Please help!!!!!!!!!!!!!
Answers
Answer:the answer is d
Explanation:
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Answers
Answer:amplitude
Explanation:
2. An object with a height of 0.3 meter is placed at a distance of 0.4 meter from a concave
spherical mirror. An image with a height of 0.1 meter is formed in front of the mirror.
How far from the mirror is the image located?
3. When an object with a height of 0.10 meter is placed at a distance of 0.20 meter from a
convex spherical mirror, the image will appear to be 0.06 meter behind the mirror.
What's the height of the image?
4. Compare and contrast the properties of the images formed by each mirror type in the
table.
Answers
Answer:
1. 12 cm
2. 0.133 m
3. 0.03 m
4. Plane mirror
Virtual image
Upright
Behind the mirror
The same size as the object
Concave mirror when the object is located a distance greater than the focal length from the mirror's surface
Real image
Inverted image
In front of the the mirror
Diminished when the object is beyond the center of curvature
Same size as object when the object is placed at the center of curvature
Enlarged when the object is placed between the center of curvature of the mirror and the focus of the mirror
Concave mirror when the object is located a distance less than the focal length from the mirror's surface
Virtual image
Upright image
Behind the the mirror
Enlarged
Convex mirror
Type = Virtual image
Appearance = Upright image
Placement = Behind the mirror
Size = Smaller than the object
Explanation:
1. For plane mirror, since there is no magnification, the virtual image distance from the mirror = object distance from the mirror = 12 cm behind the mirror
2. The height of the object = 0.3 m
The distance of the object from the mirror = 0.4 meters
Height of image formed = 0.1 meter
We have;
Image distance from the mirror = 0.1/0.3×0.4 = 2/15 = 0.133 m
Image distance from the mirror = 0.133 m
3.
The image height = 0.06/0.2×0.1 = 3/100 = 0.03 meter
The image height = 0.03 meter
4. Plane mirror
Type = Virtual image
Appearance = Upright image with the left transformed to right
Placement = Behind the mirror
Size = The same size as the object
Concave mirror when the object is located a distance greater than the focal length from the mirror's surface
Type = Real image
Appearance = Inverted image
Placement = In front of the the mirror
Size = Diminished when the object is beyond the center of curvature
Same size as object when the object is placed at the center of curvature
Enlarged when the object is placed between the center of curvature of the mirror and the focus of the mirror
Concave mirror when the object is located a distance less than the focal length from the mirror's surface
Type = Virtual image
Appearance = Upright image
Placement = Behind the the mirror
Size = Enlarged
Convex mirror
Type = Virtual image
Appearance = Upright image
Placement = Behind the mirror
Size = Smaller than the object.
Answer:
1. The mirror is 12 centimeters away from the image. This is a plane mirror with a flat reflecting surface. The distance between the object and the mirror surface is equal to the distance between the mirror surface and the image.
2. hiho=siso
0.1 m0.3 m=si0.4 m
Multiply each side of this equation by 0.4.
0.4×(0.10.3=si0.4)×0.4
si=0.40.3
si = 0.133 m
3. hiho=siso
hi0.10 m=0.06 m0.02 m
Multiply each side of this equation by 0.10.
0.10×(hi0.10=0.060.20)×0.10
hi=0.0060.20
hi = 0.03 m
4.
Image Formation
Mirror Type Appearance Placement Size
Plane Virtual Erect (Upright); Appears to have left and right reversed Behind the mirror; the distance between the mirror and the image is equal to the distance between the mirror and the object Depends on the size of the mirror and placement of the object
Concave (when the object is located a distance greater than a focal length from mirror's surface) Real Inverted In front of the mirror Smaller than the object
Concave (when object is located a distance less than the focal length of the mirror) Virtual Erect (Upright) Behind the mirror Enlarged
Convex Virtual Erect (Upright) Behind the mirror Smaller than the object
Explanation:
PENN
b. it has mass
c. it is changing position
d. it has a reference point
Answers
because an object need to be in motion to make any kind of movement
The acceleration of gravity is 9.81 m/s?.
If the upward acceleration of the bucket is
2.8 m/s², find the force exerted by the rope
on the bucket of water.
1. 69.658
2. 52.84
3. 85.318
4. 70.86
5. 84.429
Answers
Explanation:
Draw a free body diagram of the bucket. There are two forces:
Tension force T pulling up.
Weight force mg pulling down.
Sum of forces in the y direction:
∑F = ma
T − mg = ma
T = mg + ma
T = m (g + a)
Given m = 4.0 kg, g = 9.81 m/s², and a = 2.8 m/s²:
T = (4.0 kg) (9.81 m/s² + 2.8 m/s²)
T = 50.44 N
Answers
Answers
Answer:
F=10.8N
Explanation:
In the picture above.
Hope this helps.