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DiKsa [7]
3 years ago
11

What components of the atom are found outside of the nucleus? protons electrons neutrons all are found in the nucleus

Physics
2 answers:
Montano1993 [528]3 years ago
6 0
The components of an atom found outside the nucleus are electrons.
AlladinOne [14]3 years ago
5 0

the answer is electrons. this is proof checked and I guarantee this is the correct answer. hope this helped :)

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A Ping-Pong ball with mass 2.5 g is attached by a thread to the bottom of a beaker. When the beaker is filled with water so that
dimaraw [331]

Answer:

0.022m or 2.2cm

Expxlanation:

Step 1:

Data obtained from the question. This includes:

Mass (m) = 2.5g = 2.5/1000 = 2.5x10^-3Kg

Tension (T) = 0.029 N

Density (ρ) = 1000 kg/m3

Acceleration due to gravity (g) = 9.81 m/s2

Diameter (d) =?

Step 2:

Finding an expression to calculate the diameter of the ball. This is illustrated below:

Tension = weight displaced - weight of the ball

Weight displaced = Mass of water x acceleration due to gravity

Mass of water = Density x volume

Mass of water = ρxV

Weight displaced = ρxVxg = ρVg

Weight of the ball = Mass of the ball x acceleration due to gravity

Weight of the ball = mg

Therefore,

Tension = weight displaced - weight of the ball

T = ρVg - mg

Make V the subject of the formula

T = ρVg - mg

T + mg = ρVg

Divide both side by ρg

V = ( T + mg) /ρg. (1)

Recall that the ball is spherical in shape and the Volume of a sphere is given by

V = 4/3πr^3

Radius (r) = diameter (d) /2

V = 4/3π(d/2)^3

V = 4/3πd^3/8

V = πd^3 /6

Substituting the value of V into equation 1, we have

V = ( T + mg) /ρg

πd^3 /6 = ( T + mg) /ρg.

Making d the subject of the formula, we have:

πd^3 /6 = (T + mg) /ρg.

d^3 = 6(T + mg) /πρg.

Taking the cube root of both sides

d = [6(T + mg) /πρg]^1/3

Step 3:

Determination of the diameter of the ball. This is illustrated below:

T = 0.029 N

m = 2.5x10^-3Kg

g = 9.81 m/s2

ρ = 1000 kg/m3

d =?

d = [6(T + mg) /πρg]^1/3

d = [6(0.029 + 2.5x10^-3x9.81)/ πx1000x9.81]^1/3

d = 0.022m

Therefore, the diameter of the ball is 0.022m or 2.2cm

6 0
3 years ago
Please help me with this question guys.
katen-ka-za [31]

Answer:

<em>The average speed is 22.2 km/h</em>

Explanation:

<u>Average Speed</u>

Given an object travels a total distance d and took a total time t, then the average speed is:

\displaystyle \bar v=\frac{d}{t}

The mailman first drives d1=7 km at v1=15 km/h. The time taken to drive is:

\displaystyle t1=\frac{d1}{v1}=\frac{7}{15}=0.467\ h

Then he drives d2=7 km at v2=43 km/h taking a time of:

\displaystyle t2=\frac{d2}{v2}=\frac{7}{43}=0.163\ h

The total time is

t=0.467 h + 0.163 h = 0.63 h

The total distance is

d = 7 km + 7 km = 14 km

The average speed is:

\displaystyle \bar v=\frac{14}{0.63}=22.2\ km/h

The average speed is 22.2 km/h

7 0
2 years ago
Will mark brainliest
Genrish500 [490]

The blue line with triangles are a cold front. The red line with half circles are a warm front. The black H is high pressure. The black L is low pressure. The black dot is overcast skies. Finally the white circle with a line down and 4 lines off of that line is 40-knot wind

idk and dont understand

8 0
3 years ago
I just wanted to say I hope everyone has the best of days! And always remember that you are loved not just by me but many people
erastova [34]

Answer:

okay......

Explanation:

thanks bro/gal

5 0
2 years ago
Read 2 more answers
A proton moves with a velocity of v with arrow = (4î − 6ĵ + k) m/s in a region in which the magnetic field is B with arrow = (î
gavmur [86]

Answer:

F = [(6.4 × 10⁻¹⁹)î + (8.0 × 10⁻¹⁹)ĵ + (22.4 × 10⁻¹⁹)k] N

Magnitude of F = (2.466 × 10⁻¹⁸) N

Explanation:

The magnetic force, F, on a given charge, q, moving with velocity, v, in a magnetic field, B, is given as the vector product

F = qv × B

where v = (4î − 6ĵ + k) m/s

B = (î + 2ĵ − k) T

The particle is a proton, hence,

q = (1.602 × 10⁻¹⁹) C

F = qv × B = q (v × B)

(v × B) is given as (4î − 6ĵ + k) × (î + 2ĵ − k)

The cross product is evaluated as a determinant of

| î ĵ k |

|4 -6 1 |

|1 2 -1 |

î [(-6)(-1) - (2)(1)] - ĵ [(4)(-1) - (1)(1)] + k [(4)(2) - (-6)(1)]

î (6 - 2) - ĵ (-4 - 1) + k (8 + 6) = (4î + 5ĵ + 14k)

(v × B) = (4î + 5ĵ + 14k)

F = q (v × B) = (1.6 × 10⁻¹⁹) (4î + 5ĵ + 14k)

F = [(6.408 × 10⁻¹⁹)î + (8.01 × 10⁻¹⁹)ĵ + (22.428 × 10⁻¹⁹)k] N

Magnitude of F =

√[(6.408 × 10⁻¹⁹)² + (8.01 × 10⁻¹⁹)² + (22.428 × 10⁻¹⁹)²]

Magnitude of F = (2.466 × 10⁻¹⁸) N

Hope this Helps!!!

8 0
3 years ago
Read 2 more answers
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