268.6567 mph is its velocity when it crosses the finish line
d=(v1+v2 /2) x t
.25=(0+v2 /2) x 6.7/3600 hours
900=v2/2 x 6.7
v2=268.6567 mph as the speed with which the dragster crosses the finish
<h3>When acceleration is not zero, can speed remain constant?</h3>
The answer is that an accelerated motion can have a constant speed. Consider a particle travelling uniformly around a circle; it experiences acceleration since the motion's direction is changing, but it maintains a constant speed along the tangential axis throughout the motion.
Acceleration is the frequency of a change in velocity. Acceleration is a vector with magnitude and direction, much as velocity. For instance, if a car is moving in a straight path and speeding up, it is said to have forward (positive) acceleration, and if it is slowing down, it is said to have backward (negative) acceleration.
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<span>By algebra, d = [(v_f^2) - (v_i^2)]/2a.
Thus, d = [(0^2)-(15^2)]/(2*-7)
d = [0-(225)]/(-14)
d = 225/14
d = 16.0714 m
With 2 significant figures in the problem, the car travels 16 meters during deceleration.</span>
Answer:
10259.6 m
Explanation:
We are given that
Radius of small wheel,r=0.17 m
Radius of large wheel,r'=0.92 m
Initial velocity,u=0
Time,t=2.7 minutes=162 s
1 min=60 s
Velocity,v=10m/s
Time,t'=13.7 minutes=822 s
Time,t''=4.1 minutes=246 s

Substitute the values



Substitute the values




Total distance traveled by rider=s+s'+s''=809.6+8220+1230=10259.6 m
D = 1/f, where D is the power in diopters and f is the focal length in meters.
D=1/20
<u>D=0.05</u>
Given: Mass m = 44 Kg; Velocity v = 10 m/s
Required: Kinetic energy K.E = ?
Formula: K.E = 1/2 mv²
K.E 1/2 (44 Kg)(10 m/s)²
K.E = 2,200 Kg.m²/s²
K.E = 2,200 J Answer is A