Please answer this question correctly

Kenny and Candy decided to sit on a see-saw while visiting a local play park. Candy, of mass

pochemuha

Answer:

(i) 208 cm from the pivot

(ii) Move further from the pivot

Explanation:

(i) Sum of the moments about the pivot of the seesaw is zero.

∑τ = Iα

(50 kg) (10 N/kg) (2.5 m) + (60 kg) (10 N/kg) x = 0

1250 Nm + 600 N x = 0

x = -2.08 m

Kenny should sit 208 cm on the other side of the pivot.

(ii) To increase the torque, Kenny should move away from the pivot.

4 0

4 years ago

PLEASE HELP BRAINLIEST!!! 15 MIN!!!

N76 [4]

D.as a magnet is being thrust through the coil

5 0

3 years ago

Read 2 more answers

Cheryl is riding on the edge of a merry-go-round, 2m from the center, which is rotating with an increasing angular speed. Cheryl

WINSTONCH [101]

In circular motion we know that there is two type of acceleration that Cheryl experience

1. Tangential acceleration

2. Centripetal acceleration

here given that

$a_t = 3 m/s^2$

for centripetal acceleration we know that

$a_c = \frac{v^2}{R}$

$a_c = \frac{4^2}{2} = 8 m/s^2$

now we know that both centripetal acceleration and tangential acceleration is perpendicular to each other

so total acceleration is vector sum of both and given as

$a_{net} = \sqrt{8^2 + 3^2} = 8.54 m/s^2$

so total acceleration is 8.54 m/s^2

8 0

3 years ago

What is the best definition for work?

olga55 [171]

Work is done when a force is in the same direction as the object moves.

Explanation:

The work done by a force in moving an object is given by the equation

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the directions of F and d

From the equation, we notice the following:

Work is done by a force when the force is in the same direction as the object moves. In fact, in this case, $\theta=0^{\circ}$ , and so $cos \theta = 1$ , which means that the work done simply becomes $W=Fd$
When the force is perpendicular to the motion of the object, $\theta=90^{\circ}$ and $cos \theta = 0$ , which means that no work done
Technically, work is also done when the force is not parallel to the motion of the object: in this case, the factor $cos \theta$ means that only the component of the force parallel to the direction of motion contributes to do work on the object.