Answer:
4.11 m/s
Explanation:
Given:
v₀ₓ = 1.25 m/s
aₓ = 0 m/s²
v₀ᵧ = 0 m/s
aᵧ = -9.8 m/s²
t = 0.4 s
Find: v
First, in the x direction:
vₓ = aₓt + v₀ₓ
vₓ = (0 m/s²) (0.4 s) + 1.25 m/s
vₓ = 1.25 m/s
And in the y direction:
vᵧ = aᵧt + v₀ᵧ
vᵧ = (-9.8 m/s²) (0.4 s) + 0 m/s
vᵧ = -3.92 m/s
The net speed is found with Pythagorean theorem:
v² = vₓ² + vᵧ²
v² = (1.25 m/s)² + (-3.92 m/s)²
v = 4.11 m/s
Answer:
<h3>Newton's 2nd law states acceleration is proportional to the net force acting on an object. The net force is the vector sum of all the forces applied to the object. ... In this case the acceleration (slowing down) of the puck is proportional to the amount of friction.</h3>
Explanation:
<h3>mark as brainliast</h3>
Answer:
421.83 m.
Explanation:
The following data were obtained from the question:
Height (h) = 396.9 m
Initial velocity (u) = 46.87 m/s
Horizontal distance (s) =...?
First, we shall determine the time taken for the ball to get to the ground.
This can be calculated by doing the following:
t = √(2h/g)
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 396.9 m
Time (t) =.?
t = √(2h/g)
t = √(2 x 396.9 / 9.8)
t = √81
t = 9 secs.
Therefore, it took 9 secs fir the ball to get to the ground.
Finally, we shall determine the horizontal distance travelled by the ball as illustrated below:
Time (t) = 9 secs.
Initial velocity (u) = 46.87 m/s
Horizontal distance (s) =...?
s = ut
s = 46.87 x 9
s = 421.83 m
Therefore, the horizontal distance travelled by the ball is 421.83 m
Answer:
\alpha = \frac{2F}{3m} \ \frac{1}{r}
maximmun x =r
Explanation:
In this exercise we are asked for the maximum angular acceleration, let's start by writing the second law of / newton for rotational motion. Let's fix our reference system at the midpoint of the bar that has a length 2r
Σ τ = (I₁ + I₂) α
where I₁ and I₂ moment of inertia of the capsule with masses m and 2m, respectively. Let's treat these capsules as point particles
I₁ = m r²
I₂ = 2m r²
the troque of a pair of force is the force times the distance perpendicular to the point of application of the force which is the same for both forces, we will assume that the counterclockwise rotation is positive
Στ = F x + F x
the angular acceleration is the same because they are joined by the bar of negligible mass, let us substitute
2 F x = (m r² + 2m r²) α
α =
α =
let's analyze this expression
* for the application point in the center (x = 0) at acceleration is zero
* for the point of application of the torque at the ends the acceleration is
this being its maximum value