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3 years ago

In an atom, the electrons travel outside the____.

Physics

1 answer:

STatiana [176]3 years ago

4 0

Nucleus, made up of protons and neutrons and can be found in the center of the atom

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A mountain climber weighs 42.0 N. If he climbs a hill 100m high. Calculate the work done in joules

quester [9]

Answer:

The answer is 4200 J.

Explanation:

The formula of work done is, W = F×D where F is the force of an object and D is the distance. Then you just substitute the values into the equation :

W = F×D

F = 42N

D = 100m

W = 42 × 100

= 4200 J

5 0

2 years ago

The observation deck of a skyscraper is 420 m above

Reil [10]

2e min :)) pls park braliest

5 0

2 years ago

Martina has a sample of an unknown substance. She measures the substance. It’s mass is 13.5 grams, and it’s volume is 5 cm3 .Whi

Dominik [7]

Cm^3 is same as mL

13.5 g / 5 mL = 2.7 g/mL
look up densities of metals
aluminum has a density of 2.7 g/mL

6 0

2 years ago

A projector is placed on the ground 22 ft. away from a projector screen. A 5.2 ft. tall person is walking toward the screen at a

Stella [2.4K]

Answer:

y = 67.6 feet, y = 114.4/ (22 - 3t)

Explanation:

For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram

Large triangle Projector up to the screen

tan θ = y / L

For the small triangle. Projector up to the person

tan θ = y₀ / (L-d)

The angle is the same, so we equate the two equations

y₀ / (L -d) = y / L

y = y₀ L / (L-d)

The distance from the screen (d), we look for it with kinematics

v = d / t

d = v t

we replace

y = y₀ L / (L - v t)

y = 5.2 22 / (22 - 3 t)

y = 114.4 (22 - 3t)⁻¹

This is the equation of the shadow height change as a function of time

For the suggested distance the shadow has a height of

y = 114.4 / (22-13)

y = 67.6 feet

7 0

2 years ago

Ignoring air resistance and the little friction from the plastic tube, the magnet was a freely-falling object in each trial. If

horsena [70]

Answer:

emf will also be 10 times less as compared to when it has fallen $40 \mathrm{m}$

Explanation:

We know, from faraday's law-

$e m f=-N \frac{\Delta \Phi}{\Delta T}$

and $\Phi=B . A$

So, as the height increases the velocity with which it will cross the ring will also increase. $(v=\sqrt{2 g h})$

Given

$\mathrm{V} 1(\text { Speed at } 40 \mathrm{m})=2 \mathrm{x} \mathrm{V} 2(\text { speed at } 10 \mathrm{m})$

$\sqrt{2 g h_{2}}=2 \times \sqrt{2 g h_{1}}=28.28 \mathrm{m} / \mathrm{s}$

Now, from $40 \mathrm{cm}$

$V_{3}=\sqrt{2 g h_{3}}=\sqrt{2 \times 10 \times 0.4}=2.82 \mathrm{m} / \mathrm{s}$

From equation a and b we see that velocity when dropped from $40 \mathrm{m}$ is 10 times greater when height is 40 $\mathrm{cm}$ so, emf will also be 10 times less as compared to when it has fallen $40 \mathrm{m}$

4 0

2 years ago