Answer:
ωf = 8.8 rad/s
v = 2.2 m/s
Explanation:
We will use the third equation of motion to find the maximum angular velocity of the wheel:

where,
α = angular acceleration = 6 rad/s²
θ = angular displacemnt = 1 rev = 2π rad
ωf = max. final angular velocity = ?
ωi = initial angular velocity = 1.5 rad/s
Therefore,

<u>ωf = 8.8 rad/s</u>
Now, for linear velocity:
v = rω = (0.25 m)(8.8 rad/s)
<u>v = 2.2 m/s</u>
<span>82.0 kg
I am going to assume that there is a typo for the number of joules of energy. Doing a google search for this exact question showed this question multiple times with a value of 4942 joules which makes sense given how close the "o" key is to the "9" key. Because of this, I will assume that the correct value for the number of joules is 4942. With that in mind, here's the solution.
The gravitational potential energy is expressed as the mass multiplied by the height, multiplied by the local gravitational acceleration. So:
E = MHA
Solving for M, the substituting the known values and calculating gives:
E = MHA
E/(HA) = M
4942/(6.15*9.8) = M
4942/60.27 = M
81.99767712 = M
Rounding to 3 significant figures gives 82.0 kg</span>
<h3><u>Answer;</u></h3>
<u>No-</u>the equation is not balanced.
<h3><u>Explanation;</u></h3>
- The balanced equation will be;
<em>2Al + 3NiBr2→ 2AlBr3 + 3Ni</em>
- <em><u>According to the law of conservation of mass, the mass of reactants should always be the same as the mass of the products in a chemical equation. </u></em>
- Therefore, the number of atoms of each element in a chemical equation should always be the same on both sides of the equation, that is the side of reactants and side of products.
- <em><u>Balancing of chemical equations ensures that the number of atoms of each element is equal in both sides of the equation</u></em>.
Answer:
Electrostatics is a branch of physics that studies electric charges at rest. Since classical physics, it has been known that some materials, such as amber, attract lightweight particles after rubbing. The Greek word for amber, or electron, was thus the source of the word 'electricity'.