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2 years ago

Can anybody write a short poem about friction

1 answer:

8 0

Answer:

you will be the clouds
and I will be the sky.
you will be the ocean
and I will be the shore.
you will be the trees
and I will be the wind.

whatever we are, you and I will always collide.

There you go! Let me know if it helped.
:)

You might be interested in

It is necessary to develop alternative sources of energy because _____.

Leona [35]

Because we wish to have a sustainable future. The current dependence on fossil fuel means that there will be a point when humans have no fuel to use because fossil fuels are a non-renewable resource.

3 0

3 years ago

Projectile <br> SHOW WORK<br> WILL MARK BRANLIEST <br> (Draw Picture and Label)

m_a_m_a [10]

a) The horizontal distance covered by the projectile is 600 m

b) The projectile reaches its maximum height after 3.00 s

c) The altitude of the highest point is 44.1 m

Explanation:

The motion of the projectile consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

In this part A, we just need to analyze the horizontal motion. We know that:

The projectile travels horizontally with a constant velocity ov $v_x = 100 m/s$
The total time of flight of the projectile is $t=6.00 s$

Therefore, the horizontal distance covered by the projectile is given by

$x=v_x t$

And substituting, we find

$x=(100)(6.0) = 600 m$

For this part, we need to analyze the vertical motion of the projectile.

First, we want to find the initial vertical velocity. We can do it by using the equation for the vertical displacement:

$s=u_y t + \frac{1}{2}at^2$

where:

$u_y$ is the initial vertical velocity

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

t is the time

s is the vertical displacement

We know that the total time of flight is t = 6.00 s: this means that when t=6, the projectile returns to its initial vertical position, so s = 0. Substituting and solving for $u_y$ , we get

$u_y = - \frac{1}{2}at=-\frac{1}{2}(-9.8)(6)=29.4 m/s$

The vertical velocity then as a function of t is given by

$v_y = u_y + at$

And at the maximum height, it becomes zero: $v_y = 0$ . Substituting and solving for t, we find the time at which the projectile reaches the maximum height:

$t=-\frac{u_y}{a}=-\frac{29.4}{-9.8}=3.00 s$

To find the altitude of the highest point in the path, we use again the equation:

$s=u_y t + \frac{1}{2}at^2$

where

$u_y = 29.4 m/s$ is the initial vertical velocity

t = 3.00 s is the time at which the projectile reaches the highest point

$a=g=-9.8 m/s^2$ is the acceleration of gravity

Substituting the values, we find

$s=(29.4)(3.00)+\frac{1}{2}(-9.8)(3.00)^2=44.1 m$

So, the highest point is at 44.1 m above the ground.

Learn more about projectiles:

brainly.com/question/8751410

#LearnwithBrainly

8 0

3 years ago

Question 20 (4 points)

irinina [24]

Explanation:

C is correct.

Newton second law states that force is directly proportional to acceleration with m being the constant of variation.

$f _{net} = m(a)$

$250 = 50(a)$

$a = 5$

A is wrong, the constant g only happens in free fall or in vertical direction

B and D are wrong due to the mathematical error or equation error

3 0

2 years ago

Imagina que compras una placa rectangular de metal de 2mm de alto, 10mm de ancho x 50mm de largo, y una masa de 0.02kg. El vende

shusha [124]

Answer:

Densidad de la placa = 20 g/cm³.

La placa no es de oro.

Explanation:

Para encontrar la densidad de la placa rectangular primero debemos hallar su volumen:

$V = 2 mm*10 mm*50 mm = 1000 mm^{3}*\frac{1 cm^{3}}{(10 mm)^{3}} = 1 cm^{3}$

Ahora, encontremos al densidad de la placa:

$d = \frac{m}{V} = \frac{20 g}{1 cm^{3}} = 20 g/cm^{3}$

Dado que la densidad del oro es 19.32 g/cm³ y que la densidad de la placa rectangular calculada es 20 g/cm³, podemos decir que dicha placa no es de oro.

Espero que te sea de utilidad!

6 0

3 years ago

Jason and Guy are throwing watermelons straight up from the ground. (They’re making fruit salad.) Jason throws his watermelon at

NISA [10]

Answer:2t

Explanation:

Given

Jason throwing watermelon at speed v and it spend t time in air

time to reach max height

$v_f=u_i+at'$