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iris [78.8K]
3 years ago
6

the acceleration due to gravity on the moon is 1.6 m/s^2 what is the gravitational potential energy of a 1200 kg lander resting

on top of a 350m hill
Physics
1 answer:
Mumz [18]3 years ago
5 0

<u>Answer</u>

672,000 Joules

<u>Explanation</u>

Gravitational potential energy (P.E) is the energy possessed by a body that is at a potential height from the ground.

IT is calculated by the formula;

P.E = mgh

Where m ⇒ mass

            g ⇒ acceleration due to gravity

            h ⇒ height from trhe ground.

P.E = 1200 × 1.6 × 350

      = 672,000 Joules.

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What is not an essential part of isolated soldier guidance
mel-nik [20]

Answer:

Option B, visual sightings

Explanation:

Options for the question are

A) accountability mechanism

B) visual sightings

C)  intelligence

D) surveillance

E) reconnaissance  operations, or communications

Solutions

The Fundamentals of Army Personnel Recovery (PR) outlines certain circumstances under which a person has to undergo survival situation thereby taking the necessary steps to avoid capture and return safely to their respective unit.

An isolated soldier is expected to know where they are, upcoming route and rally points. They are supposed to know the near and far recognition signals, recovery site protocols, challenge and password etc. A proper preparation is to be done for this including planning, medicines, kits, etc.  

Visual sightings is not an essential part of isolated PR

Hence, option B is correct

4 0
3 years ago
What is the purpose of the background research step of the scientific method?
Lilit [14]
Background research is used to familiarize yourself with the topic or problem you're conducting your experiment on and help you form an educated guess explaining or solving it that will become your hypothesis. 
3 0
2 years ago
a nonrelativistic proton is confined to a length of 2.0 pm on the x-axis. what is the kinetic energy of the proton if its speed
Elina [12.6K]

Answer:

K = 1.29eV

Explanation:

In order to calculate the kinetic energy of the proton you first take into account the uncertainty principle, which is given by:

\Delta x \Delta p\geq \frac{h}{4\pi}      (1)

Δx : uncertainty of position = 2.0pm = 2.0*10^-12m

Δp: uncertainty of momentum = ?

h: Planck's constant = 6.626*10^-34 J.s

You calculate the minimum possible value of Δp from the equation (1):

\Delta p=\frac{h}{4\pi \Delta x}=\frac{6.626*10^{-34}J.s}{4\pi(2.0*10^{-12}m)}\\\\\Delta p=2.63*10^{-23}kg.\frac{m}{s}

The minimum kinetic energy is calculated by using the following formula:

k=\frac{(\Delta p)^2}{2m}       (2)

m: mass of the proton = 1.67*10^{-27}kg

k=\frac{(2.63*10^{-23}kgm/s)^2}{2(1.67*10^{-27}kg)}=2.08*10^{-19}J

in eV you have:

2.08*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.29eV

The kinetic energy of the proton is 1.29eV

7 0
3 years ago
A 2000 kg car experiences a constant braking
bezimeni [28]
Solving for acceleration:

F = ma
10000 = 2000*a
a = 5 m/s^2

Solving for velocity:

Vf = Vi + a(t)
0 = Vi - (5)(6)
Vi = 30 m/s

Solving for displacement or distance:

S = Vi*t - 1/2 * a * t^2
S = 30(6) - 1/2 (5)(6)^2
S = 90m

The car would have traveled 90m before coming to rest.
5 0
3 years ago
A meter stick swinging from one end oscillates with a frequency f0. What would be the frequency, in terms of f0 , if the bottom
snow_lady [41]

To solve this problem we will proceed to define the Period of a stick, then we will define the frequency, which is the inverse of the period. We will compare the change suffered by the new length and replace that value. The Time period of meter stick is

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

Here,

L = Length

g = Gravity

At the same time the frequency is

f = \frac{1}{T}

Therefore the frequency in Terms of the Period is

f_0 = \frac{1}{2\pi}\sqrt{\frac{g}{L}}

If bottom third were cut off then the new length is

L' = \frac{2}{3} L

Replacing this value at the new frequency we have that,

f ' = \frac{1}{2\pi} \sqrt{\frac{3g}{2L}}

f' = \sqrt{\frac{3}{2}}(\frac{1}{2\pi}\sqrt{\frac{g}{L}})

Finally,

\therefore f' = \sqrt{\frac{3}{2}}f_0

3 0
3 years ago
Read 2 more answers
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