<span>The question is 'a centre seeking force related to acceleration is ............... force. The answer is centripetal force. Motion in a curved path is an accelerated motion and it requires a force that will direct the moving object towards the centre of curvature of the path of motion. This centre seeking force is known as centripetal force.</span>
a una velocidad de
22 m/s, quien lo golpea y devuelve en la misma
dirección con una velocidad de 14 m/s. Si el
tiempo de contacto del balón con la jugadora es
de 0,03 s, ¿con qué fuerza golpeó la jugadora el
balón?
21 Una bala de 0,8 g, está en la recámara de un rifl e
cuando se g
(a) The free body of all the forces include, frictional force, weight of the box acting perpendicular and another acting parallel to the plane.
(b) When the box is sliding down, the frictional force acts towards the right.
(c) When the box slides up, the direction of the frictional force changes, it acts towards the left.
<h3>
Free body diagram</h3>
The free body diagram of all the forces on the box is obtained by noting the upward force and downward forces on the box as shown below;
/ W2
Ф → Ff
↓W1
where;
- Ff is the frictional force resisting the down motion of the box
- W1 is the perpendicular component of the box weight = Wcos(33)
- W2 is the parallel component of the box weight = Wsin(33)
(b) When the box is sliding down, the frictional force acts towards the right.
(c) When the box slides up, the direction of the frictional force changes, it acts towards the left.
Learn more about free body diagram of inclined objects here: brainly.com/question/4176810
The object takes 0.5 seconds to complete one rotation, so its rotational speed is 1/0.5 rot/s = 2 rot/s.
Convert this to linear speed; for each rotation, the object travels a distance equal to the circumference of its path, or 2<em>π</em> (1.2 m) = 2.4<em>π</em> m ≈ 7.5 m, so that
2 rot/s = (2 rot/s) • (2.4<em>π</em> m/rot) = 4.8<em>π</em> m/s ≈ 15 m/s
thus giving it a centripetal acceleration of
<em>a</em> = (4.8<em>π</em> m/s)² / (1.2 m) ≈ 190 m/s².
Then the tension in the rope is
<em>T</em> = (50 kg) <em>a</em> ≈ 9500 N.
Answer:
answer a: a large front gear with a small back gear
answer b: a small front gear with a large back gear
Explanation:
just simple gearing ratios
