<span>The image formed is larger than the object, right-side up, and virtual.
hope this helps.
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When an object moves its length contracts in the direction of motion. The faster it moves the shorter it gets in the direction of motion.
The object in this question moves and then stops moving. So it's length first contracts and then expands to its original length when the motion stops.
The speed doesn't have to be anywhere near the speed of light. When the object moves its length contracts no matter how fast or slow it's moving.
Because if weight is being applied to an object, the object will have some sort of effect to counter react due to the weights primary objective.
Here I =0.4 amp and t=2hr or 2×60×60 sec
We known I=nQ/t where Q is charge and n is no of charge.
n=It/Q
0.4x(2x60x60)/1.6x10^-19
Answer:
a = -8.912 m/s²
Explanation:
Given,
The initial velocity of the car, u = 28 m/s
The final velocity of the car, v = 0
The distance traveled by car, d = 88 m
The velocity displacement relation is given by the formula
v = d/t
∴ t = d/v
Substituting in the above values in the given equation
t = 88/28
= 3.142 s
The acceleration is given by the formula
a = (v-u)/t
= (0 - 28)/3.142
= -8.912 m/s²
The negative sign is that the car is decelerating.
Hence, acceleration a = -8.912 m/s²
