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Tasya [4]
3 years ago
10

Which image is observe by seeing a distant thing using magnifying glass?

Physics
2 answers:
Snezhnost [94]3 years ago
8 0

Answer: D) Big - straight

<h2>Brainliest pls if correct!</h2>

lesya692 [45]3 years ago
7 0

Answer:

1/f = 1/o + 1 /i       thin lens equation

1 / i = 1/f - 1/o = (o - f) / (o f)    rearranging

or i = o f / (o - f)   for the image distance

This image will be positive (real) with values given

m = -i / o      magnification

m = - f / (o - f)      shows m is negative and the image inverted

Also, since o here is large m will be small and the image is small

I think they mean A as the right answer

Different texts use different terms but here i is image distance and o the object distance

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A golf ball traveling 3m/s to the right collides in a head-on collision with a stationary bowling ball in a friction-free enviro
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Answer: equal to 3 m/s

Explanation:

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Initial Kinetic Energy =Final Kinetic Energy

Considering Bowling ball remains at rest after collision other wise some energy will also be acquired by bowling ball which automatically decreases the amount of Kinetic Energy of golf ball resulting its speed to decrease by some extent.  

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A 0.25 kg hockey puck is sliding along the ice with an initial velocity of 5 m/s when a player strikes it with his stick, causin
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5 0
3 years ago
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12 N to the right (based on your POV)
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6 0
3 years ago
The magnitude of the velocity of a projectile when it is at its maximum height above ground level is 14 m/s. (a) What is the mag
Alecsey [184]

When the projectile is at its maximum height above ground, it's at the point
of changing from rising to falling.  At that exact point, its vertical speed is zero,
so the 14 m/s must be all horizontal velocity.  That's not going to change.

Since we need to consider changes in vertical speed now, we need to make
some assumption about where this is all happening, so that we know the
acceleration of gravity.  I'll assume that it's all happening on or near the Earth,
and the acceleration of gravity is 9.8 m/s².

I'm also going to neglect air resistance.

a). 1.2 sec before it reaches its maximum height, the projectile is rising
at a vertical speed of (1.2 x 9.8) = 11.76 m/s. 
The magnitude of its velocity is

the square root of (14² + 11.76²) = 18.28 m/s, directed about 40° above horizontal.

b).  1.2 sec after it reaches its maximum height, the projectile is falling
at a vertical speed of (1.2 x 9.8) = 11.76 m/s. 
The magnitude of its velocity is

the square root of (14² + 11.76²) = 18.28 m/s, directed about 40° below horizontal.

===========================

In 1.2 second before or after zero vertical speed, an object in free fall moves

         (1/2) (g) (t²) = (4.9) (1.2²) = 7.06 meters .

c). & d).
1.2 seconds before it reaches maximum height, the projectile is located at

           x = -14 m
           y = -7.06 m

e). & f).
1.2 seconds after it reaches maximum height, the projectile is located at

           x = +14 m
           y = -7.06 m .


I hope you recognize that 6 answers, plus a little bit of explanation,
all for 5 points, ain't too shabby.  You made out well.

5 0
3 years ago
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