Recall this equation for a device in a direct current circuit:
P = IV
P is the power dissipated by the device, I is the current through the device, and V is the voltage drop of the device.
If we choose to use the ampere as the unit of current and the volt as the unit of voltage, then the product of the current and the voltage will give the power with watts as the unit.
Newton's third law says:
"<span>For every action, there is an equal and opposite reaction. ".
So, the force that Tom does on the sister is equal to force the sister applies on Tom:
</span>

<span>where the label "t" means "on Tom", while the label "s" means "on the sister".
From Newton's second law, we also know
</span>

where m is the mass and a the acceleration. <span>so we can rewrite the first equation as
</span>

<span>And find Tom's acceleration:
</span>

<span>
</span>
Explanation:
The given data is as follows.
Mass of small bucket (m) = 4 kg
Mass of big bucket (M) = 12 kg
Initial velocity (
) = 0 m/s
Final velocity (
) = ?
Height
= 2 m
and,
= 0 m
Now, according to the law of conservation of energy
starting conditions = final conditions

235.44 =
+ 78.48
= 4.43 m/s
Thus, we can conclude that the speed with which this bucket strikes the floor is 4.43 m/s.
Answer:
acceleration, a = 9.8 m/s²
Explanation:
'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.
u = 0 m/s
After 2 seconds, velocity of the ball is 19.6 m/s.
t = 2s, v = 19.6 m/s
Using
v = u + at
19.6 = 0 + 2a
a = 9.8 m/s²
Question
Initially, the baton is spinning about a line through its center at angular velocity 3.00 rad/s. What is its angular momentum? Express your answer in kilogram meters squared per second.
Answer:

Explanation:
The angular momentum L of the baton moving about an axis perpendicular to it, passing through the center of the baton is,

Here, l is the length of the baton.
Substitute 0.120 kg for m, 3 rads/s for ![\omega[\tex] and 0.8 m for l [tex]\begin{array}{c}\\L = \frac{1}{{12}}m{l^2}\omega \\\\ = \frac{1}{{12}}\left( {0.120{\rm{ kg}}} \right){\left( {{\rm{80}}{\rm{.0 cm}}} \right)^2}{\left( {\frac{{1 \times {{10}^{ - 2}}{\rm{m}}}}{{1{\rm{ cm}}}}} \right)^2}\left( {{\rm{3}}{\rm{.00 rad/s}}} \right)\\\\ = 0.0192{\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\\\end{array}](https://tex.z-dn.net/?f=%5Comega%5B%5Ctex%5D%20and%200.8%20m%20for%20l%20%5Btex%5D%5Cbegin%7Barray%7D%7Bc%7D%5C%5CL%20%3D%20%5Cfrac%7B1%7D%7B%7B12%7D%7Dm%7Bl%5E2%7D%5Comega%20%5C%5C%5C%5C%20%3D%20%5Cfrac%7B1%7D%7B%7B12%7D%7D%5Cleft%28%20%7B0.120%7B%5Crm%7B%20kg%7D%7D%7D%20%5Cright%29%7B%5Cleft%28%20%7B%7B%5Crm%7B80%7D%7D%7B%5Crm%7B.0%20cm%7D%7D%7D%20%5Cright%29%5E2%7D%7B%5Cleft%28%20%7B%5Cfrac%7B%7B1%20%5Ctimes%20%7B%7B10%7D%5E%7B%20-%202%7D%7D%7B%5Crm%7Bm%7D%7D%7D%7D%7B%7B1%7B%5Crm%7B%20cm%7D%7D%7D%7D%7D%20%5Cright%29%5E2%7D%5Cleft%28%20%7B%7B%5Crm%7B3%7D%7D%7B%5Crm%7B.00%20rad%2Fs%7D%7D%7D%20%5Cright%29%5C%5C%5C%5C%20%3D%200.0192%7B%5Crm%7B%20kg%7D%7D%20%5Ccdot%20%7B%7B%5Crm%7Bm%7D%7D%5E%7B%5Crm%7B2%7D%7D%7D%7B%5Crm%7B%2Fs%7D%7D%5C%5C%5Cend%7Barray%7D)