Which of the following accurately describes how scientists use models?

PLEASE HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

frozen [14]

A. Since it does not say the ball is moving your answer is A.

4 0

3 years ago

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

4 years ago

A light-rail train going from one station to the next on a straight section of track accelerates from rest at 1.1m/s^2 for 20s.

SSSSS [86.1K]

Answer:

A.) 1430 metres

B.) 80 seconds

Explanation:

Given that the train accelerates from rest at 1.1m/s^2 for 20s. The initial velocity U will be:

U = acceleration × time

U = 1.1 × 20 = 22 m/s

It then proceeds at constant speed for 1100 m

Then, time t will be

Time = distance/ velocity

Time = 1100/22

Time = 50 s

before slowing down at 2.2m/s^2 until it stops at the station.

Deceleration = velocity/time

2.2 = 22/t

t = 22/2.2

t = 10s

Using area under the graph, the distance between the two stations will be :

(1/2 × 22 × 20) + 1100 + (1/2 × 22 × 10)

220 + 1100 + 110

1430 m

The time taken between the two stations will be

20 + 50 + 10 = 80 seconds

6 0

3 years ago

PLEASE HURRY AND HELP<br> What is the correct unit for acceleration?<br> A. m<br> B. m/s<br> C. m/s2

Natalka [10]

The answer is C. m/s^2

4 0

4 years ago

What does energy and Newton's Laws (all three) have to do with roller coasters?

Kipish [7]

Answer:

Potential Energy is the energy that is waiting to be released like at the top of the roller coaster, Kinetic energy is the energy that is moving the rollercoaster downhill fast.

Three Laws:

The rollercoaster will rest at the top, and then the gears will make it move downhill, and make it stop.

The gears make the rollercoaster move, and in reaction, it pushes back on the gears.

7 0

3 years ago

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