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aleksklad [387]
3 years ago
11

Sarah and her friends the swim team are playing on the high dive. One of Sarah's friends throws her back pack, containing her sw

im supplies, into the pool from the high dive platform. The back pack weighs 50 N. When the back pack hits the water, it encounters a 50.0 N upward resistance force. The swim coach, who also happens to be the physics teacher says: "This is a great opportunity to discuss the forces exerted on Sarah's back pack!"
Several students in the group made comments to the Coach about the back pack. Which student has the correct analysis of the forces and motion of the back pack?


A) Once the back pack hits the water, the forces are balanced and the back pack will stop moving.
B) The upward force of the water on the back pack is less than the downward pull of gravity. The back pack accelerates until it reaches the bottom of the pool.
C) Once the back pack hits the water, the forces are balanced (50 N down and 50 N up). But when the back pack hits the water, it stops accelerating; it does not stop moving.
Eliminate
D) Upon hitting the water, the back pack will accelerate upwards because the water applies an upward force. The forces are unbalanced; the upward force is greater than gravity.
Physics
1 answer:
katovenus [111]3 years ago
8 0

Answer: The answer is C:Once the back pack hits the water, the forces are balanced (50 N down and 50 N up). But when the back pack hits the water, it stops accelerating; it does not stop moving.

Explanation:

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b) The size of the image is -2 cm (inverted)

Explanation:

a)

We can solve this first part of the problem by applying the mirror equation:

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

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

For a mirror, the focal length is half the radius of curvature, R:

f=\frac{R}{2}

For this mirror, R = 20 cm, so its focal length is

f=\frac{20}{2}=+10 cm (positive for a concave mirror)

Here we also know:

p = 30 cm is the distance of the object from the mirror

So, by applying the equation, we can find q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15} \rightarrow q = 15 cm

b)

We can solve this part by using the magnification equation:

M=-\frac{y'}{y}=\frac{q}{p}

where

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

Here we have:

q = 15 cm

p = 30 cm

y = 4 cm

Solving for y', we find the size of the image:

y'=-y\frac{q}{p}=-(4)\frac{15}{30}=-2 cm

and the negative sign means that the image is inverted.

#LearnwithBrainly

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Explanation:

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