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:
x-component=-9.3 m
Magnitude of A=17.7m
Explanation:
We are given that
We have to find the x-component of A and magnitude of A.
According to question
Substitute the values then we get
m
The value of x-component of A is negative because the vector A lie in second quadrant.
Hence, the x- component of A=-9.3 m
1)Kenetic Energy is defined as energy which a body possesses by virtue of being in motion. 2)KE) is KE = 0.5 x mv2. Here m stands for mass, the measure of how much matter is in an object, and v stands for the velocity of the object, or the rate at which the object changes its position..
And I hope this helped :)
Answer:
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The bearing could be the below:
oppositely charged, same initial direction
same charge, opposite initial direction
You can decide by utilizing your correct hand and put your fingers toward the attractive field (North to South). Thumb toward present or charged molecule. The course of your palm will demonstrate the heading of compelling set on a decidedly charged molecule and the bearing of the back of your hand will demonstrate the bearing of a contrarily charged molecule.
