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3 years ago

A person just supports a mass of 20kg suspended from a rope.

Physics

1 answer:

Georgia [21]3 years ago

7 0

Answer:

F = 200 N

Explanation:

Given that,

The mass suspended from the rope, m = 20 kg

We need to find the resultant force acting on the rope. The resultant force on the rope is equal to its weight such that,

F = mg

Where

g is acceleration due to gravity

Put all the values,

F = 20 kg × 10 m/s²

F = 200 N

So, the resultant force on the mass is 200 N.

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Lucy took 3 hours to cover 2/3 of a journey. She covered the

Grace [21]

Deleted this answer, sorry for any inconveniences that it may have caused.

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3 years ago

In his novel From the Earth to the Moon (1866), Jules Verne describes a spaceship that is blasted out of 12,000 yards/s. the Col

saw5 [17]

Answer:

The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

Explanation:

Let suppose that spaceship is accelerated uniformly. A yard equals 0.914 meters. A feet equals 0.304 meters. If air viscosity and friction can be neglected, then acceleration ( $a$ ), measured in meters per square second, is estimated by this kinematic formula:

$a = \frac{v^{2}-v_{o}^{2}}{2\cdot \Delta s }$ (1)

Where:

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$ , $v$ - Initial and final speeds of the spaceship, measured in meters.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $v = 10968\,\frac{m}{s}$ and $\Delta s = 212.8\,m$ , then the acceleration experimented by the spaceship is:

$a = \frac{\left(10968\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (212.8\,m)}$

$a = 282652.782\,\frac{m}{s^{2}}$

The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

5 0

3 years ago

When devising a model, scientists can only use the information available during their lifetime. This means that the current mode

11111nata11111 [884]

no, it not useless. we still learn Bohr's model in HS n dats almost 200 yr old! while there may be new models, previous one is good for explaining the basics. it is also useful to learn previous model n see how our understanding improves over time.

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3 years ago

Read 2 more answers

. If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect tim

inn [45]

Answer:

The appropriate response will be "Length must be increased by 0.012%".

Explanation:

The given values is:

ΔT = 5 s/day

Now,

⇒ $\frac{\Delta T}{T} =\frac{5}{24\times 60\times 60}$