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3 years ago

Different _ of an element have different numbers of neutrons?

Physics

2 answers:

natali 33 [55]3 years ago

4 0

Answer:

Isotopes

Explanation:

Your welcome

uranmaximum [27]3 years ago

4 0

Different isotopes of an element have different numbers of neutrons.

<u>Explanation: </u>

There are various elements existing in the Universe. Each element is made up of atoms and there can be multiple types of atoms of an element based on the numbers and configurations of its sub-particles i.e. protons, electrons and neutrons. These are known as Isotopes.

Isotopes are the specific set of atoms of an element that have same proton and electron counts but different neutron counts. The isotopes have same protons and electrons count, thus having the same atomic number but, different atomic mass number because of different number of neutrons.

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. An iron cube (ρ = 7860 kg/m3) having sides of 1 cm is submerged in open fresh water (ρ = 1000kg/m3) with its top surface at a

Korolek [52]

Answer:

$\Delta P = 98.07\,Pa$

Explanation:

The difference between top and bottom surfaces is computed by the following hydrostatic equivalence:

$\Delta P = \rho_{w}\cdot g \cdot \Delta h$

$\Delta P = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.01\,m)$

$\Delta P = 98.07\,Pa$

5 0

3 years ago

A cable is 100-m long and has a cross-sectional area of 1.0 mm2. A 1000-N force is applied to stretch the cable. Young's modulus

Blizzard [7]

Answer:

1 m

Explanation:

L = 100 m

A = 1 mm^2 = 1 x 10^-6 m^2

Y = 1 x 10^11 N/m^2

F = 1000 N

Let the cable stretch be ΔL.

By the formula of Young's modulus

$Y=\frac{F\times L}{A\times\Delta L}$

$\Delta L=\frac{F\times L}{A\times\Y}$

$\Delta L=\frac{1000\times 100}{10^{-6}\times10^{11}}$

ΔL = 1 m

Thus, the cable stretches by 1 m.

4 0

3 years ago

Which of the following is most useful to determine how much energy is being used by a circuit in a given amount of time?

user100 [1]

Answer:

The answer is A.

Explanation:

5 0

3 years ago

A steady beam of alpha particles (q = + 2e, mass m = 6.68 × 10-27 kg) traveling with constant kinetic energy 22 MeV carries a cu

cluponka [151]

Answer:

Explanation:

q = 2e = 3.2 x 10^-19 C

mass, m = 6.68 x 10^-27 kg

Kinetic energy, K = 22 MeV

Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A

(a) time, t = 2.8 s

Let N be the alpha particles strike the surface.

N x 2e = q

N x 3.2 x 10^-19 = i t

N x 3.2 x 10^-19 = 0.27 x 10^-6 x 2.8

N = 2.36 x 10^12

(b) Length, L = 16 cm = 0.16 m

Let N be the alpha particles

K = 0.5 x mv²

22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²

v² = 1.054 x 10^15

v = 3.25 x 10^7 m/s

So, N x 2e = i x t

N x 2e = i x L / v

N x 3.2 x 10^-19 = 2.7 x 10^-7 x 0.16 / (3.25 x 10^7)

N = 4153.85

Kinetic energy = Potential energy

K = q x V

22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V

V = 1.17 x 10^7 V

5 0

3 years ago

An artificial satellite is moving

Leto [7]

Explanation:

Distance covered by the satellite in 24 hours

s=2πr

=2×3.14×42250=265464.58 km

Therefore speed of satellite,

time taken

distance travelled

24×60×60

265464.58

=3.07 km s

−1

5 0

3 years ago

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