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3 years ago

15

The acceleration due to gravity on the Moon's surface is

1 answer:

Molodets [167]3 years ago

7 0

Answer:

50 lb

Explanation:

Given,

The weight of astronaut's life support backpack on Earth (w) = 300 lb

Acceleration due to gravity on Earth (g) = 9.8 m/s²

Acceleration due to gravity on Moon = g'

$g'=\frac{g}{6}$

We know that weight of an object on Earth is,

$w = m\times g$

$m = \frac{w}{g}$

Similarly, weight on Moon will be

$w' = m\times g'$

$w' = \frac{w}{g}\times\frac{g}{6}$

$w' = \frac{300}{6}$

$w' = 50$

Thus the astronaut's life support backpack will weigh 50 lb on Moon.

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List two animals that migrate to the clearing

nalin [4]

Sea animals and birds

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3 years ago

An object falls from rest on a high tower and takes 5.0 s to hit the ground. Calculate the object's position from the top of the

Lena [83]

Answer:

After 1 sec = 4.9 m

After 2 sec = 19.6 m

After 3 sec = 44.1 m

After 4 sec = 78.4 m

After 5 sec = 122.5 m

Explanation:

After 1 sec:

<em>u=0m/s t=1 s a=9.8m/s²</em>

s = ut + (1/2)at²

=0(1) + (1/2)(9.8)(1²) = 4.9m

After 2 sec:

<em>u=0m/s t=2 s a=9.8m/s²</em>

s = ut + (1/2)at²

=0(2) + (1/2)(9.8)(2²) = 19.6m

After 3 sec:

<em>u=0m/s t=3 s a=9.8m/s²</em>

s = ut + (1/2)at²

=0(3) + (1/2)(9.8)(3²) = 44.1m

After 4 sec:

<em>u=0m/s t=4 s a=9.8m/s²</em>

s = ut + (1/2)at²

=0(4) + (1/2)(9.8)(4²) = 78.4m

After 5 sec:

<em>u=0m/s t=5 s a=9.8m/s²</em>

s = ut + (1/2)at²

=0(5) + (1/2)(9.8)(5²) = 122.5m

7 0

3 years ago

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

$Emf = -N\frac{\Delta \phi}{\Delta t}$

Where $N$ is the number of turns

$\Delta \phi$ is the change in magnetic flux

$\Delta t$ is the time interval

The change in magnetic flux is given by

$\Delta \phi = \phi _{f} - \phi _{i}$

Where $\phi _{f}$ is the final magnetic flux

and $\phi _{i}$ is the initial magnetic flux

Magnetic flux is given by the formula

$\phi = BAcos(\theta)$

Where $B$ is the magnetic field

$A$ is the area

and $\theta$ is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, $\theta = 0^{o}$

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

$A = \pi r^{2}$

∴ $A = \pi (0.15)^{2}$

$A = 0.0225\pi$

∴ $\phi_{i} = (1.5)(0.0225\pi)cos(0^{o} )$

$\phi_{i} = (1.5)(0.0225\pi)$

( NOTE: $cos (0^{o}) = 1$ )

$\phi_{i} = 0.03375\pi$ Wb

For $\phi_{f}$

The field pointed upwards, that is $\theta = 90^{o}$ . Since $cos (90^{o}) = 0$

Then

$\phi_{f} = 0$

Hence,

$\Delta \phi = 0- 0.03375\pi$

$\Delta \phi = - 0.03375\pi$

From the question

$\Delta t = 2.8 ms = 2.8 \times 10^{-3} s$

Here, $N$ = 1

Hence,

$Emf = -N\frac{\Delta \phi}{\Delta t}$ becomes

$Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }$

$Emf = 37.9 V$

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

5 0

3 years ago

What causes the formation of ocean currents?

aliina [53]

Ocean currents can be generated by wind, density differences in watermasses caused by temperature and salinity variations, gravity, and events such as earthquakes

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3 years ago

Potassium is a crucial element for the healthy operation of the human

Degger [83]

Answer:

1

The mass of the Potassium-40 is $m_{40}} = 2.88*10^{-6} kg$

2

The Dose per year in Sieverts is $Dose_s = 26.4 *10^{-10}$

Explanation:

From the question we are told that

The isotopes of potassium in the body are Potassium-39, Potassium-40, and Potassium- 41

Their abundance is 93.26%, 0.012% and 6.728%

The mass of potassium contained in human body is $m = 3.0 g = \frac{3}{1000} = 0.0003 \ kg$ per kg of the body

The mass of the first body is $m_1 = 80 \ kg$

Now the mass of potassium in this body is mathematically evaluated as

$m_p = m * m_1$

substituting value

$m_p = 80 * 0.0003$

$m_p =0.024 kg$

The amount of Potassium-40 present is mathematically evaluated as

$m_{40}} =$ 0.012% * 0.024

$m_{40}} = \frac{0.012}{100} * 0.024$

$m_{40}} = 2.88*10^{-6} kg$

The dose of energy absorbed per year is mathematically represented as

$Dose = \frac{E}{m_1}$

Where E is the energy absorbed which is given as $E = 1.10 MeV = 1.10 * 10^6 * 1.602*10^{-19}$

Substituting value

$Dose = \frac{ 1.10 * 10^6 * 1.602*10^{-19}}{80}$

$Dose = 22*10^{-10} J/kg$

The Dose in Sieverts is evaluated as

$Dose_s = REB * Dose$

$Dose_s = 1.2 * 22*10^{-10}$

$Dose_s = 26.4 *10^{-10}$

3 0

3 years ago