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2 years ago

Which of the following statements describes nucleons?

2 answers:

7 0

The correct answer is D.

A nucleon<span> is one of either of the two types of subatomic particles (neutrons and protons) which are located in the nucleus of atoms.
</span>

The total number of nucleon in the nucleus of an atom gives you an idea about the mass of that atom. In fact, one may refer mass number as nucleon number.

Simply put, nucleons are the particles that make nucleus of an atom and are held up together inside the nucleus due to nuclear force.

labwork [276]2 years ago

3 0

Answer:

D. Nucleons include only neutrons and protons.

Explanation:

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A 15 kg bucket of water is suspended by a very light ropewrapped around a solid uniform cylinder 0.300 m in diamter withmass 12.

matrenka [14]

Answer:

Part a)

$T = 42 N$

Part b)

$v_f = 11.8 m/s$

Part c)

$t = 1.7 s$

Part d)

$F = 159.7 N$

Explanation:

Part a)

While bucket is falling downwards we have force equation of the bucket given as

$mg - T = ma$

for uniform cylinder we will have

$TR = I\alpha$

so we have

$T = \frac{1}{2}MR^2(\frac{a}{R^2})$

$T = \frac{1}{2}Ma$

now we have

$mg = (\frac{M}{2} + m)a$

$a = \frac{mg}{(\frac{M}{2} + m)}$

$a = \frac{15 \times 9.81}{(6 + 15)}$

$a = 7 m/s^2$

now we have

$T = \frac{12 \times 7}{2}$

$T = 42 N$

Part b)

speed of the bucket can be found using kinematics

so we have

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(7)(10)$

$v_f = 11.8 m/s$

Part c)

now in order to find the time of fall we can use another equation

$v_f - v_i = at$

$11.8 - 0 = 7 t$

$t = 1.7 s$

Part d)

as we know that cylinder is at rest and not moving downwards

so here we can use force balance

$F = T + Mg$

$F = 42 + (12 \times 9.81)$

$F = 159.7 N$

5 0

2 years ago

A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a di

siniylev [52]

Answer:

1408.685 KN/C

Explanation:

Given:

R = 0.45 m

σ = 175 μC/m²

P is located a distance a = 0.75 m

k = 8.99*10^9

The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

$E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })\\$

part a)

Electric Field strength at point P: a = 0.75 m

$E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C$

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

$E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o$

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

6 0

3 years ago

Which of the following is an example of a lever? knife ramp pencil sharpener wheelbarrow

Kay [80]

I believe it is a knife

5 0

2 years ago

Read 2 more answers

Megan was doing time trials on her bike around a 400m horizontal track. She took 32 seconds to travel 400m. What was her average

Mars2501 [29]

To find average speed, we divide the distance of travel (in this case, 400 metres) by the time she took, 32 seconds. Therefore: 12.5 seconds is her average speed.

5 0

2 years ago

Read 2 more answers

An exoplanet with one half of Earth's mass and 50% of Earth's radius is discovered.

Georgia [21]

Answer:

The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet

Explanation:

The given parameters are;

The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M

The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R

The weight of the cadet on Earth = 800 N

$The \ weight, W =G\dfrac{M \times m}{R^{2}} = 800 \ N$

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

$W_1 =G\dfrac{\dfrac{M}{2} \times m}{ \left ( \dfrac{R}{2} \right ) ^{2}} = G\dfrac{\dfrac{M}{2} \times m \times 4}{ R ^{2}} = 2 \times G \times \dfrac{M \times m}{R^{2}} = 2 \times 800 \, N = 1,600 \, N$

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.

7 0

2 years ago