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2 years ago

9

Why are the parts of an atom that electrons occupy called electron clouds?

1 answer:

Anna [14]2 years ago

8 0

Because it's literally impossible to tell exactly where something that size is
located at any particular time.

And that's NOT because it's so small that we can't see it. It's because any
material object behaves as if it's made of waves, and the smaller the object is,
the more the size of its waves get to be like the same size as the object.
When you get down to things the size of subatomic particles, it doesn't make
sense any more to try and talk about where the particle actually "is", and we only
talk about the waves that define it, and how the waves all combine to become a
cloud of <em><u>probability</u></em> of where the particle is.

I know it sounds weird. But that's the way it is. Sorry.

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Which objects would be considered to be in free fall?

SVEN [57.7K]

Answer:

Any object that is being acted upon only by the force of gravity is said to be in a state of free fall.

Explanation:

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2 years ago

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A 600 N astronaut travels to an asteroid where the gravitational force is one-hundredth that of Earth. What is the astronaut's w

Ganezh [65]

Answer:

A) 6N

Explanation:

the weight of the astronaut on earth can be calculated with the formula

Weight = mass * gravity of earth

Since the gravitational force is one-hundredth of Earth, the formula should be

Weight = mass * (gravity of earth / 100)

Weight = (mass * gravity of earth) / 100

Since you already know the weight, you only need to divide by 100

Weight = (mass * gravity of earth) / 100

Weight = 600N / 100

Weight = 6N

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2 years ago

1. Encontrar la densidad de una sustancia, Si pesa 459 g y ocupa un volumen de 34.5 cmɜ.

ryzh [129]

Answer:

your language is to different to mine sorry

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2 years ago

Two waiters are trying to get through a single door of a kitchen. One pushes on one side of a door 0.567 m from the hinge with a

NISA [10]

Answer:

$275.5 N$

Explanation:

$F_{1}$ = Force on one side of the door by first waiter = 257 N

$F_{2}$ = Force on other side of the door by second waiter

$r_{1}$ = distance of first force by first waiter from hinge = 0.567 m

$r_{2}$ = distance of second force by second waiter from hinge = 0.529 m

Since the door does not move. hence the door is in equilibrium

Using equilibrium of torque by force applied by each waiter

$r_{1} F_{1} = r_{2} F_{2} \\(0.567) (257) = (0.529) F_{2}\\F_{2} = 275.5 N$

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2 years ago

It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

$v^{2}=u^{2}+2\cdot a \cdot s$

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/ $Sec^{2}$

s= Distance traveled before stop m

Case 1

u= 13 m/sec, v=0, s= 57.46 m, a=?

$0^{2} = 13^{2} + 2 \cdot a \cdot57.46$

a = -1.47 m/ $Sec^{2}$ (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/ $Sec^{2}$ (since same friction force is applied)

$v^{2} = 29^{2} - 2 \cdot 1.47 \cdot S$

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

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2 years ago