All categories

3 years ago

Describe each class of lever and explain to characteristics of each

Physics

2 answers:

Nataly [62]3 years 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]3 years 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:

You might be interested in

A planet's distance from the sun is 2.0x10^11 m. what is the orbital period?

Korolek [52]

Answer:

The square of the orbital period of a planet is directly proportional to the cube of the semimajor axis of its orbit.

Explanation:

hope this helps.

4 0

2 years ago

A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d

prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

$V_{t}= V_{o}e^{\frac{-t}{R*C} }$

Voltage in this case is the energy dissipated so

$E_{t}= E_{o}e^{\frac{-t}{R*C} }$

$\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }$

$\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }$

Using the equation to find capacitance

$ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }$

$C= 5.32x10^{-6}$ F

C= 5.32 uF because u is the symbol for micro that is equal to $10^{-6}$

8 0

3 years ago

Based on the most widely accepted scientific explanation, elements began to form shortly after the big bang. Which of the follow

worty [1.4K]

Big Bang is a theory that, as stated above, has grown to become widely acceptable scientific explanation of the origin of things. Research would say that according to this theory, the Earth originated billion years ago through an explosive of a highly energized point.

Through this explosion, the elements were formed. However, only the lightest of the elements including hydrogen and helium, with traces of lithium and beryllium. From the choices, the answer would have to be the second choice, helium.

Answer: B. helium

5 0

3 years ago

Read 2 more answers

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit

Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

$M=8.7\cdot 10^{13}kg$ is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

$F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N$

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

$m\frac{v^2}{r}=G\frac{Mm}{r^2}$