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Nataliya [291]
2 years ago
13

Dimas reads that it takes 270 N to lift a firefighter's equipment. If a firefighter does 5400 J of work carrying the equipment u

p a staircase, how high does the firefighter climb? (Work: W = Fd) 20 m 200 m 5130 m 1,458,000 m
Physics
2 answers:
Alex787 [66]2 years ago
7 0

Answer:

h = 20 m

Explanation:

Given that,

Force to lift a firefighter's equipment, F = 270 N

Work done, W = 5400 J

We need to find the height climbed by the firefighter. We know that,

Work done, W = Fd

So,

d=\dfrac{W}{F}\\\\d=\dfrac{5400}{270}\\\\d=20\ m

So, the firefighter climb from a height of 20 m.

hodyreva [135]2 years ago
7 0

Answer:

Its A

Explanation: I got it right on Edge *lip bite*

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Un coche inicia un viaje de 450 km a las ocho de la mañana con una velocidad media de 90 km/h. ¿A qué hora llegará a su destino?
Artyom0805 [142]

Answer:

Llegara a su destino a la 1:00 pm

Explanation:

Si el coche va a 90 km/h buscamos un numero q al multiplicarlo por 90 nos de 450. Entonces 90×5 = 450, si hacemos la cuenta desde las ocho de la mañana mas las 5 horas del viaje terminaria llegando a su destino a la 1:00 pm.

5 0
2 years ago
:What will be the value of the refractive index of the medium? Critical angle of that medium is 30 degree
earnstyle [38]

Answer:

Let the second medium be air (n₁=1)

The refractive index n₂ of the medium where first medium is air is found (a)

(a) n₂ = 2

Explanation:

Critical angle can be defined as the angle of incidence that provides the angle of refraction of 90°.

Refractive index of a medium can be defined as a number that describes that how fast a light will travel through that medium.

Critical angle and Refractive index are related by:

\theta_{critical}= sin^{-1}(\frac{n_1}{n_2})

sin \theta_{critical}=\frac{n_1}{n_2}

To find refractive index of medium with respect to air, substitute n₁=1 (Refractive index of air is 1)

Also θ(critical)=30°

Find n₂ :

sin30= \frac{1}{n_2}\\0.5=\frac{1}{n_2}\\n_2=\frac{1}{0.5}\\n_2=2

8 0
2 years ago
A stuntman with a mass of 80.5 kg swings across a moat from a rope that is 11.5 m. At the bottom of the swing the stuntman's spe
goldenfox [79]

Answer:

  • No
  • 5.49 m/s

Explanation:

The net force required to accelerate the stuntman in a circular arc of radius 11.5 m will be ...

  F = mv²/r . . . . where this m is the mass being accelerated, v is the tangential velocity, and r is the radius.

Here, the net force needs to be ...

  F = (80.5 kg)(8.45 m/s)²/(11.5 m) . . . . . where this m is meters

  ≈ 499.8175 kg·m/s² = 499.8 N

Gravity exerts a force on the stuntman of ...

  F = mg = (80.5 kg)(9.8 m/s²) = 788.9 kg·m/s² = 788.9 N

Then the tension required in the rope/vine is ...

  499.8 N+788.9 N= 1288.7 N

This is more than the capacity of the rope, so we do not expect the stuntman to make it across the moat.

_____

The allowed net force for centripetal acceleration is ...

  1000 N -788.9 N = 211.1 N

Then the allowed velocity is ...

  211.1 = 80.5v²/11.5

  30.16 = v² . . . .  multiply by 11.5/80.5

  5.49 = v . . . . . . take the square root

The maximum speed the stuntman can have is 5.49 m/s.

_____

<em>Comment on crossing the moat</em>

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

5 0
3 years ago
What is one positive outcome of the recent<br>events?​
fenix001 [56]

Answer:

Analyze the positive and negative consequences of catastrophic events of the last 40-50 years and the individuals that had an impact on these events. Meaning.

Explanation:

5 0
3 years ago
Read 2 more answers
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WINSTONCH [101]
Check the current weather map for 2 different times, and see where the center of the storm is. That tells you what direction it's moving. With its speed and direction, you have its velocity.
6 0
3 years ago
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