All categories

3 years ago

What is simple machines and it type.I need it today please

Physics

1 answer:

raketka [301]3 years ago

4 0

Simple machines are tools, devices or objects that are used to make work easier.

TYPES OF SIMPLE MACHINES.

Levers
Inclined Plane
Wedges
Screw
Pulley
Wheel & Axle

THE LEVERS

A lever is simply a plank or ridge beam that is free to rotate on a pivot.

PARTS OF A LEVER

Load
Effort
Fulcrum or pivot

CLASSES OF LEVERS

THE FIRST CLASS LEVER

In this class, the FULCRUM is between the EFFORT and the LOAD. The mechanical advantage is more if the load is closer to the FULCRUM. Examples: seesaws, boat oars and crowbars.

THE SECOND CLASS LEVER

In this class, the LOAD is between the EFFORT and the FULCRUM. The mechanical advantage is more if the LOAD is closer to the FULCRUM. Example: wheelbarrow.

THE THIRD CLASS LEVER

In this class, the EFFORT is between the LOAD and the FULCRUM. The mechanical advantage is more if the EFFORT is closer to the LOAD. Example: garden shovel.

THE INCLINED PLANE OR RAMP

An inclined plane is a simple machine with a sloping surface. It makes it easier for us to move objects from lower grounds to higher grounds.

WEDGES

A wedge is simply a triangular tool, often made of wood, stone, metal or plastic. It is thick at one end and tapers to a thin or sharp edge at the other end. Examples: knife, axe, doorstopper, nail, blade on the snow plough or farmer grader.

PULLEY

A pulley is simply a wheel with a groove in it, and a rope in the groove. It is used to lift heavy objects from the ground to high places.

WHEEL & AXLE

Involves two circular objects joined at the center. It works when force is applied to the wheel or when force is applied to the axle. Example: door knob.

You might be interested in

What process modifies light waves to vibrate in a single plane

AnnyKZ [126]

This process is called polarization.

8 0

3 years ago

A single-phase 60-Hz overhead power line is symmetrically supported on a horizontal cross arm. Spacing between the centers of th

pentagon [3]

Complete question is;

A single-phase 60-Hz overhead power line is symmetrically supported on a horizontal cross arm. Spacing between the centers of the conductors acing between the centers of the conductors (say, a and b) is 2.5 m. A telephone line is also symmetrically supported on a horizontal cross arm 1.8 m directly below the power line. Spacing between the centers of these conductors (say, c and d) is 1.0 m.

The mutual inductance per unit length between circuit a-b and circuit c-d is given as 4 x 10^(-7) ln √((D_ad × D_bc)/(D_ac × D_bd)) H/m

where, for example, D_ad denotes the distance in meters between conductors a and d.

a. Hence, compute the mutual inductance per kilometer between the power line and the telephone line.

b. Find the 60-Hz voltage per kilometer induced in the telephone line when the power line carries 150 A

Answer:

A) M = 1.01 × 10^(-4) H/km

B) v_cd = 5.712 V/km

Explanation:

A) From the distances given in the question, we can deduce that;

D_ac = √(((2.5/2) - (1/2))² + 1.8²)

D_ac = 1.95 m

Also;

D_ad = √(((2.5/2) + (1/2))² + 1.8²)

D_ad = 2.51 m

I_a and I_b are put of phase by 180°. Thus, due to a and b, the flux linkages to c and d is given as;

φ_cd = 4 x 10^(-7)I_a( ln (2.51/1.95))

Mutual inductance per km is given as;

M = φ_cd/I_a

Thus;

M = 4 x 10^(-7)( ln (2.51/1.95))

M = 1.01 × 10^(-7) H/m

Per km;

M = 1.01 × 10^(-7) × 1000

M = 1.01 × 10^(-4) H/km

B) voltage per km is gotten by;

v_cd = ωMI

Now, ω = 2πf = 2π × 60 = 377 rad/s

Thus;

v_cd = 377 × 1.01 × 10^(-4) × 150

v_cd = 5.712 V/km

5 0

3 years ago

Do it in order. from smallest to largest

Sholpan [36]

Answer:

The earth, The sun, the solar system and the milky way.

7 0

3 years ago

A particle executes simple harmonic motion with an amplitude of 2.18 cm.

Bogdan [553]

Answer:

The positive displacement from the midpoint of its motion at the speed equal one half of its maximum speed is 3.56 cm.

Explanation:

Maximum speed is :

v (max) = Aω

Speed v at any displacement y is given by

$v^{2}$ = $w^{2}$ ( $A^{2}$ - $y^{2}$ ) ........................................................ i

And,

v = $\frac{1}{2}$ v (max)

or, 2 × v = Aω .................................................... ii

Eliminating ω from equations i and ii,

$\frac{1}{4}$ $A^{2}$ $w^{2}$ = $w^{2}$ ( $A^{2}$ - $y^{2}$ )

or, $y^{2}$ = ( $\frac{3}{4}$ ) $A^{2}$ =( $\frac{3}{4}$ ) $2.18^{2}$

or, y = 3.56 cm.

3 0

3 years ago

A racecar driver steps on the gas, and his racecar travels 20 meters in 2 seconds starting from rest. The acceleration of the ra

givi [52]

Answer:

2.5m/s^2

Explanation:

Step one:

given

distance = 20meters

time = 2 seconds

initial velocity u= 0m/s

let us solve for the final velocity

velocity = distance/time

velocity= 20/2

velocity= 10m/s

$v^2=u^2+2as\\\\10^2=0^2+2*a*20\\\\100=40a$

divide both sides by 40

$a= 100/40\\\\a=10/4\\\\a= 5/2\\\\a= 2.5m/s^2$

5 0

3 years ago

What is simple machines and it type.I need it today please​

What is simple machines and it type.I need it today please