- Mass=0.25kg
- Force=8N
Acceleration toward the center of a curved path is called
1 answer:
Answer:
Centripetal acceleration.
Explanation:
Centripetal acceleration is a property of a body moving in a uniform circular path and it is directed radially towards the center of the circle in which body is rotating.
The force which causes this acceleration is centripetal force which is also directed towards the center of the circle and pulls the body towards its center.
It is calculated through following formula
where v is velocity and r is the radius of the circle.
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Answer:
A) 16.6 m/s i -17.2 m/s j B) 23.9 m/s c) 46º below horizontal.
Explanation:
A) Once released, the football is not under the influence of any external force in the horizontal direction, so it continues moving at a constant speed equal to the initial velocity, i.e., 16.6 m/s.
If we choose the horizontal direction to be coincident with the x-axis, and make positive the direction towards the right (assuming that this was the direction along which the football was thrown), we can write the horizontal component of the veelocity vector, as follows:
vₓ = 16.6 m/s i
In the vertical direction, the football, once released, is in free fall, starting from rest.
So, we can find the vertical component of the velocity vector, at a given point in time, applying the definition of acceleration, as follows:
vy = a*t = -g*t = -9.81 m/s²*1.75 s = -17.2 m/s
Assuming that the upward direction is the positive for the y-axis (perpendicular to the chosen x-axis), we can write the vertical component of the velocity vector, at t=1.75 s, as follows:
vy = -17.2 m/s j
So, the velocity vector, in terms of the unit vectors i and j, can be written in this way:
v = 16.6 m/s i -17.2 m/s j
b) The magnitude of this vector can be found applying trigonometry, as the magnitude is the hypotenuse of a triangle with sides equal to vx and vy, as follows:
v = 23.9 m/s
c) The direction of the vector (below the horizontal) can be found as the angle which tangent is given by the quotient between vy and vx, as follows:
tg θ =
⇒ θ = tg⁻¹ (-1.036) = 46º below horizontal.