Answer:
the shooting angle ia 18.4º
Explanation:
For resolution of this exercise we use projectile launch expressions, let's see the scope
R = Vo² sin (2θ) / g
sin 2θ = g R / Vo²
sin 2θ = 9.8 75/35²
2θ = sin⁻¹ (0.6)
θ = 18.4º
To know how for the arrow the tree branch we calculate the height of the arrow at this point
X2 = 75/2 = 37.5 m
We calculate the time to reach this point since the speed is constant on the X axis
X = Vox t
t2 = X2 / Vox = X2 / (Vo cosθ)
t2 = 37.5 / (35 cos 18.4)
t2 = 1.13 s
With this time we calculate the height at this point
Y = Voy t - ½ g t²
Y = 35 sin 18.4 1.13 - ½ 9.8 1,13²
Y = 6.23 m
With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch
Q = mcθ
Where m = mass of water in kg.
c = specific heat capacity in kJ/kg⁰C, c for water = 4200 kJ/kg⁰C
θ = temperature rise in ⁰C
Q = 100*4200* 20 Note here the temperature rise is 20 ⁰C
Q = 8 400 000 J
In calories, 4.2 J = 1 Calorie
= 8 400 000 / 4.2 = 200 000
Q = 200 000 Calories
Answer:
375 m.
Explanation:
From the question,
Work done by the frictional force = Kinetic energy of the object
F×d = 1/2m(v²-u²)..................... Equation 1
Where F = Force of friction, d = distance it slide before coming to rest, m = mass of the object, u = initial speed of the object, v = final speed of the object.
Make d the subject of the equation.
d = 1/2m(v²-u²)/F.................. Equation 2
Given: m = 60.0 kg, v = 0 m/s(coming to rest), u = 25 m/s, F = -50 N.
Note: If is negative because it tends to oppose the motion of the object.
Substitute into equation 2
d = 1/2(60)(0²-25²)/-50
d = 30(-625)/-50
d = -18750/-50
d = 375 m.
Hence the it will slide before coming to rest = 375 m
Answer:
48.16 %
Explanation:
coefficient of restitution = 0.72
let the incoming speed be = u
let the outgoing speed be = v
kinetic energy = 0.5 x mass x 
- incoming kinetic energy = 0.5 x m x
- coefficient of restitution =

0.72 =
v = 0.72u
therefore the outgoing kinetic energy = 0.5 x m x 
outgoing kinetic energy = 0.5 x m x 
outgoing kinetic energy = 0.5184 (0.5 x m x
)
recall that 0.5 x m x
is our incoming kinetic energy, therefore
outgoing kinetic energy = 0.5184 x (incoming kinetic energy)
from the above we can see that the outgoing kinetic energy is 51.84 % of the incoming kinetic energy.
The energy lost would be 100 - 51.84 = 48.16 %