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3 years ago

Second and third class levers both have __________.

Physics

2 answers:

Zolol [24]3 years ago

5 0

Answer:

i think its d on edgenuity

Explanation:

nataly862011 [7]3 years ago

4 0

The answer would be:

<u><em>D. the fulcrum at one end of the lever.</em></u>

Both second class and third class levers have their fulcrum located at the end of the lever. They differ in the location of the load and effort. For third class levers the effort is in between the load and the fulcrum. For second class levers, the load is in between the effort and fulcrum. You can look at the attached picture as reference.

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A child whose weight is 287 N slides down a 7.20 m playground slide that makes an angle of 31.0Â° with the horizontal. The coeff

natulia [17]

Answer:

$H =212.6 \ J$

$v = 7.647 \ m/s$

Explanation:

From the question we are told that

The child's weight is $W_c = 287 \ N$

The length of the sliding surface of the playground is $L = 7.20 \ m$

The coefficient of friction is $\mu = 0.120$

The angle is $\theta = 31.0 ^o$

The initial speed is $u = 0.559 \ m/s$

Generally the normal force acting on the child is mathematically represented as

=> $N = mg * cos \theta$

Note $m * g = W_c$

Generally the frictional force between the slide and the child is

$F_f = \mu * mg * cos \theta$

Generally the resultant force acting on the child due to her weight and the frictional force is mathematically represented as

$F =m* g sin(\theta) - F_f$

Here F is the resultant force and it is represented as $F = ma$

=> $ma = m* g sin(31.0) - \mu * mg * cos (31.0)$

=> $a = g sin(31.0)- \mu * g * cos (31.0)$

=> $a = 9.8 * sin(31.0) - 0.120 * 9.8 * cos (31.0)$

=> $a = 4.039 \ m/s^2$

$F_f = 0.120 * 287 * cos (31.0)$

=> $F_f = 29.52 \ N$

Generally the heat energy generated by the frictional force which equivalent tot the workdone by the frictional force is mathematically represented as

$H = F_f * L$