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3 years ago

What is a type of pulley that increases the size and effort of force?

Physics

1 answer:

Scilla [17]3 years ago

4 0

Fixed pulley. But it has to make multiple strands so you can get a higher mechanical advantage

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What type of force causes your car to move?

amid [387]

Answer:

frictional force

Explanation:

3 0

3 years ago

The schematic below represents an electrical circuit containing Five resistors.

Marta_Voda [28]

Answer: the answer is in the file

Explanation:

5 0

3 years ago

A cannonball is catapulted toward a castle. The cannonball’s velocity when it leaves the catapult is 51.6 m/s at an angle of 37.

Andrei [34K]

Answer:

$v_{y}=35.21m/s$

Explanation:

From the exercise we know the cannonball's initial velocity, the angle which its released with respect to the horizontal and its initial height

$v_{o}=51.6m/s\\\beta =37.0º\\y_{o}=7m$

If we want to know whats the y-component of velocity we need to use the following formula:

$v_{y}^2=v_{oy}^2+ag(y-y_{o})$

Knowing that $g=-9.8m/s^2$

$v_{y}=\sqrt{((51.6m/s)sin(37))^2-2(9.8m/s^2)(0m-7m)}=35.21m/s$

So, the cannonball's y-component of velocity is $v_{y}=35.21m/s$

6 0

3 years ago

Using a rope that will snap if the tension in it exceeds 387 N, you need to lower a bundle of old roofing material weighing 449

Anni [7]

Answer:

(a) $a\approx1.4 m.s^{-2}$

(b) $v\approx 4.133 m.s^{-1}$

Explanation:

Given:

Limiting tension of snapping of the rope, T= 387 N

Weight of the object to be lifted, $w=449 N$

∴mass, $\Rightarrow m= 45.8163 kg$

height of letting down the weight, h = 6.1 m

(a)

Now,

The force to be compensated for being on the verge of snapping:

(T-w) = 62 N

Therefore, we need to produce and acceleration equivalent to the above force.

∵ $F=m.a$

$62=45.8163\times a$

$a= \frac{62}{45.8163}$

$a\approx 1.4 m.s^{-2}$

(b)

From the equation of motion ,we have:

$v^{2} =u^{2} +2a.s$ ....................(2)

where:

u= initial velocity= 0 (here, starting from rest)

v= final velocity = ?

$a= 1.4 m.s^{-2}$

s= displacement =h =6.1 m

Now, putting the values in eq. (2)

$v^2= 0^2 + 2\times 1.4\times 6.1$

$v\approx 4.133 m.s^{-1}$ is the velocity with which the body will hit the ground in the given conditions.

7 0

4 years ago

Un movil pasa por el punto A en direccion hacia B (350cm más adelante) y, luego, sigue hasta el punto C. Sabiendo que pasa por B

xeze [42]

Explanation:

PRIMERO HACES EL RECUENTO DEL TIEMPO Y LO CONVIERTES EN

SEGUNDOS Y ENTONCES

t = 227 s $t_{AB}$ = 227 S - 38 s = 189 s

$t_{BC}$ = 38 s

LUEGO USANDO LA ECUACIÓN DE GALILEO GALILEI SSUPONIENDO

QUE EL MOVIL VIAJA A VELOCIDAD CONSTANTE

v = 3.50 m/189 s = 0.0185 m/s

PARA LA DISTANCIA NTRE B Y C

$x_{BC}$ = 0.0185 m/S( 38 s) = 0.703 m

LA HORA EN QUE EL MOVIL PASA POR A ES

11:43:15 - 38 s - 189 s = 11:39:29

6 0

3 years ago