I would go with salt in ocean water as when you heat it and the water begins to evaporate it will leave the salt behind
Also water in mud as the water would evaporate and leave the mud residue behind
GOOD LUCK
BRAINLIEST IF HELPED
Answer:
a) 1.3 rad/s
b) 0.722 s
Explanation:
Given
Initial velocity, ω = 0 rad/s
Angular acceleration of the wheel, α = 1.8 rad/s²
using equations of angular motion, we have
θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²
where
θ2 - θ1 = 53.2 rad
t2 - t1 = 7s
substituting these in the equation, we have
θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²
53.2 =ω(0) * 7 + 1/2 * 1.8 * 7²
53.2 = 7.ω(0) + 1/2 * 1.8 * 49
53.2 = 7.ω(0) + 44.1
7.ω(0) = 53.2 - 44.1
ω(0) = 9.1 / 7
ω(0) = 1.3 rad/s
Using another of the equations of angular motion, we have
ω(0) = ω(i) + α*t1
1.3 = 0 + 1.8 * t1
1.3 = 1.8 * t1
t1 = 1.3/1.8
t1 = 0.722 s
According to the law of conservation of momentum:
m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision
Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.
m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?
Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.
The velocity of the 2nd car after the collision is
0.03m/s.
Answer: 0.42 Amperes
Explanation:
Given that:
Current, I = ?
Electric charge Q = 100 coulomb
Time, T = 4.0 minutes
(The SI unit of time is seconds. so, convert 4.0 minutes to seconds)
If 1 minute = 60 seconds
4.0 minutes = 4.0 x 60 = 240 seconds
Since electric charge, Q = current x time
i.e Q = I x T
100 coulomb = I x 240 seconds
I = 100 coulomb / 240 seconds
I = 0.4167 Amperes (round to the nearest hundredth which is 0.42 amperes)
Thus, 0.42 Amperes of current flows in the circuit.