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3 years ago

How many neutrons does element X have if its atomic number is 59 and its mass number is 164?

Physics

1 answer:

Y_Kistochka [10]3 years ago

4 0

The mass is the number of n + p if you subtract p from mass you will find n
164 - 59 = 105

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Who was the man who lived from 460B.C.–370B.C. and was among the first to suggest the idea of atoms?

ch4aika [34]

C. Democritus was a Greek philosopher who defined the concept of atoms and is the only one of the answer choices who lived during that time period.

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3 years ago

Halley’s Comet appears around the sun about once every 76 years. It passes through the part of its orbit nearest the sun in just

Scilla [17]

Answer:

There are actually three, Kepler’s laws that is, of planetary motion: 1) every planet’s orbit is an ellipse with the Sun at a focus; 2) a line joining the Sun and a planet sweeps out equal areas in equal times; and 3) the square of a planet’s orbital period is proportional to the cube of the semi-major axis of its orbit. As it’s the third which is most often used, Kepler’s law usually means Kepler’s third law (of planetary motion).

Explanation:

Kepler's third law would tell us that Halley's comet has an average distance much greater than that of the Earth. However, there is a time in Halley's comet's orbit that brings it closer to the Sun than the Earth. Kepler's third law is a mathematical relation between a planet's period and its average distance.

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3 years ago

10.88 = in scientific notation

rewona [7]

Answer:

1.088x10^1

Explanation:

8 0

3 years ago

A planet of mass 7.00 1025 kg is in a circular orbit of radius 6.00 1011 m around a star. The star exerts a force on the planet

BARSIC [14]

Answer:

A) 1.88 * 10^17 m

B) 1.22 * 10^34 J

C) 1.95 * 10^34 J

Explanation:

Parameters given:

Mass of planet = 7.00 * 10^25 kg

Radius of orbit = 6.00 * 10^11 m

Force exerted on planet = 6.51 * 10^22 N

Velocity of planet = 2.36 * 10^4 m/s

A) The distance traveled by the planet is half of the circumference of the orbit (which is circular).

The circumference of the orbit is

C = 2 * pi * R

R = radius of orbit

C = 2 * 3.142 * 6.0 * 10¹¹

C = 3.77 * 10¹² m

Hence, distance traveled will be:

D = 0.5 * 3.77 * 10¹²

D = 1.88 * 10 ¹² m/s

B) Work done is given as:

W = F * D

W = 652 * 10²² * 1.88 * 10¹¹

W = 1.22 * 10³⁴ J

C) Change in Kinetic energy is given as:

K. E. = 0.5 * m * v²

K. E. = 0.5 * 7 * 10^25 * (2.36 * 10^4)²

K. E. = 1.95 * 10³⁴ J

7 0

3 years ago

Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth

zalisa [80]

Refer to the diagram shown below.

Given:
R = 6.37 x 10⁶ m, the radius of the earth
h = 3.58 x 10⁷ m, the height of the satellite above the earth's surface.
Therefore
R + h = 4.217 x 10⁷ m

In geosynchronous orbit, the period of rotation is 1 day.
Therefore the period is
T = (24 h)*(60 min/h)*(60 s/min) = 86400 s

The angular velocity is
ω = (2π rad)/(86400 s) = 7.2722 x 10⁻⁵ rad/s

Part (a)
The tangential speed is
v = (R+h)*ω
= (4.217 x 10⁷ m)*(7.2722 x 10⁻⁵ rad/s)
= 3066.7 m/s
= 3.067 km/s

Part (b)
The centripetal acceleration is
a = v²/(R+h)
= (3066.7 m/s)²/(4.217 x 10⁷ m)
= 0.223 m/s²

Answers:
(a) The speed is 3.067 km/s
(b) The acceleration is 0.223 m/s²

7 0

3 years ago