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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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Answer:

40 km/h

Explanation:

First...

Look at the formula speed is equal to the distance over time or s = d/t.

Next...

Use the formula: 240/6.0

Finally...

Solve: 40

So the answer: 40 km/h

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Which conclusion, based on measurements of the ball in the lab in this lesson, is correct? a. The density of the ball was 1.48 g
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I need help with this someone help me?
Brilliant_brown [7]
Graph A so answer B also why isn’t answer a with graph A and B with graph B etc like that’s just confusing lol
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3 years ago
An object is 15 cm in front of a diverging lens with a
Rainbow [258]

A) See ray diagram in attachment (-6.0 cm)

By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f = -10 cm is the focal length (negative for a diverging lens)

p = 15 cm is the distance of the object from the lens

Solving for q,

\frac{1}{q}=\frac{1}{-10 cm}-\frac{1}{15 cm}=-0.167 cm^{-1}

q=\frac{1}{-0.167 cm^{-1}}=-6.0 cm

B) The image is upright

As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:

h_i = - h_o \frac{q}{p}

where h_i, h_o are the size of the image and of the object, respectively.

Since q < 0 and p > o, we have that h_i >0, which means that the image is upright.

C) The image is virtual

As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.

This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual

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What force must the worker exert to get the box moving &amp; what force must the worker exert to accelerate the box at 0.1 meter
photoshop1234 [79]

Since static friction is the minimum force required to just start the motion of a stationary object.

Here if we need to start an object from rest then we required F = 700 N

So for the first part of the above problem Force will be F = 700 N

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now we can write the equation of net force as

F - F_k = m*a

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F_k = kinetic friction = 220 N

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now we will have

F - 220 = 500* 0.1

F = 220 + 50 = 270 N

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