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VLD [36.1K]
3 years ago
9

A 68 kg runner exerts a force of 59 N. what is the acceleration of the runner

Physics
2 answers:
LuckyWell [14K]3 years ago
8 0
<span>.87 m/s^2 ,hope this helps!!!!!!</span>
romanna [79]3 years ago
5 0

Answer: 0.87m/s²

Explanation: Recall from Newton's second law

Force = mass*acceleration

So,

Acceleration = Force/mass

Acceleration = 59N/68kg

= 0.868N/kg

Which is also 0.87m/s² in standard unit.

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Helicopters rotor blades, could spin at high speed of 510 rpm. Find the angular displacement in radian for 3 hour(s) operation
malfutka [58]

Answer:

The angular displacement of the blade is 576,871.2 radians

Explanation:

Given;

angular speed of the Helicopters rotor blades, ω = 510 rpm (revolution per minute)

time of motion, t = 3 hours

The angular speed of the Helicopters rotor blades in radian per second is given as;

\omega = \frac{510 \ rev}{mins} *\frac{2 \pi \ rad}{1 \ rev} *\frac{1 \ min}{60 \ s}\\\\\omega =  53.414 \ rad/s

The angular displacement in radian is given as;

θ = ωt

where;

t is time in seconds

θ = (53.414)(3 x 60 x 60)\\

θ = 576,871.2 radians

Therefore, the angular displacement of the blade is 576,871.2 radians

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A battery with an internal resistance ofrand an emf of 10.00 V is connected to a loadresistorR=r. As the battery ages, the inter
yanalaym [24]

Answer:

The current is reduced to half of its original value.

Explanation:

  • Assuming we can apply Ohm's Law to the circuit, as the internal resistance and the load resistor are in series, we can find the current I₁ as follows:

        I_{1} = \frac{V}{R_{int} +r_{L} }

  • where Rint = r and RL = r
  • Replacing these values in I₁, we have:

       I_{1} = \frac{V}{R_{int} +r_{L} } = \frac{V}{2*r} (1)

  • When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:

       I_{2} = \frac{V}{R_{int} +r_{L} } = \frac{V}{(3*r) +r} = \frac{V}{4*r}  (2)

  • We can find the relationship between I₂, and I₁, dividing both sides, as follows:

        \frac{I_{2} }{I_{1} } = \frac{V}{4*r} *\frac{2*r}{V} = \frac{1}{2}

  • The current when the internal resistance triples, is half of the original value, when the internal resistance was r, equal to the resistance of the load.  
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