Answer:
friction is when two objects tough with force basically. when an object is rubbed against something. when an object comes in contact with something else. or ig u can call it hot air rubbing off the inevitable force lol that works too , bud
whatever helps u rememeber
Explanation:
Make a table, listing the x and y coordinates of each square's center of gravity and its mass. Multiply the coordinates by the mass, add the results for each x and y, then divide by the total mass.
The x-coordinate of the center of gravity is 15/14 a.
The y-coordinate of the center of gravity is 47/42 a.
Answer:
θ=108rad
t =10.29seconds
α=-8.17rad/s²
Explanation:
Given that
At t=0, Wo=24rad/sec
Constant angular acceleration =30rad/s²
At t=2, θ=432rad as it try to stop because the circuit break
Angular motion
W=Wo+αt
θ=Wot+1/2αt²
W²=Wo²+2αθ
We need to find θ between 0sec to 2sec when the wheel stop
a. θ=Wot+1/2αt²
θ=24×2+1/2×30×2²
θ=48+60
θ=108rad.
b. W=Wo+αt
W=24+30×2
W=84rad/s
This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.
Wo=84rad/sec
W=0rad/s, because the wheel stop at θ=432rad
Using W²=Wo²+2αθ
0²=84²+2×α×432
-84²=864α
α=-8.17rad/s²
It is negative because it is decelerating
Now, time taken for the wheel to stop
W=Wo+αt
0=84-8.17t
-84=-8.17t
Then t =10.29seconds.
a. θ=108rad
b. t =10.29seconds
c. α=-8.17rad/s²
Answer:
96.21 ft/s
Explanation:
To solve this, you only need to use one expression which is:
Vf² = Vo² + 2gh
g = 9.8 m/s²
However, this exercise is talking in feet, so convert the gravity to feet first:
g = 9.8 * 3.28 = 32.15 ft/s²
Vo is zero, because it's a free fall and in free fall the innitial speed is always zero. With this, let's calculate the speed at 2 seconds, with a height of 64 ft, and then with the 256 ft:
V1 = √2*32.15*64
V1 = 64.15 ft/s
V2 = √2*32.15*256
V2 = 128.3 ft/s
So the average rate is:
V = 128.3 + 64.15 / 2
V = 96.22 ft/s
Answer:
T'=70.92°C
Explanation:
Given that
V= 100 L=0.1 m³
P=400 KPa
T=25°C
Work done on the air = 15 KJ
W= -15 KJ
If we assume that air is ideal gas
P V = m R T
R=0.287 KJ/kg.K for air
T= 273 + 25 = 298 K
By putting the values
P V = m R T
400 x 0.1 = m x 0.287 x 298
m=0.46 kg
From first law of thermodynamics
Q= ΔU +W
Insulated piston–cylinder , Q=0
ΔU = m Cv ΔT
ΔU = - W
Cv = 0.71 KJ/kg.k for air
0.46 x 0.71 x (T' -25) = 15
T'=70.92 °C
So the final temperature of air is T'=70.92 °C
