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
 
        
             
        
        
        
At t =0, the velocity of A is greater than the velocity of B.
We are told in the question that the spacecrafts fly parallel to each other and that for the both  spacecrafts, the velocities are described as follows;
A: vA (t) = ť^2 – 5t + 20 
B: vB (t) = t^2+ 3t + 10
Given that t = 0 in both cases;
 vA (0) = 0^2 – 5(0) + 20 
vA = 20 m/s
For vB
 vB (0) = 0^2+ 3(0) + 10
vB = 10 m/s
We can see that at t =0, the velocity of A is greater than the velocity of B.
Learn more: brainly.com/question/24857760
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?
 
        
             
        
        
        
This would be typical of an elastic collision.
        
                    
             
        
        
        
Answer:
m = 35 g 
Explanation:
The specific heat of a material can be calculated by the following formula:

where,
C = Specific Heat of Wax = 220 J/g
Q = Amount of Heat Supplied by the Heater = 7700 J
m = mass of wax melted = ?
Therefore,

<u>m = 35 g </u>