Voltage = Current (I) × Resistance (R)
V = 10 × 28.5 = 285v
Your equation is:

An equation is balanced only if there are the same number of atoms of each element on both sides of the arrow - aka same number of atoms of each element in both reactants (left of the arrow) and products (right of the arrow).
It'll be easiest to tackle this by counting up the number of atoms of each element on the left and on the right and comparing those numbers. If there is a number in front of the entire compound, that means that number applies to all elements in the compound. If the number is a subscript (little number to the right of the element), that means that number only applies to the element that the subscript is attached to:
1) On the left, you have:

2) On the right, you have:

You can see that the number of oxygen and hydrogen atoms aren't equal on both the left (reactants) and the right (products), so the equation is unbalanced.
Your final answer is "T<span>he equation is
unbalanced because the number of hydrogen atoms and
oxygen is
not equal in the reactants and in the products."</span>
Momentum is conserved when carts are collided on a slanting plane.
To find the answer, we need to know about the conversation of momentum.
<h3>What's the conversation of momentum?</h3>
- Conservation of linear momentum says the total momentum before the collision and after the collision remains the same.
- Mathematically, m1u1+m2u2 = m1v1+m2v2
<h3>How is the momentum conserved when collision occurs on a slanting plane?</h3>
- On a slanting plane, the velocity has two components,
- horizontal component
- horizontal component Vertical component
- So, its momentum has also similar two components.
- The momentum is conserved along horizontal direction and vertical direction separately.
Thus, we can conclude that the momentum is conserved when carts are collided on a slanting plane.
Learn more about the conversation of momentum here:
brainly.com/question/7538238
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A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.
A car experiences a deceleration (a) of -41.62 m/s² and comes to a stop (final velocity = v = 0 m/s) after 10.99 m (s).
We can calculate the initial velocity of the car (u) using the following kinematic equation.
![v^{2} = u^{2} + 2as\\\\u = \sqrt[]{v^{2}-2as} = \sqrt[]{(0m/s)^{2}-2(-42.61m/s^{2} )(10.99m)} = 30.60m/s](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%3D%20u%5E%7B2%7D%20%2B%202as%5C%5C%5C%5Cu%20%3D%20%5Csqrt%5B%5D%7Bv%5E%7B2%7D-2as%7D%20%3D%20%5Csqrt%5B%5D%7B%280m%2Fs%29%5E%7B2%7D-2%28-42.61m%2Fs%5E%7B2%7D%20%29%2810.99m%29%7D%20%3D%2030.60m%2Fs)
A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.
Learn more: brainly.com/question/14851168