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3 years ago

Which body have a period of revolution of 29.6 days?

2 answers:

6 0

It would be C the moon because the total of all positions (new moon, first quarter, third quarter, full moon, waxing crescent, waning crescent, waxing gibbous, waning gibbous) of the moon would take about 29.6 days on a regular calendar. I hope this helps. :)

adelina 88 [10]3 years ago

4 0

The moon is the closest one on the list. Its period of revolution around the Earth is 27.32 days. (But the length of the cycle of phases, like from one Full Moon to the next one, is 29.53 days.)

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A physics student uses a spring loaded launcher to fire a 58 gram projectile horizontally. The projectile leaves the launcher at

timofeeve [1]

Answer:

no answer

Explanation:

the question does not sk what to find

3 0

3 years ago

What speed would a fly with a mass of 0.55g need in order to have a kinetic energy of 7.6 •10^4 j?

masya89 [10]

Answer:

16613 m/s

Explanation:

Given that

mass of the fly, m = 0.55 g = 0.55*10^-3 kg

Kinetic Energy of the fly, E = 7.6*10^4 J

Speed of the fly, v = ? m/s

We know that the Kinetic Energy is that energy that an object, in this case, the fly, possesses due to its motion.

The Kinetic Energy, KE of any object is represented by the formula

KE = 1/2 * m * v²

If we substitute the values in the relation, we have,

7.6*10^4 = 1/2 * 0.55*10^-3 * v²

v² = (15.2*10^4) / 0.55*10^-3

v² = 2.76*10^8

v = √2.76*10^8

v = 16613 m/s

Thus, the fly would need a speed of 16.6 km/s in order to have a Kinetic Energy of 7.6*10^4 J

7 0

3 years ago

The radioactive 60co isotope is used in nuclear medicine to treat certain types of cancer. Calculate the wavelength and frequenc

Ivanshal [37]

1. Frequency: $3.23\cdot 10^{20} Hz$

The energy given is the energy per mole of particles:

$E=1.29\cdot 10^{11} J/mol$

1 mole contains a number of Avogadro of particles, $N_A$ , equal to

$N_A=6.022\cdot 10^{23}$ particles

So, by setting the following proportion, we can calculate the energy of a single photon:

$1.29 \cdot 10^{11} J/mol : 6.022 \cdot 10^{23} ph/mol = E_1 : 1 ph\\E_1 = \frac{(1.29\cdot 10^{11} J/mol)(1 ph)}{6.022\cdot 10^{23} ph/mol}=2.14\cdot 10^{-13} J$

This is the energy of a single photon; now we can calculate its frequency by using the formula:

$E_1 = hf$

where

$h=6.63\cdot 10^{-34} Js$ is the Planck's constant

f is the photon frequency

Solving for f, we find

$f=\frac{E_1}{h}=\frac{2.14\cdot 10^{-13} J}{6.63\cdot 10^{-34} Js}=3.23\cdot 10^{20} Hz$

2. Wavelength: $9.29\cdot 10^{-13} m$

The wavelength of the photon is given by the equation:

$\lambda=\frac{c}{f}$

where

$c=3\cdot 10^8 m/s$

is the speed of the photon (the speed of light). Substituting,

$\lambda=\frac{3 \cdot 10^8 m/s}{3.23\cdot 10^{20} Hz}=9.29\cdot 10^{-13} m$

6 0

3 years ago

How do you find the oscillation period in seconds for different pendulum lengths?

GalinKa [24]

To find:

The equation to find the period of oscillation.

Explanation:

The period of oscillation of a pendulum is directly proportional to the square root of the length of the pendulum and inversely proportional to the square root of the acceleration due to gravity.

Thus the period of a pendulum is given by the equation,

$T=2\pi\sqrt{\frac{L}{g}}$

Where L is the length of the pendulum and g is the acceleration due to gravity.

On substituting the values of the length of the pendulum and the acceleration due to gravity at the point where the period of the pendulum is being measured, the above equation yields the value of the period of the pendulum.

Final answer:

The period of oscillation of a pendulum can be calculated using the equation,

$T=2\pi\sqrt{\frac{L}{g}}$

3 0

1 year ago

HELP I RLLY NEED HELP WITH THIS PLEASE HELP ME WHAT IS PHYSICAL SCIENCE ANYBODY???

Paul [167]

The sciences concerned with the study of inanimate natural objects, including physics, chemistry, astronomy, and related subjects.

7 0

3 years ago