Answer:
No, i disagree.
Explanation:
If the car is moving, it only has a velocity with a component in the horizontal direction. If we use galilean relativity, the velocity of the ball observed by my friend standing in the ground should only be affected in the horizonal direction, while the vertical stays the same for both observers.
Answer:
a) v₃ = 19.54 km, b) 70.2º north-west
Explanation:
This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition
vector 1 moves 26 km northeast
let's use trigonometry to find its components
cos 45 = x₁ / V₁
sin 45 = y₁ / V₁
x₁ = v₁ cos 45
y₁ = v₁ sin 45
x₁ = 26 cos 45
y₁ = 26 sin 45
x₁ = 18.38 km
y₁ = 18.38 km
Vector 2 moves 45 km north
y₂ = 45 km
Unknown 3 vector
x3 =?
y3 =?
Vector Resulting 70 km north of the starting point
R_y = 70 km
we make the sum on each axis
X axis
Rₓ = x₁ + x₃
x₃ = Rₓ -x₁
x₃ = 0 - 18.38
x₃ = -18.38 km
Y Axis
R_y = y₁ + y₂ + y₃
y₃ = R_y - y₁ -y₂
y₃ = 70 -18.38 - 45
y₃ = 6.62 km
the vector of the third leg of the journey is
v₃ = (-18.38 i ^ +6.62 j^ ) km
let's use the Pythagorean theorem to find the length
v₃ = √ (18.38² + 6.62²)
v₃ = 19.54 km
to find the angle let's use trigonometry
tan θ = y₃ / x₃
θ = tan⁻¹ (y₃ / x₃)
θ = tan⁻¹ (6.62 / (- 18.38))
θ = -19.8º
with respect to the x axis, if we measure this angle from the positive side of the x axis it is
θ’= 180 -19.8
θ’= 160.19º
I mean the address is
θ’’ = 90-19.8
θ = 70.2º
70.2º north-west
The mass of the puck is
m = 0.15 kg.
The diameter of the puck is 0.076 m, therefore its radius is
r = 0.076/2 = 0.038 m
The sliding speed is
v = 0.5 m/s
The angular velocity is
ω = 8.4 rad/s
The rotational moment of inertia of the puck is
I = (mr²)/2
= 0.5*(0.15 kg)*(0.038 m)²
= 1.083 x 10⁻⁴ kg-m²
The kinetic energy of the puck is the sum of the translational and rotational kinetic energy.
The translational KE is
KE₁ = (1/2)*m*v²
= 0.5*(0.15 kg)*(0.5 m/s)²
= 0.0187 j
The rotational KE is
KE₂ = (1/2)*I*ω²
= 0.5*(1.083 x 10⁻⁴ kg-m²)*(8.4 rad/s)²
= 0.0038 J
The total KE is
KE = 0.0187 + 0.0038 = 0.0226 J
Answer: 0.0226 J
Remember your kinematic equations for constant acceleration. One of the equations is

, where

= final position,

= initial position,

= initial velocity, t = time, and a = acceleration.
Your initial position is where you initially were before you braked. That means

= 100m. You final position is where you ended up after t seconds passed, so

= 350m. The time it took you to go from 100m to 350m was t = 8.3s. You initial velocity at the initial position before you braked was

= 60.0 m/s. Knowing these values, plug them into the equation and solve for a, your acceleration:
Your acceleration is approximately
.
Emily throws the ball at 30 degree below the horizontal
so here the speed is 14 m/s and hence we will find its horizontal and vertical components


vertical distance between them

now we will use kinematics in order to find the time taken by the ball to reach at Allison

here acceleration is due to gravity

now we will have

now solving above quadratic equation we have

now in order to find the horizontal distance where ball will fall is given as

here it shows that horizontal motion is uniform motion and it is not accelerated so we can use distance = speed * time

so the distance at which Allison is standing to catch the ball will be 5.33 m