What class of lever is shown below?

An 97 kg climber climbs to the top of Mount Everest, which has a peak height of 8850 m above sea level. What is the climber’s po

Kisachek [45]

Eg=mgh
Eg=(97)(9.8)(8850)
Eg=8412810J

4 0

3 years ago

A 1.60-kg object is held 1.05 m above a relaxed, massless vertical spring with a force constant of 330 N/m. The object is droppe

pentagon [3]

Answer:

(A) l = 0.39 m

(B) l =0.38 m

(C) l = 0.14 m

Explanation:

Answer:

Explanation:

Answer:

Explanation:

from the question we are given the following values:

mass (m) = 1.6 kg

height (h) = 1.05 m

compression of spring (l) = ?

spring constant (k) = 330 N/m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) initial potential energy of the object = final potential energy of the spring

potential energy of the object = mg(1.05 + l)

potential energy of the spring = 0.5 x k x l^{2} (k= spring constant)

therefore we now have

mg(1.05 + l) = 0.5 x k x l^{2}

1.6 x 9.8 x (1.05 + l) = 0.5 x 300 x l^{2}

15.68 (1.05 + l) = 150 x l^{2}

16.5 + 15.68l = 150l^{2}

l = 0.39 m

(B) with constant air resistance the equation applied in part A above becomes

initial P.E of the object - air resistance = final P.E of the spring

mg(1.05 + l) - 0.750(1.05 + l) = 0.5 x k x l^{2}

1.6 x 9.8 x (1.05 + l) - 0.750(1.05 + l) = 0.5 x 300 x l^{2}

(16.5 + 15.68l) - (0.788 + 0.75l) = 150l^{2}

16.5 + 15.68l - 0.788 - 0.75l = 150l^{2}

15.71 + 14.93l = 150^{2}

l =0.38 m

(C) where g = 1.63 m/s^{2} and neglecting air resistance

the equation mg(1.05 + l) = 0.5 x k x l^{2} now becomes

1.6 x 1.63 x (1.05 + l) = 0.5 x 300 x l^{2}

2.608 (1.05 +l) = 0.5 x 300 x l^{2}

2.74 + 2.608l = 150 x l^{2}

l = 0.14 m

6 0

2 years ago

Si tenemos tres cubos del mismo tamaño (hierro, madera e icopor). ¿Qué diferencias puede encontrar entre ellos?

cricket20 [7]

Answer: La diferencia es el peso (o la masa), siendo que el cubo de hierro es el mas pesado, después viene el de madera y después el de icopor.

Explanation:

Ok, los 3 cubos tienen el mismo tamaño, lo que implica que tienen el mismo volumen.

Ahora es útil recordar la relación:

Densidad = masa/volumen.

Masa = densidad*volumen.

Nosotros sabemos que la densidad del hierro es mas grande que la de la madera, y la densidad de la madera es mas grande que la del icopor.

Entonces, por la relación anterior, y sabiendo que todos los cubos tienen el mismo volumen, podemos reconocer que el cubo de hierro tiene mayor masa, después viene el de madera y después viene el de icopor.

Y sabiendo que:

masa*gravedad = peso

podemos saber que el cubo mas pesado es el de hierro, después el de madera y después el de icopor.

Además de esta diferencia, también hay otras que no dependen tanto del tamaño del objeto, como pueden ser las capacidades caloríficas, el como reaccionan a campos eléctricos y cosas así que son triviales, pues son diferentes para casi todos los materiales.

5 0

3 years ago

A person in a kayak starts paddling, and it accelerates from 0 to 0.61 m/s in a distance of 0.39 m. If the combined mass of the

Iteru [2.4K]

Answer:

35.3 N

Explanation:

U = 0, V = 0.61 m/s, s = 0.39 m

Let a be the acceleration.

Use third equation of motion

V^2 = u^2 + 2 as

0.61 × 0.61 = 0 + 2 × a × 0.39

a = 0.477 m/s^2

Force = mass × acceleration

F = 74 × 0.477 = 35.3 N

6 0

3 years ago

A rubber ball and a lump of clay have equal mass. They are thrown with equal speed against a wall. The ball bounces back with ne

gregori [183]

Answer:

The ball experiences the greater momentum change

Explanation:

The momentum change of each object is given by:

$\Delta p = m \Delta v= m (v-u)$

where

m is the mass of the object

v is the final velocity

u is the initial velocity

Both objects have same mass m and same initial velocity u. So we have:

- For the ball, the final velocity is

$v=-u$

Since it bounces back (so, opposite direction --> negative sign) with same speed (so, the magnitude of the final velocity is still u). So the change in momentum is

$\Delta p=m(v-u)=m((-u)-u)=-2mu$

- For the clay, the final velocity is

$v=0$

since it sticks to the wall. So, the change in momentum is

$\Delta p = m(v-u)=m(0-u)=-mu$

So we see that the greater momentum change (in magnitude) is experienced by the ball.

3 0

3 years ago