Answer:
Current for 12 ohms = 2.5
Explanation:
Ohm's Law:
Voltage = IR
Current = V/R
Resistance = V/I
Therefore, current = 30V/12 Ohms
30/12 = 2.5
So current for 12 ohms = 2.5
I hope this helps :>
Answer:
0.81 m/s^2
Explanation:
A child rides a bicycle from rest
The child reaches a velocity of 3.8m/s
The time is 4.7 secs
Therefore the acceleration can be calculated as follows
v= u + at
3.8= 0 + a(4.7)
3.8= 0 + 4.7a
3.8-0= 4.7a
3.8= 4.7a
a= 3.8/4.7
= 0.81 m/s^2
Hence the acceleration is 0.81 m/s^2
Answer:
In an elastic collision, the momentum is conserved and the mechanical energy is conserved too.
Explanation:
There are two types of collisions:
- Elastic collision: in an elastic collision, the total momentum before and after the collision is conserved; also, the total mechanical energy before and after the collision is conserved.
- Inelastic collision: in an inelastic collision, the total momentum before and after the colllision is conserved, while the total mechanical energy is not conserved (in fact, part of the energy is converted into other forms of energy such that thermal energy, due to the presence of frictional forces)
However, there is one type of double-replacement reaction that we can predict: the precipitation reaction. A precipitation reaction occurs when two ionic compounds are dissolved in water and form a new ionic compound that does not dissolve; this new compound falls out of solution as a solid precipitate.
If an icy surface means no friction, then Newton's second law tells us the net forces on either block are
• <em>m</em> = 1 kg:
∑ <em>F</em> (parallel) = <em>mg</em> sin(45°) - <em>T</em> = <em>ma</em> … … … [1]
∑ <em>F</em> (perpendicular) = <em>n</em> - <em>mg</em> cos(45°) = 0
Notice that we're taking down-the-slope to be positive direction parallel to the surface.
• <em>m</em> = 0.4 kg:
∑ <em>F</em> (vertical) = <em>T</em> - <em>mg</em> = <em>ma</em> … … … [2]
<em />
Adding equations [1] and [2] eliminates <em>T</em>, so that
((1 kg) <em>g</em> sin(45°) - <em>T </em>) + (<em>T</em> - (0.4 kg) <em>g</em>) = (1 kg + 0.4 kg) <em>a</em>
(1 kg) <em>g</em> sin(45°) - (0.4 kg) <em>g</em> = (1.4 kg) <em>a</em>
==> <em>a</em> ≈ 2.15 m/s²
The fact that <em>a</em> is positive indicates that the 1-kg block is moving down the slope. We already found the acceleration is <em>a</em> ≈ 2.15 m/s², which means the net force on the block would be ∑ <em>F</em> = <em>ma</em> ≈ (1 kg) (2.15 m/s²) = 2.15 N directed down the slope.
