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Ganezh [65]
4 years ago
11

How much work is requried to uniformly accelerate a merry-go-round?

Physics
1 answer:
tiny-mole [99]4 years ago
5 0
Do not worry if you don't recognize both parts of the problem at this point. If you recognize the dynamics problem,<span> On the other hand, if you recognize this as a kinematics problem you will quickly see that you need to find angular acceleration before you can begin and so will need to do that pre-step first.</span>
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What is the momentum of a 50 kg object traveling 200 m/s
garik1379 [7]
Momentums equation is just p=mv
mass times velocity so 50x200
p=10,000
3 0
3 years ago
A very long nonconducting cylinder of diameter 10.0 cm carries charge distributed uniformly over its surface. Each meter of leng
Murrr4er [49]

Answer:

Explanation:

The concept of electric field, force acting on proton is applied and appropriate derivations were made to calculate the distance from the surface as shown in the attached file.

7 0
3 years ago
A characteristic controlled by one or more genes is called a what
Rama09 [41]
I believe the answer would be polygenic. The word translates to "multiple genes". This occurs when many genes make up one specific thing such as our hair color or eye color.
4 0
4 years ago
Solutions with ions that react with acids or bases to lessen their effects are fubrefs
Softa [21]

Answer:

Buffers

Explanation:

A buffer solution is a solution containing weak acids and their salts or weak bases and their salts.

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4 0
3 years ago
Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 9 m into 32 cm. of water. Assu
baherus [9]

Answer:

823.46 kgm/s

Explanation:

At 9 m above the water before he jumps, Henri LaMothe has a potential energy change, mgh which equals his kinetic energy 1/2mv² just as he reaches the surface of the water.

So, mgh = 1/2mv²

From here, his velocity just as he reaches the surface of the water is

v = √2gh

h = 9 m and g = 9.8 m/s²

v = √(2 × 9 × 9.8) m/s

v = √176.4 m/s

v₁ = 13.28 m/s

So his velocity just as he reaches the surface of the water is 13.28 m/s.

Now he dives into 32 cm = 0.32 m of water and stops so his final velocity v₂ = 0.

So, if we take the upward direction as positive, his initial momentum at the surface of the water is p₁ = -mv₁. His final momentum is p₂ = mv₂.

His momentum change or impulse, J = p₂ - p₁ = mv₂ - (-mv₁) = mv₂ + mv₁. Since m = Henri LaMothe's mass = 62 kg,

J = (62 × 0 + 62 × 13.28) kgm/s = 0 +  823.46 kgm/s = 823.46 kgm/s

So the magnitude of the impulse J of the water on him is 823.46 kgm/s

6 0
3 years ago
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