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Murljashka [212]
3 years ago
14

A 65.0-kg woman steps off a 11.0-m diving platform and drops straight down into the water. 1) If she reaches a depth of 4.60 m,

what is the average resistance force exerted on her by the water
Physics
1 answer:
Ganezh [65]3 years ago
8 0

Answer:

2160.26 N.

Explanation:

From the question,

The total energy of the woman = work done by the water.

mg(h+d) = F'×d.......................... Equation 1

m = mass of the the woman, g= acceleration due to gravity, h = height of the platform, d = depth of water reached by the woman, F'= Resistance force exerted on her by the water.

make F' the subject of the equation

F' = mg(h+d)/d............................ Equation 2

Given: m = 65.0 kg, g = 9.8 m/s², h = 11 m, d = 4.6 m.

Substitute into equation 2

F' = 65(9.8)(11+4.6)/4.6

F' = 9937.2/4.6

F' = 2160.26 N.

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A heater has a resistance of 10.0 Ω. It operates on a 12.0 V. What is the current through the resistor?
irakobra [83]

Answer:

1.2 amps :)

Explanation:

A heater has a resistance of 10.0 Ω. It operates on a 12.0 V. What is the current through the resistor?

         Known:

  • R = 10.0 Ω                            
  • V = 12.0 V

       Unknown:

  • I = ???

I = V/R

= 12.0 V / 10.0 Ω

= 1.2 amps

6 0
2 years ago
A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the
andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and  how far away from the overpass is the roadrunner when the coyote drops the net.

d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

5 0
2 years ago
A car accelerates from rest at 3.6 m/s 2 . How much time does it need to attain a speed of 5 m/s?
Olenka [21]

car starts from rest

v_i = 0

final speed attained by the car is

v_f = 5 m/s

acceleration of the car will be

a = 3.6 m/s^2

now the time to reach this final speed will be

t = \frac{v_f - v_i}{a}

t = \frac{5 - 0}{3.6}

t = 1.39 s

so it required 1.39 s to reach this final speed

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3 years ago
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Answer:

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5 0
2 years ago
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