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fomenos
3 years ago
15

TAKE 68 Points!!!please Help Me.Find the cut and slope just look the picture.​

Physics
1 answer:
zavuch27 [327]3 years ago
6 0

Answer:i think it is 5.00x10^2              +68.0

Explanation:

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A convex mirror, like the passenger-side rearview mirror on a car, has a focal length of -3.0 m . An object is 6.0 m from the mi
Artist 52 [7]

A) -2.0 m

Look at the ray diagram attached in the picture, where:

p identifies the location of the object

q identifies the location of the image

F identifies the focus of the mirror

Each tick represents 1 m

We have

p = 6.0 m is the distance of the object from the mirror

f = -3.0 m is the focal length

From the ray diagram, we see that q has a distance of 2.0 m from the mirror, and it's on the other side of the mirror compared to the object, so

q = -2.0 m

This can also be verified by using the mirror equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-3.0 m}-\frac{1}{6.0 m}=-\frac{3}{6.0 cm}\\q = \frac{-6.0 cm}{3}=-2.0 cm

B) Upright and virtual

As we see from the picture, the image is upright, since it has same orientation as the object.

Also, we notice that the image is on the other side of the mirror, compared to the object. For a mirror,

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- An image is said to be virtual if it is on the opposite side of the mirror

Therefore, this means that the image is virtual.

8 0
3 years ago
What is the theoretical volume of 10mL of an ideal gas at absolute zero?
irinina [24]
PV=nRT
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(v1/t1)=(v2/t2)
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4 years ago
Where can volcanoes form? A. Where there are cracks in the crust B. Along fault lines C. Where the crust is thin and can be rupt
schepotkina [342]
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8 0
4 years ago
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A 72.9-kg base runner begins his slide into second base when moving at a speed of 4.02 m/s. The coefficient of friction between
elena-14-01-66 [18.8K]

Answer:

-589.05 J

Explanation:

Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner

So, W = ΔK

W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)

So, substituting the values of the variables into the equation, we have

W = 1/2m(v₁² - v₀²)

W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)

W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)

W = 1/2 × 72.9 kg(-16.1604 m²/s²)

W = 1/2 × (-1178.09316 kgm²/s²)

W = -589.04658 kgm²/s²

W = -589.047 J

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3 years ago
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your answer is phenomena which are too large


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