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2 years ago

This is a substance that cannot be broken down further by chemical means.

Physics

2 answers:

viktelen [127]2 years ago

5 0

The answer is ELEMENT

Serggg [28]2 years ago

4 0

I think its an element

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If the is the absense of light is the dark. thennnnnnnnnn since there is light particles. is there dark particles?if not then im

TEA [102]

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2 years ago

Read 2 more answers

Packages having a mass of 6 kgkg slide down a smooth chute and land horizontally with a speed of 3 m/sm/s on the surface of a co

nalin [4]

Answer:

t = 1.02 s

Explanation:

The computation of the time required is shown below:

The package speed for belt is

= 3 - 1

= 2 m/s

Moreover, the decelerative force would be acted on the block i.e u.m.g

So, the decelerative produced

= 0.2 × 9.81

= 1.962 m/s^2

And, final velocity = 0

v = u - at

here

V = 0 = final velocity

u = 2 m/s

so,

0 = 2 - 1.962 × t

t = 1.02 s

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2 years ago

Part A Determine the magnitude of the x component of F using scalar notation. Fx F x = nothing lb Request Answer Part B Determin

maksim [4K]

As we know that force F makes an angle of 60 degree with X axis

so the X component is given as

$cos60 = \frac{F_x}{F}$

now we have

$F_x = F cos60$

$F_x = 0.50 F$

Similarly we know that force F makes an angle of 45 degree with Y axis

so the X component is given as

$cos45 = \frac{F_y}{F}$

now we have

$F_y = F cos45$

$F_y = 0.707 F$

Now for the component along z axis we know that

$F_x^2 + F_y^2 + F_z^2 = F^2$

now plug in all components

$(0.707 F)^2 + (0.50 F)^2 + F_z^2 = F^2$

$0.5 F^2 + 0.25 F^2 + F_z^2 = F^2$

$F_z^2 = F^2(1 - 0.75)$

$F_z^2 = 0.25 F^2$

$F_z = 0.5 F$

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3 years ago

20 POINTSSS!!!!!!!!

Soloha48 [4]

Answer: A
Hope this help you!!

7 0

2 years ago

A shot-putter released the shot at an angle of 41.5 degrees and a height of 1.9 m with an initial velocity of 13.3 m/s. How far

liq [111]

Answer:

x = 17.88[m]

Explanation:

We can find the components of the initial velocity:

$(v_{x})_{o} = 13.3*cos(41.5)=9.96[m/s]\\(v_{y})_{o} = 13.3*sin(41.5)=8.81[m/s]$

We have to remember that the acceleration of gravity will be worked with negative sign, since it acts in the opposite direction to the movement in direction and the projectile upwards.

g = - 9.81[m/s^2]

Now we must find the time it takes for the projectile to hit the ground, as the problem mentions that it does not impact on the board.

$y=y_{o} +(v_{y} )_{o} *t-0.5*g*(t)^{2} \\0=1.9+(8.81*t)-(4.905*t^{2})\\-1.9=8.81*t*(1-0.5567*t)\\t=0\\t=1.796[s]$

With this time we can calculate the horizontal distance:

$x=(v_{x})_{o} *t\\x=9.96*1.796\\x=17.88[m]$

3 0

3 years ago