All categories

3 years ago

On a planet whose radius is 1.1 x 10^7 m, the acceleration due to gravity is 41 m/s^2. What is the mass of the planet (in kg)?

Physics

1 answer:

MA_775_DIABLO [31]3 years ago

3 0

Answer:

M = 7.43×10^25 kg

Explanation:

let the mass of the planet be M and r be the radius of the planet and g be the gravitational acceleretion.

then:

the gravitational acceleration is given by:

g = G×M/(r^2)

M = g×r^2/G

= [(41)×(1.1×10^7)^2]/(6.67408×10^-11)

= 7.43×10^25 kg

therefore, the mass of the planet is 7.43×10^25 kg.

You might be interested in

If upthrust and weight of a liquid displaced when a solid is immersed in it are U and W respectively, then

Stels [109]

Definitely 4 because it’s u for I hi ub u

5 0

2 years ago

Define a dipole. Hence, write the expression for calculating the electric moment of a dipole

Aleksandr-060686 [28]

Answer:

Explanation:

An electric dipole is formed by two point charges +q and −q connected by a vector a. The electric dipole moment is defined as p = qa

4 0

1 year ago

An atom with seven valence electrons would most likely lose an electron to become stable true or fasle

mel-nik [20]

Answer:

false

Explanation:

since it has seven valence electrons it means it's a non metal and non metals always gain electrons.

5 0

2 years ago

If electrons of energy 12.8 ev are incident on a gas of hydrogen atoms in their ground state, what are the energies of the photo

timama [110]

A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs.

4 0

3 years ago

) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note tha

vaieri [72.5K]

(a) 0.0021 s, 2926.5 rad/s

The frequency of the B note is

$f= 466 Hz$

The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{466 Hz}=0.0021 s$

The angular frequency instead is given by

$\omega = 2\pi f$

And substituting

f = 466 Hz

We find

$\omega = 2\pi (466 Hz)=2926.5 rad/s$

(b) 20 Hz, 125.6 rad/s

In this case, the period of the sound wave is

T = 50.0 ms = 0.050 s

So the frequency is equal to the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{0.050 s}=20 Hz$

While the angular frequency is given by:

$\omega = 2\pi f = 2 \pi (20 Hz)=125.6 rad/s$

(c) $4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s$

The minimum angular frequency of the light wave is

$\omega_1 = 2.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s$

The maximum angular frequency of the light wave is

$\omega_2 = 4.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{7.48\cdot 10^{14}Hz}=1.34\cdot 10^{-15}s$

(d) $2.0\cdot 10^{-7}s, 3.14\cdot 10^{7} rad/s$

In this case, the frequency is

$f=5.0 MHz = 5.0 \cdot 10^6 Hz$

So the period in this case is

$T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6 Hz}=2.0 \cdot 10^{-7} s$

While the angular frequency is given by

$\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s$

7 0

3 years ago