Answer:
The ratio of the man's kinetic energy to that of the woman's kinetic energy is 0.629.
Explanation:
Given;
weight of the man, W = 700 N
Weight of the woman, W = 440 N
momentum is given by;
Kinetic energy of the man;
Momentum of the man is calculated as;
The kinetic energy of the woman is given by;
The momentum of the woman is given;
Since, momentum of the man = momentum of the woman
mass of the mas = 700 / 9.8 = 71.429
mass of the woman is = 440 / 9.8 = 44.898
Therefore, the ratio of the man's kinetic energy to that of the woman's kinetic energy is 0.629.
Aaron's car is moving at speed of 30 m/s
His reaction time is given as 0.7 s
but when he is tired the reaction time is doubled
Now we need to find the distance covered by his car when he is tired during the time when he react to apply brakes
So here since during this time speed is given as constant so we can say that distance covered can be product of speed and time
So here we can use
So the car will move to 42 m during the time when he apply brakes
Answer:
d) -4.0
Explanation:
The magnification of a lens is given by
where
M is the magnification
q is the distance of the image from the lens
p is the distance of the object from the lens
In this problem, we have
p = 50 cm is the distance of the object from the lens
q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct
Also, q is positive since the image is real
So, the magnification is
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45 )
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = - 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct