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3 years ago

14

A ship was traveling across the Atlantic Ocean. It traveled 2,000 kilometers at a rate of 40 kilometers per hour. How long did i

t take the ship to travel this distance?
A.
0.02 hr
B.
40 hr
C.
50 hr
D.
80,000 hr

1 answer:

8_murik_8 [283]3 years ago

6 0

The correct answer is D

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If a ball is dropped from rest and falls 8 m to the ground, what is the speed just before it

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12.5m/s

Explanation:

(Assuming the question was asking for the speed just before it hit the ground)

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Vf^2 = Vi^2+2aΔd

Vf^2 = 0 + 2(9.8)(8) (plugged in values, initial velocity is 0 since the ball was at rest)

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The acceleration due to gravity vector is always in the ____ direction.

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. A student times a car traveling a distance of 2 m. She finds that it takes the car 5 s to

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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

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(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).

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4 years ago