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1 year ago

Describe each class of lever and explain to characteristics of each

Physics

2 answers:

Nataly [62]1 year ago

8 0

-- Class I lever

The fulcrum is between the effort and the load.

The Mechanical Advantage can be anything, more or less than 1 .

Example: a see-saw

-- Class II lever

The load is between the fulcrum and the effort.

The Mechanical Advantage is always greater than 1 .

Example: a nut-cracker, a garlic press

-- Class III lever

The effort is between the fulcrum and the load.

The Mechanical Advantage is always less than 1 .

I can't think of an example right now.

harina [27]1 year ago

6 0

Answer:

First Class of Lever: In this, Fulcrum is always changes the direction of the input force and can be used to increase the force or the distance. Second Class of Lever: In this, Fulcrum does not change direction of the input force & Output force is greater than the input force. Third Class of Lever: In this, the input force is between the fulcrum and the load does not change the direction of the input force. Here, Output force is less than input force.

Explanation:

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An elephant's legs have a reasonably uniform cross section from top to bottom, and they are quite long, pivoting high on the ani

PilotLPTM [1.2K]

Answer:

t = 6,485 s , t_step = 25.94 s

the elephant gives 2.3 step very minute

Explanation:

Let's approximate this system to a simple pendulum that has angular velocity

w = √L / g

Angular velocity and period are related

w = 2π / T

T = 2π √g / L

Let's find the period

T = 2π √9.8 / 2.3

T = 12.97 s

Stride time is

t = T / 2

t = 12.97 / 2

t = 6,485 s

Frequency is inversely proportional to period

f = 1 / t

f = 1 / 6,485

f = 0.15 Hz

Since the elephant has 4 legs and each uses a time t, the total time for one step is

t_step = 4 t

t_step = 4 6.485

t_step = 25.94 s

f_step = 1/t_step =0.0385 s-1

Now let's use a proportion rule to find the number of steps in 60 s

#_step = 60 / t_step

#__step = 60 / 25.94

#_step = 2.3 steps

So the elephant gives 2.3 step very minute

4 0

1 year ago

A 5.0 kg cannonball is dropped from the top of a tower. It falls for 1.6 seconds before slamming into a sand pile at the base of

stepan [7]

Answer:

15.7 m/s

Explanation:

The motion of the cannonball is a accelerated motion with constant acceleration g = 9.8 m/s^2 towards the ground (gravitational acceleration). Therefore, the velocity of the ball at time t is given by:

$v(t)=u + gt$

where

u = 0 is the initial velocity

g = 9.8 m/s^2 is the acceleration

t is the time

If we substitute t=1.6 s into the equation, we find the final velocity of the cannonball:

$v(1.6 s)=0+(9.8 m/s^2)(1.6 s)=15.7 m/s$

4 0

2 years ago

Two resistors of resistances R1 and R2, with R2>R1, are connected to a voltage source with voltage V0. When the resistors are

ehidna [41]

Answer

The Value of r = 0.127

Explanation:

The mathematical representation of the two resistors connected in series is

$R_T = R_1 +R_2$

And from Ohm law

$I_s =\frac{ V}{R_T}$

$I_s = \frac{V_0}{R_1 +R_2} ---(1)$

The mathematical representation of the two resistors connected in parallel is

$R_T = \frac{1}{R_1} +\frac{1}{R_2}$

$= \frac{R_1 R_2}{R_1 +R_2}$

From the question $I_p =10I_s$

=> $I_p =10I_s = \frac{V_0 }{\frac{R_1R_2}{R_1 +R_2} } = \frac{V_0 (R_1 +R_2)}{R_1 R_2}---(2)$

Dividing equation 2 with equation 1

=> $\frac{10I_s}{I_s} =\frac{\frac{V_0 (R_1 +R_2)}{R_1 R_2}}{\frac{V_0}{R_1 +R_2}}$

$10 = \frac{(R_1+R_2)^2}{R_1 R_2}----(3)$

We are told that $r = \frac{R_1}{R_2} \ \ \ \ \ = > R_1 = rR_2$

From equation 3

$10 = \frac{(1-r)^2}{r}$

$=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1+r^2 + 2r = 10r$

$=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r^2 -8r+1 = 0$

Using the quadratic formula

$r =\frac{-b\pm \sqrt{(b^2 - 4ac)} }{2a}$

a = 1 b = -8 c =1

$= \frac{8 \pm\sqrt{((-8)^2- (4*1*1))} }{2*1}$

$r= \frac{8+ \sqrt{60} }{2} \ or \ r = \frac{8 - \sqrt{60} }{2}$

$r = \ 7.87\ or \ r \ = \ 0.127$

Now r = 0.127 because it is the least value among the obtained values

4 0

1 year ago

Imagine you’re driving along a road and you approach a bridge. You notice a sign that reads, “Bridge freezes before road.” Why d

nydimaria [60]

<u>Answer:</u>

<h3>During wet and freezing temperatures, ice is able to form at a faster pace on bridges because freezing winds blow from above and below and both sides of the bridge, causing heat to quickly escape. The road freezes slower because it is merely losing heat through its surface.</h3>

<u>Sources:</u>

-- https://intblog.onspot.com/en-us/why-do-bridges-become-icy-before-roads

and

-- https://www.accuweather.com/en/accuweather-ready/why-bridges-freeze-before-roads/687262

I hope this helps you! ^^

6 0

1 year ago

A ball is dropped from the top of a building.After 2 seconds, it’s velocity is measured to be 19.6 m/s. Calculate the accelerati

zlopas [31]

Answer:

acceleration, a = 9.8 m/s²

Explanation:

'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.

u = 0 m/s

After 2 seconds, velocity of the ball is 19.6 m/s.

t = 2s, v = 19.6 m/s

Using

v = u + at

19.6 = 0 + 2a

a = 9.8 m/s²

6 0

2 years ago

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