Answer:
The following factors could be responsible for size of an Impact Crater:
- Geography of the region where it impacted.
- Size of the Impactor
- Velocity of the Impactor
Hope it helps!<3
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
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Answer:
1) d
2) 5 m/s
3) 100
Explanation:
The equation of position x for a constant acceleration a and an initial velocity v₀, initial position x₀, time t is:
(i) 
The equation for velocity v and a constant acceleration a is:
(ii) 
1) Solve equation (ii) for acceleration a and plug the result in equation (i)
(iii) 
(iv) 
Simplify equation (iv) and use the given values v = 0, x₀ = 0:
(v) 
2) Given v₀= 3m/s, a=0.2m/s², t=10 s. Using equation (ii) to get the final velocity v:
3) Given v₀=0m/s, t₁=10s, t₂=1s and x₀=0. Looking for factor f = x(t₁)/x(t₂) using equation(i) to calculate x(t₁) and x(t₂):
Answer:
a. 8.96 m/s b. 1.81 m
Explanation:
Here is the complete question.
a) A long jumper leaves the ground at 45° above the horizontal and lands 8.2 m away.
What is her "takeoff" speed v
0
?
b) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 m away horizontally and 2.5 m, vertically below.
If she long jumps from the edge of the left bank at 45° with the speed calculated in part a), how long, or short, of the opposite bank will she land?
a. Since she lands 8.2 m away and leaves at an angle of 45 above the horizontal, this is a case of projectile motion. We calculate the takeoff speed v₀ from R = v₀²sin2θ/g. where R = range = 8.2 m.
So, v₀ = √gR/sin2θ = √9.8 × 8.2/sin(2×45) = √80.36/sin90 = √80.36 = 8.96 m/s.
b. We use R = v₀²sin2θ/g to calculate how long or short of the opposite bank she will land. With v₀ = 8.96 m/s and θ = 45
R = 8.96²sin(2 × 45)/9.8 = 80.2816/9.8 = 8.192 m.
So she land 8.192 m away from her bank. The distance away from the opposite bank she lands is 10 - 8.192 m = 1.808 m ≅ 1.81 m
Answer:
The answer would be Igneous rock
Explanation:
