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1 year ago

compare the numbers of each subatomic particle found in both diagrams and then list what makes hydrogen diffrent from hellium

Physics

1 answer:

Cloud [144]1 year ago

7 0

The helium atom is found to contain more particles than the hydrogen atom.

<h3>What is a subparticle?</h3>

An atom is composed of particles and these particles are electrons protons and neutrons. The collective name for these particles is the subatomic particles.

Now we know that the helium atom contains more sub atomic particles than the atom of hydrogen. This is because, the helium atom contains two neutrons and two protons as well as two electrons. On the other hand, the hydrogen atom contains one neutron and one proton as well as one electron.

Thus, the helium atom is found to contain more particles than the hydrogen atom.

Learn more about helium atom:brainly.com/question/4945478

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A 12,500 N alien UFO is hovering about the surface of Earth. At time , its position can be given as () = ((0.24 m/s^3)^3 + 25 m)

stepladder [879]

a) $F=(3675i-4543k)N$

b) 5843 N

Explanation:

The position of the UFO at time t is given by the vector:

$r(t)=(0.24t^3+25)i+(4.2t)j+(-0.43t^3+0.8t^2)k$

Therefore it has 3 components:

$r_x=0.24t^3+25\\r_y=4.2t\\r_z=-0.43t^3+0.8t^2$

We start by finding the velocity of the UFO, which is given by the derivative of the position:

$v_x=r'_x=\frac{d}{dt}(0.24t^3+25)=3\cdot 0.24t^2=0.72t^2\\v_y=r'_y=\frac{d}{dt}(4.2t)=4.2\\v_x=r'_z=\frac{d}{dt}(-0.43t^3+0.8t^2)=-1.29t^2+1.6t$

And then, by differentiating again, we find the acceleration:

$a_x=v'_x=\frac{d}{dt}(0.72t^2)=1.44t\\a_y=v'_y=\frac{d}{dt}(4.2)=0\\a_z=v'_z=\frac{d}{dt}(-1.29t^2+1.6t)=-2.58t+1.6$

The weight of the UFO is W = 12,500 N, so its mass is:

$m=\frac{W}{g}=\frac{12500}{9.8}=1276 kg$

Therefore, the components of the force on the UFO are given by Newton's second law:

$F=ma$

So, Substituting t = 2 s, we find:

$F_x=ma_x=(1276)(1.44t)=(1276)(1.44)(2)=3675 N\\F_y=ma_y=0\\F_z=ma_z=(1276)(-2.58t+1.6)=(1276)(-2.58(2)+1.6)=-4543 N$

So the net force on the UFO at t = 2 s is

$F=(3675i-4543k)N$

The magnitude of a 3-dimensional vector is given by

$|v|=\sqrt{v_x^2+v_y^2+v_z^2}$

where

$v_x,v_y,v_z$ are the three components of the vector

In this problem, the three components of the net force are:

$F_x=3675 N\\F_y=0\\F_z=-4543 N$

Therefore, substituting into the equation, we find the magnitude of the net force:

$|F|=\sqrt{3675^2+0^2+(-4543)^2}=5843 N$

7 0

3 years ago

What is the rate of vibration measured in

jek_recluse [69]

Answer:

I'm not 100% sure tbh but the only thing I think makes sense to represent vibration would be frequency which is measure in Hertz (Hz)

Explanation:

7 0

3 years ago

Read 2 more answers

A stone dropped in a pond sends out a circular ripple whose radius increases at a constant rate of 4 ft/sec. After 12 seconds, h

Natali5045456 [20]

Answer:

After 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec

Explanation:

Area of a circle = πr²

where;

r is the circle radius

Differentiate the area with respect to time.

$\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$

dr/dt = 4 ft/sec

after 12 seconds, the radius becomes = $\frac{dr}{dt} X 12 = 4 \frac{ft}{sec} X 12 sec = 48 ft$

To obtain how rapidly is the area enclosed by the ripple increasing after 12 seconds, we calculate dA/dt

$\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$

$\frac{dA}{dt} = 2\pi (48)(4)$

dA/dt = 1206.528 ft²/sec

Therefore, after 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec

7 0

3 years ago

An object is stopped at point ‘a’, and travels to point ‘c’ in 8s. What was its acceleration?

lutik1710 [3]

In SI units, its acceleration is (the distance from A to C) / 32 m/s^2 .

7 0

3 years ago

Kevin is refinishing his rusty wheelbarrow. He moves his sandpaper back and forth 45 times over a rusty area, each time moving a

dmitriy555 [2]

W = _|....F*dx*cos(a)........With F=force, x=distance over which force acts on object,
.......0.............................and a=angle between force and direction of travel.

Since the force is constant in this case we don't need the equation to be an integral expression, and since the force in question - the force of friction - is always precisely opposite the direction of travel (which makes (a) equal to 180 deg, and cos(a) equal to -1) the equation can be rewritted like so:

W = F*x*(-1) ............ or ............. W = -F*x

The force of friction is given by the equation: Ffriction = Fnormal*(coeff of friction)

Also, note that the total work is the sum of all 45 passes by the sandpaper. So our final equation, when Ffriction is substituted, is:

W = (-45)(Fnormal)(coeff of friction)(distance)
W = (-45)...(1.8N).........(0.92).........(0.15m)
W = ................-11.178 Joules

5 0

3 years ago