Which reaction would produce a solution that conducts electricity extremely well? Please help
1 answer:
You might be interested in
(a) 196 N
At terminal speed, the velocity of the box is constant: this means that its acceleration is zero, so according to Newton's second Law, the resultant of the forces acting on the box is zero. Since there are only two forces acting on the box:
- The weight, acting downward:
- The air resistance, acting upward:
It means that at terminal speed, the two forces are balanced:
So we have:
(b) 637 N
The exercise is exactly identical as before, but this time the mass of the box is different: m = 65 kg. Therefore, the air resistance in this case will be:
Answer:
a. Since h = 5.74 m > 5.7 m, the height of the cliff, the block of ice will successfully launch to the top of the cliff.
b. -2.81 m/s²
c. 3.78 s
d. -351.25 N
Explanation:
a. After landing on the flat land above, each block of ice travels 20 meters while slowing to a stop.
For the block of ice to reach the top of the cliff, its maximum height, h should be greater than or equal to 5.7 m. That is, h ≥ 5.7 m.
The maximum height of a projection h, projected with an initial velocity v at an angle Ф is h = v²sin²Ф/2g where g = acceleration due to gravity = 9.8 m/s².
For the block of ice, v = 15 m/s and Ф = 45°. So,
h = v²sin²Ф/2g
= (15 m/s)²sin²45/(2 × 9.8 m/s²)
= 225 (m/s)²(1/√2)²/19.6 m/s²
= 225 (m/s)²(1/2)/19.6 m/s²
= 112.5 (m/s)²/19.6 m/s²
= 5.74 m
Since h = 5.74 m > 5.7 m, the height of the cliff, the block of ice will successfully launch to the top of the cliff.
The graph is in the attachment.
b. What is the rate of acceleration while the blocks slow to a stop?
Using v² = u² + 2as where u = initial horizontal velocity of block = 15m/scos45° = 10.61 m/s, v = final velocity of block = 0 m/s since it stops, a = acceleration and s = distance block moves = 20 m
So, a = (v² - u²)/2s
substituting the variables into the equation, we have
a = ((0 m/s)² - (10.61 m/s)²)/2(20 m)
= - 112.57 (m/s)²)/40 m
= -2.81 m/s²
c. How long do the blocks take to slow to a stop?
Using v = u + at where u = initial horizontal velocity of block = 10.61 m/s v = final velocity of block = 0 m/s since it stops, a = acceleration = -2.81 m/s² and t = time it takes block of ice to stop
So, making t subject of the formula,
t = (v - u)/a
substituting the values of the variables, we have
t = ( 0 m/s - 10.61 m/s)/-2.81 m/s²
= -10.61 m/s/-2.81 m/s²
= 3.78 s
d. What is the amount of friction between the ice and the snowy ground?
The frictional force, f = net force on block of ice
f = ma where m = mass of bock = 125 kg and a = acceleration of block = -2.81 m/s²
f = ma
= 125 kg(2.81 m/s²)
= -351.25 N
To solve the problem it is necessary to apply the concepts related to Voltage in an Inductor. By general definition the voltage in in an inductor is defined as
Where,
Voltage induced (emf)
L = Inductance
di = Rate of change of current flow
dt = rate of change of time
Our values are given by
Replacing
Therefore the magnitude of average induced emf is 8.6mV