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2 years ago

A train travels 600 km in one hour. What is the trains velocity in meters/second

Physics

2 answers:

Nookie1986 [14]2 years ago

6 0

Answer:

Speed of the train, v = 166.66 m/s

Explanation:

Given that,

Distance traveled by the train in one hour, d = 600 km = 600000 m

Time, t = 1 hour = 3600 seconds

We need to find the velocity of the train. Total distance covered in total time is called the velocity of an object. It is given by :

$v=\dfrac{d}{t}$

$v=\dfrac{600000\ m/s}{3600\ s}$

v = 166.66 m/s

So, the velocity of the train is 166.66 m/s. Hence, this is the required solution.

eimsori [14]2 years ago

4 0

v= s/t

600 ÷ 60

10 m/s

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2 years ago

Susie walks 3 blocks north to the local CVS store, then 4 blocks east to her grandmother’s house. She then walks 2 blocks west a

Slav-nsk [51]

Answer:

Suzie is 3 blocks north of where she started

Explanation:

Displacement is the minimum distance between the initial and final point of motion.

Here, Suzie first walks 3 blocks north. From there she walks 4 blocks east. Then 2 blocks to the east then 2 blocks north and then 2 blocks east. She covered 4 blocks east toward west. This is the same distance she covered traveling east. But she is 2 blocks north. From there she traveled a block south to the pizzeria and another block to her friends house. She covered the two block she had traveled north.

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2 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).

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2 years ago

You are a particle physicist at the Large Hadron Collider who is tasked with designing an apparatus to separate annihilation pro

JulijaS [17]

Find solution in the attachments

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2 years ago

A 60.0-kg mass person wishes to push a 120-kg mass box across a level floor. The coefficient of static friction between the pers

Aneli [31]

Answer:

μ = 0.350

Explanation:

For the person to able to move the box, the force exerted by the person on the box must equal the force exerted by the box:

$F_{p} = F_{b}$

In this case, force can be calculated as a product of mass (m) by the acceleration of gravity (g) and the coefficient of static friction (μ):

$m_{p}*g*\mu_{p}=m_{b}*g*\mu_{b}\\m_{p}*\mu_{p}=m_{b}*\mu_{b}\\60*0.7=120*\mu_{b}\\\mu_{b}= 0.35$

Therefore, for the person to be able to push the box horizontally, the coefficient of static friction between the box and the floor should not be higher than 0.350.

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2 years ago