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MariettaO [177]
2 years ago
12

Use the diagram below to answer the following question:

Physics
1 answer:
d1i1m1o1n [39]2 years ago
5 0

Answer:

3.0 cm

Explanation:

We can solve this problem by using the mirror equation:

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

where

f is the focal length of the mirror

p is the distance of the object from the mirror

q is the distance of the image from the mirror

In this problem we have:

f = 1.5 cm is the focal length of the mirror (positive for a concave mirror)

p = 3.0 cm is the distance of the object from the mirror

Therefore, the distance of the image is:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{1.5}-\frac{1}{3.0}=\frac{1}{3.0}\\\rightarrow q=3.0 cm

And the positive sign means that the image is real.

(The second part of the exercise is just the description of the image of the first exercise).

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a 77 N net force is applied to a box which slides horizontally across a floor for 6.7 m. What amount of work is done on the box
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W=fd
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"When Baby Ian is held horizontally on his back, he sometimes makes an embracing motion by arching his back, extending his legs,
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Answer:

Moro reflex  

Explanation:

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Lab: newton's laws of motion assignment: lab report
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Newton's motion laws state that if an object is at rest or in movement, it will tend to maintain its basal state.

<h3>What are Newton's motion laws?</h3>

Newton's motion laws are a set of scientific statements aimed at explaining the physical property of movement.

These laws explain why objects in movement tend to maintain the same velocity for a short period of time.

In conclusion, Newton's motion laws state that if an object is at rest or in movement, it will tend to maintain its basal state.

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8 0
1 year ago
Projectile's horizontal range on level ground is R=v20sin2θ/g. At what launch angle or angles will the projectile land at half o
seraphim [82]

Answer:

\theta = 15^o \: or\: 75^o

Explanation:

As we know that the formula of range is given as

R = \frac{v^2sin2\theta}{g}

now we know that

maximum value of the range of the projectile is given as

R_{max} = \frac{v^2}{g}

now we need to find such angles for which the range is half the maximum value

so we will have

\frac{R}{2} = \frac{v^2}{2g} = \frac{v^2sin(2\theta)}{g}

sin(2\theta) = \frac{1}{2}

2\theta = 30 or 150

\theta = 15^o \: or\: 75^o

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What factors might affect how fast a balloon falls to the ground?
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2 years ago
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