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Leno4ka [110]
3 years ago
9

Please help me!!!!!!

Physics
1 answer:
Semenov [28]3 years ago
5 0

Answer:

the answer will be radio because has the longest wavelength .

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An unknown element is found to have two naturally occurring isotopes, 200X and 210X with atomic masses of 200.028 and 210.039 re
Kruka [31]

Answer:

The average atomic mass of X is 206.0346

Explanation:

Atomic mass of 200X = 200.028

% abundance of 200X = 40% = 40/100 = 0.4

Atomic mass of 210X = 210.039

% abundance of 210X = 100% - 40% = 60% = 60/100 = 0.6

Average atomic mass of X = (0.4×200.028) + (0.6×210.039) = 80.0112 + 126.0234 = 206.0346

5 0
4 years ago
An atomic nucleus is composed of A) protons. B) protons and neutrons. C) protons and electrons. D) protons, neutrons, and electr
8090 [49]

Answer:

B

Explanation:

The nucleus is made up of protons and neutrons, and electrons float around the neucleus in the electron cloud.

3 0
3 years ago
Read 2 more answers
Which is a disadvantage of using wind as an energy source?
lesantik [10]

Answer:

it produces

Explanation:

a lot of waste

5 0
3 years ago
Read 2 more answers
The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5
nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

G\frac{mM}{R^2}=m\frac{v^2}{R}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational constant

m=11,100 kg is the mass of the telescope

M=5.97\cdot 10^{24} kg is the mass of the Earth

R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s

6 0
3 years ago
A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the
kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring  73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E  =  2000 N/C

Electric field due tor ring is guiven as

E = \frac{KQx}{[x^2+ R^2]^{3/2}}

E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}

E1 = 3714.672 N/C

electric field due to point charge q

E  =\frac[kq}{x^2}

E = \frac{9\times 10^9 \times q}{0.70^2}

E2 = 1.837\times 10^{10}\times q

now the eelctric charge at point P is

E = E1 + E22000 =  3714.672 + 1.837\times 10[10} \times q

solving for q

q = - 93.334 nC

7 0
4 years ago
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