Answer : The correct option is (d) 2.73 m
Explanation :
By the 2nd equation of motion,
where,
s = distance or height = ?
u = initial velocity = 3.0 m/s
t = time = 0.5 s
a = acceleration due to gravity = 
Now put all the given values in the above equation, we get:
Therefore, the correct option is (d) 2.73 m
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.
Learn more about velocity refer
brainly.com/question/24681896
#SPJ9
Answer:
F= 403429 kpa
Explanation:
Pressure is the product of force and area
Mathematically,
P=F*A -------where F is force and A is area.
A= 40 *0.1 = 4mm² -----convert to m²
A= 4e⁻⁶ m²
P= 4000000 pa
F= P/A = 4000000/4e⁻⁶
F= 403428793.493 pa
F= 403429 kpa
Answer:
Explanation:
cSep 20, 2010
well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.
Anonymous
Sep 20, 2010
First you need to solve for time by using
d=(1/2)(a)(t^2)+(vi)t
1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s
t=.45 sec
Then you find the horizontal distance traveled by using
v=d/t
1.3m/s=d/.54sec
d=.585m
Then you need to find the time of player B by using
d=(1/2)(a)(t^2)+(vi)t
1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0
t=.61 sec
Finally to find player Bs initial horizontal velocity you use the horizontal equation
v=d/t
v=.585m/.61 sec
so v=.959m/s
