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2 years ago

10

A rock weighing 98 newtons is pushed off the edge of a bridge 50 meters above the ground. What was the kinetic energy of the roc

k at the midway point of its fall?

1 answer:

Ahat [919]2 years ago

3 0

2,450 Joules, kinetic energy is 1/2 mass x velocity squared.

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A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal

Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration ( $g=9.8 m/s^2$ ) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

$u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s$

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

$v=18.9 m/s$

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

$v_y^2 - u_y^2 = 2ah$ (1)

where

$u_y = 16.4 m/s$ is the initial vertical velocity

$v_y$ is the vertical velocity at height h

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, $v_y = 0$ , so we can use the equation to find the maximum height

$h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m$

So, at half of the maximum height,

$h = \frac{13.7}{2}=6.9 m$

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

$v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s$

And so, the speed of the stone at half of the maximum height is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s$

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

$v_1 = 18.9 m/s$

while the speed at half of the maximum height (part b) is

$v_2 = 22.2 m/s$

So the difference is

$\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s$

And so, in percentage,

$\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%$

So, the stone in part (b) is moving 17.4% faster than in part (a).

4 0

2 years ago

A mover uses a ramp to load a crate of nails onto a truck. The crate, which must be lifted 1.5 m from the street to the bed of t

const2013 [10]

Answer: M.A = 3

Explanation:

A ramp is an example of an inclined plain. Where the

Height H = 1.5 m

Length L = 4.5 m

Mechanical advantage of a machine is the ratio of the load to effort. While mechanical advantage M.A of an inclined plain is the ratio of the length of the plain to the height of the plain.

M.A = L/H

Substitute the values of L and H into the formula

M.A = 4.5/1.5 = 3

The mechanical advantage of the ramp is 3

8 0

2 years ago

The maximum tension that a 0.80 m string can tolerate is 15 N. A 0.35-kg ball attached to this string is being whirled in a vert

zimovet [89]

Answer:

v=5.86 m/s

Explanation:

Given that,

Length of the string, l = 0.8 m

Maximum tension tolerated by the string, F = 15 N

Mass of the ball, m = 0.35 kg

We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :

$F=\dfrac{mv^2}{r}$

v is the maximum speed

$v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{15\times 0.8}{0.35}} \\\\v=5.86\ m/s$

Hence, the maximum speed of the ball is 5.86 m/s.

3 0

2 years ago

A roller coaster travels 200 feet horizontally and then rises 135 feet at an angle of 30 degrees above the ground. What is the m

Delvig [45]

Say the initial point is (0,0)

The final point is

x = 200 + 135*cos(30) = 200 + 135*sqrt(3)/2 = 316.91 ft
y = 135*sin(30) = 135/2 = 67.5 ft

Resultant vector = (316.91, 67.5) - (0,0) = 316.91, 67.5) ft

8 0

2 years ago

Moving water, like that of a river, carries sediment as it moves along its bed. The faster the water flows, the more sediment th

katovenus [111]

Correct option: A

An object remains at rest until a force acts on it.

As the water moves faster, it applies greater force on the sediment, which over comes the frictional forces between the bed and the sediment. So, when the river flows faster, more and larger sediment particles are carried away. When the flow slows down, the river couldn't apply enough force on the larger sediments which can overcome the frictional force between the sediment and the river bed. So, the net force on the heavier particles become zero. Hence, the heavier particles of the load will settle out.

3 0

2 years ago

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