows:
R₁ = 7 Ω
R₂ = 4 Ω
R₃ = 8 Ω
V₁ = 12 V
V₂ = 9 V
Now, we apply KVL to 1st Loop:
V₁ = I₁R₁ + (I₁ - I₂)R₂
12 = 7I₁ + (I₁ - I₂)(4)
12 = 7I₁ + 4I₁ - 4I₂
I₁ = (12 + 4 I₂)/11 ------------ equation (1)
Now, we apply KVL to 2nd Loop:
V₂ = (I₂ - I₁)R₂ + I₂R₃
9 = (I₂ - I₁)(4) + 8I₂
9 = 4I₂ - 4I₁ + 8I₂
9 = 12I₂ - 4I₁ -------------- equation (2)
using equation (1)
9 = 12I₂ - 4[(12 + 4 I₂)/11]
99 = 132 I₂ - 48 - 16 I₂
147 = 116 I₂
I₂ = 147/116
I₂ = 1.3 A
use this value in equation 2:
9 = 12(1.3 A) - 4I₁
4I₁ = 15.6 - 9
I₁ = 6.6 A/4
I₁ = 1.6 A
Hence, the currents through all resistors are:
<u>I₁ = 1.6 A (through 7 Ohm Resistor)</u>
<u>I₂ = 1.3 A (through 8 Ohm Resistor)</u>
<u>I₃ = I₁ - I₂ = 1.6 A - 1.3 A = 0.3 A (through 4 Ohm Resistor)</u>
