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2 years ago

Atomic math challenge will give brainly and thanks

Physics

2 answers:

nevsk [136]2 years ago

8 0

Answer:

1. Hydrogen

Atomic # = 1

Atomic Mass = 1.00794 ( If you round it it's 1.008 )

# of protons = 1

# of neutrons = none

# of electrons = 1

WARRIOR [948]2 years ago

8 0

Had the same exact question

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Consider a projectile launched with an initial velocity of v0 = 120 ft/s, inclined at an angle, θ with the horizontal. Let us as

Natali [406]

Answer:

How to find the maximum height of a projectile?

if α = 90°, then the formula simplifies to: hmax = h + V₀² / (2 * g) and the time of flight is the longest. ...

if α = 45°, then the equation may be written as: ...

if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that's the case of horizontal projectile motion.

Explanation:

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2 years ago

A uniform thin wire is bent into a quarter-circle of radius a = 20.0 cm, and placed in the first quadrant. Determine the coordin

Mashcka [7]

Answer:

$r_{cm}=[12.73,12.73]cm$

Explanation:

The general equation to calculate the center of mass is:

$r_{cm}=1/M*\int\limits {r} \, dm$

Any differential of mass can be calculated as:

$dm = \lambda*a*d\theta$ Where "a" is the radius of the circle and λ is the linear density of the wire.

The linear density is given by:

$\lambda=M/L=M/(a*\pi/2)=\frac{2M}{a\pi}$

So, the differential of mass is:

$dm = \frac{2M}{a\pi}*a*d\theta$

$dm = \frac{2M}{\pi}*d\theta$

Now we proceed to calculate X and Y coordinates of the center of mass separately:

$X_{cm}=1/M*\int\limits^{\pi/2}_0 {a*cos\theta*2M/\pi} \, d\theta$

$Y_{cm}=1/M*\int\limits^{\pi/2}_0 {a*sin\theta*2M/\pi} \, d\theta$

Solving both integrals, we get:

$X_{cm}=2*a/\pi=12.73cm$

$Y_{cm}=2*a/\pi=12.73cm$

Therefore, the position of the center of mass is:

$r_{cm}=[12.73,12.73]cm$

5 0

2 years ago

a projectile is shot horizontally from the edge of a cliff, 230m above the ground. the projectile lands 300m from base of the cl

bekas [8.4K]

Answer:

The time taken by the projectile to hit the ground is 6.85 sec.

Explanation:

Given that,

Vertical height of cliff = 230 m

Distance = 300 m

Suppose, determine the time taken by the projectile to hit the ground.

We need to calculate the time

Using second equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Where, s = vertical height of cliff

u = initial vertical velocity

g = acceleration due to gravity

Put the value in the equation

$230=0+\dfrac{1}{2}\times9.8\times t^2$

$t=\sqrt{\dfrac{230\times2}{9.8}}$

$t=6.85 sec$

Hence, The time taken by the projectile to hit the ground is 6.85 sec.

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Serga [27]

Answer:

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Explanation:

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1 year ago

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