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kykrilka [37]
3 years ago
12

How many days does it take for a free to grow?

Physics
1 answer:
vovikov84 [41]3 years ago
7 0
Idk what is growing but if it’s a free than c
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An electron and a 0.033 0-kg bullet each have a velocity of magnitude 495 m/s, accurate to within 0.010 0%. Within what lower li
lara31 [8.8K]

Answer:

1.170*10^-3 m

3.23*10^-32 m

Explanation:

To solve this, we apply Heisenberg's uncertainty principle.

the principle states that, "if we know everything about where a particle is located, then we know nothing about its momentum, and vice versa." it also can be interpreted as "if the uncertainty of the position is small, then the uncertainty of the momentum is large, and vice versa"

Δp * Δx = h/4π

m(e).Δv * Δx = h/4π

If we make Δx the subject of formula, by rearranging, we have

Δx = h / 4π * m(e).Δv

on substituting the values, we have

for the electron

Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 4.95*10^-2

Δx = 6.63*10^-34 / 5.67*10^-31

Δx = 1.170*10^-3 m

for the bullet

Δx = (6.63*10^-34) / 4 * 3.142 * 0.033*10^-31 * 4.95*10^-2

Δx = 6.63*10^-34 / 0.021

Δx = 3.23*10^-32 m

therefore, we can say that the lower limits are 1.170*10^-3 m for the electron and 3.23*10^-32 for the bullet

7 0
3 years ago
A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.
kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

\frac{1}{p}+\frac{1}{q}=\frac{1}{f}

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm

And the positive sign means the image is upright.

6 0
3 years ago
A device that converts high voltages to lower voltages is a(n) ________.
AysviL [449]

Answer: A) Transformer

Explanation: A Transformer is an apparatus, device or a component in a system that is used to convert high voltage into low voltage. It is used to either increase or decrease the voltage of an alternating current.

A transformer uses the basic principle of electro magnetic induction, having two or more coils, the voltage is changed from one coil to another but with thesame frequency as alternating current energy passes through them.

7 0
3 years ago
A ball falls from a shelf. Assuming there is no friction, why is the conservation of mechanical energy independent of mass?(1 po
cluponka [151]

Answer:

It's B

Explanation:

Hope this helps, tell me if im wrong!

5 0
2 years ago
Read 2 more answers
Which statement(s) is/are indicative of a covalent bond?
Over [174]
Do you have any answer options?

The most I can help is to tell you that a covalent bond is a type of bond that shares electrons between the bonded atoms.

4 0
3 years ago
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