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3 years ago

An element’s atomic number is 35. how many portions would an atom this element have

Physics

1 answer:

kotykmax [81]3 years ago

4 0

35 protons are present in an element whose atomic number is 35.

<u>Explanation: </u>

In an atom of an element, the number of protons = atomic number

Number of protons = number of electrons

Mass Number = number of protons + number of neutrons

Hence, an atom with atomic number 35 will have 35 protons.

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If you weigh 690 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun an

lana [24]

Answer:

Explanation:

Given that,

Mass of star M(star) = 1.99×10^30kg

Gravitational constant G

G = 6.67×10^−11 N⋅m²/kg²

Diameter d = 25km

d = 25,000m

R = d/2 = 25,000/2

R = 12,500m

Weight w = 690N

Then, the person mass which is constant can be determined using

W =mg

m = W/g

m = 690/9.81

m = 70.34kg

The acceleration due to gravity on the surface of the neutron star is can be determined using

g(star) = GM(star)/R²

g(star) = 6.67×10^-11 × 1.99×10^30 / 12500²

g (star) = 8.49 × 10¹¹ m/s²

Then, the person weight on neutron star is

W = mg

Mass is constant, m = 70.34kg

W = 70.34 × 8.49 × 10¹¹

W = 5.98 × 10¹³ N

The weight of the person on neutron star is 5.98 × 10¹³ N

5 0

3 years ago

An object of mass 30 kg is falling in air and experiences a force due to air resistance of 50 newtons. Calcuate the acceleration

rjkz [21]

Answer:

this answer you question

6 0

2 years ago

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical ener

defon

Complete question:

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find

(a) the force constant of the spring and (b) the amplitude of the motion.

Answer:

(a) the force constant of the spring = 47 N/m

(b) the amplitude of the motion = 0.292 m

Explanation:

Given;

mass of the spring, m = 200g = 0.2 kg

period of oscillation, T = 0.410 s

total mechanical energy of the spring, E = 2 J

The angular speed is calculated as follows;

$\omega = \frac{2\pi}{T} \\\\\omega = \frac{2\pi}{0.41} \\\\\omega = 15.33 \ rad/s$

(a) the force constant of the spring

$\omega = \sqrt{\frac{k}{m} } \\\\\omega^2 = \frac{k}{m} \\\\k = m \omega^2\\\\k = (0.2)(15.33)^2\\\\k = 47 \ N/m$

(b) the amplitude of the motion

E = ¹/₂kA²

2E = kA²

A² = 2E/k

$A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2\times 2}{47} }\\\\A = 0.292 \ m$

7 0

2 years ago

The angular momentum of an electron in each of its orbits must be some multiple of:

Nastasia [14]

The Bohr's proposal for the angular momentum of an electron in Bohr's model of the hydrogen atom is:

L=(n*h)/(2π), where n is the number of the energy level and h is the Planck's constant. This equation shows us the quantization of angular momentum of the electron. So the correct answer is the second one: Planck's constant.

7 0

2 years ago

A 0.60 kg rubber ball has a speed of 2.0 m/s at point A, and kinetic energy of 7.5 J at point

aliina [53]

<span>Let's first off calculate the kinetic energy using the formula 1/2MV^2. Where the mass, M, is 0.6Kg. And speed, V, is 2. Hence we have 1/2 * 0.6 * 2^2 = 1.2J. Since kinetic energy is energy due to motion; hence at point B the rubber has a KE of 1.2J and not 7.5J. So I would say that only the Mass and speed is actually true; While it's kinetic energy is not true.</span>

7 0

3 years ago