Answer:Central nervous system
Explanation:
A horizontal line on a position-time graph represents a stationary object or, in this case, a parked car. The car is not moving since there's no change in position over time. Speed or velocity would be depicted by the slope of the line on the position-time graph.
Your question pertains to understanding what a horizontal line on a position-time graph indicates about a car's motion. In the context of a position-time graph, a horizontal line suggests that the position of the car is not changing with respect to time, indicating option A: The car is not moving. The car being stationary means there is no change in position over time, hence the horizontal line on the graph.
Options B, C, and D suggest a change in position over time, which would be represented by a slanted line, not a horizontal one. Option B: 'moving at a constant speed' and Option C: 'moving at a constant velocity' would both produce a straight line with a slope, not a horizontal line. Option D: 'speeding up' would be represented by a curved line, showing acceleration.
In contrast, a stationary object or a car at rest is represented by a horizontal line on a position-time graph. Velocity or speed, on the other hand, is determined by the slope of the line on a position-time graph. Hence, a horizontal line, having no slope, signifies zero speed or velocity, indicating that the car is not moving.
#SPJ3
Answer:
27.5 m
Explanation:
Given:
u = 18 km/h = 5 m/s
v = 21.6 km/h = 6 m/s
a = 0.20 m/s²
Find: s
v² = u² + 2as
(6 m/s)² = (5 m/s)² + 2 (0.20 m/s²) s
s = 27.5 m
The image produced is virtual and smaller than the object.
The image produced is real and enlarged.
The image produced is real and smaller than the object.
The image produced is virtual and of the same size as the object.
Answer:
The image produced is virtual and smaller than the object.
Explanation:
For concave lens we know that
here we have
Magnification will be given as
so image will be virtual and formed behind the lens
Now the object position is shifted to new position at distance of focal length
now again we will have
here we have
Magnification will be given as
So again we will have virtual image with magnification 1/2
so here size of image is less than object size by factor of 1/2 and it is virtual
Answer:
Explanation:
given,
mean (μ) = 12.3 Kg
standard deviation (σ ) = 0.1
random sample = 25
probability between 12.25 and 12.35 kg
using z-table