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3 years ago

7.According to the periodic table, how many total particles are there in the nucleus of a Titanium, Ti, atom?

Physics

1 answer:

FromTheMoon [43]3 years ago

3 0

Answer:

there are 22 massive, positively charged, fundamental particles in the element's nucleus. Given this, the element is unquestionably titanium.

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57.0- g tennis ball is traveling straight at a player at 21.0 m/ s. The player volleys the ball straight back at 25.0 m/ s. If t

Yakvenalex [24]

Answer:43.7 N

Explanation:

Given

Mass of tennis ball is =57 gm

velocity of ball=21 m/s

The player volleys the ball straight back at 25 m/s(i.e. in opposite of initial direction)

time of contact=0.06 s

We know change in momentum $(\Delta P) =Impulse(Fdt)$

initial momentum $=57\times 10^{-3}\times 21$

final momentum $=-57\times 10^{-3}\times 25$

change in momentum

$\Delta P=57\times 10^{-3}\times 21-(-57\times 10^{-3}\times 25)$

$\Delta P=57\times 10^{-3}\times 46=2.622 kg-m/s$

Now 2.622 $=F_{avg}\times 0.06$

$F_{avg}=43.7 N$

4 0

3 years ago

You are given a vector in the XY plane that has a magnitude of 89.0 units and a y component of -70.0 units.

Bad White [126]

Hi I need help with my work can you help me please

8 0

3 years ago

A flashlight beam makes an angle of 60 degrees with the surface of the water before it enters the water. In the water what angle

vekshin1

Answer:

θ₂ = 40.5º

Explanation:

For this exercise we must use the law of refraction

n₁ sin θ₁ = n₂ sin θ₂

where index 1 is for the incident ray and index 2 is for the refracted ray

in this case the incident ray has an angle of θ₁ = 60º and the refractive index of the water is

n₂ = 1,333

sin θ₂ = $\frac{n_1}{n_2} \ sin \theta_1$

let's calculate

sin θ₂ = 1 / 1.3333 sin 60

sin θ₂ = 0.64968

θ₂ = sin⁻¹ (0.64968)

θ₂ = 40.5º

8 0

3 years ago

PLEASE HELP A student climbs to the top of a press box where the cameras are. He wonders how many meters he is off of the ground

Feliz [49]

The height of the rail on top of the press box where the ball was dropped from is 11.025 m.

The given parameters:

Time of motion of the ball, t = 1.5 s
Let the height of the rail = h

<h3>Maximum height of fall;</h3>

The maximum height through which the ball was dropped from is calculated by applying second equation of motion;

$h = v_0_y t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\h = \frac{1}{2} (9.8) (1.5)^2\\\\h = 11.025 \ m$

Thus, the height of the rail on top of the press box where the ball was dropped from is 11.025 m.

Learn more about height of projectiles here: brainly.com/question/10008919

3 0

2 years ago

Using the definition of moment of inertia, calculate icm, the moment of inertia about the center of mass, for this object.

Step2247 [10]

<span> The masses have no inertia about their own CM, and "the object" is the two masses. </span>
<span>1. Icm (at point A) = 2mr^2
hope this helps</span>

3 0

3 years ago