Answer:
34 m/s
Explanation:
Potential energy at top = kinetic energy at bottom + work done by friction
PE = KE + W
mgh = ½ mv² + Fd
mg (d sin θ) = ½ mv² + Fd
Solving for v:
½ mv² = mg (d sin θ) − Fd
mv² = 2mg (d sin θ) − 2Fd
v² = 2g (d sin θ) − 2Fd/m
v = √(2g (d sin θ) − 2Fd/m)
Given g = 9.8 m/s², d = 150 m, θ = 28°, F = 50 N, and m = 65 kg:
v = √(2 (9.8 m/s²) (150 m sin 28°) − 2 (50 N) (150 m) / (65 kg))
v = 33.9 m/s
Rounded to two significant figures, her velocity at the bottom of the hill is 34 m/s.
Answer:
26.8 seconds
Explanation:
To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:
v = final velocity
z = initial velocity
x = distance
t = time
a = acceleration


First let's find the final velocity the plane will have at the end of the runway using the first equation:


Now we can plug this into the second equation to find t:


Then using 3 significant figures we round to 26.8 seconds
Answer:

Explanation:
Given:
- mass of water,

- initial temperature of water,

- final temperature of water,

- specific heat of water,

<u>Now the amount of heat energy required:</u>



Since all of the mechanical energy is being converted into heat, therefore the same amount of mechanical energy is required.
Assuming that the angle is the same for both ropes, then D. is the answer. You have to consider also if the ropes are close together or far apart and if the force to move the object is in line with the ropes or perpendicular to them.
<span />