Answer:
0.25 m.
Explanation:
We'll begin by calculating the spring constant of the spring.
From the diagram, we shall used any of the weight with the corresponding extention to determine the spring constant. This is illustrated below:
Force (F) = 0.1 N
Extention (e) = 0.125 m
Spring constant (K) =?
F = Ke
0.1 = K x 0.125
Divide both side by 0.125
K = 0.1/0.125
K = 0.8 N/m
Therefore, the force constant, K of spring is 0.8 N/m
Now, we can obtain the number in gap 1 in the diagram above as follow:
Force (F) = 0.2 N
Spring constant (K) = 0.8 N/m
Extention (e) =..?
F = Ke
0.2 = 0.8 x e
Divide both side by 0.8
e = 0.2/0.8
e = 0.25 m
Therefore, the number that will complete gap 1is 0.25 m.
A) gases particles are far apart
' W ' is the symbol for 'Watt' ... the unit of power equal to 1 joule/second.
That's all the physics we need to know to answer this question.
The rest is just arithmetic.
(60 joules/sec) · (30 days) · (8 hours/day) · (3600 sec/hour)
= (60 · 30 · 8 · 3600) (joule · day · hour · sec) / (sec · day · hour)
= 51,840,000 joules
__________________________________
Wait a minute ! Hold up ! Hee haw ! Whoa !
Excuse me. That will never do.
I see they want the answer in units of kilowatt-hours (kWh).
In that case, it's
(60 watts) · (30 days) · (8 hours/day) · (1 kW/1,000 watts)
= (60 · 30 · 8 · 1 / 1,000) (watt · day · hour · kW / day · watt)
= 14.4 kW·hour
Rounded to the nearest whole number:
14 kWh
Answer:
kinetic energy will change by a factor of 1/2
Option C) 1/2 is the correct answer
Explanation:
Given the data in the question;
we know that;
Kinetic energy = 1/2.mv²
given that mass of the object is doubled; m1 = 2m
speed is halved; v1 = V/2
Now, New kinetic energy will be; 1/2.m1v1²
we substitute
Kinetic Energy = 1/2 × 2m × (v/2)²
Kinetic Energy = 1/2 × 2m × (v²/4)
Kinetic Energy = 1/2 × m × (v²/2)
Kinetic Energy = 1/2 [ 1/2mv² ]
Kinetic Energy = 1/2 [ KE ]
Therefore; kinetic energy will change by a factor of 1/2
Option C) 1/2 is the correct answer
Missing question:
"Determine (a) the astronaut’s orbital speed v and (b) the period of the orbit"
Solution
part a) The center of the orbit of the third astronaut is located at the center of the moon. This means that the radius of the orbit is the sum of the Moon's radius r0 and the altitude (

) of the orbit:

This is a circular motion, where the centripetal acceleration is equal to the gravitational acceleration g at this altitude. The problem says that at this altitude,

. So we can write

where

is the centripetal acceleration and v is the speed of the astronaut. Re-arranging it we can find v:

part b) The orbit has a circumference of

, and the astronaut is covering it at a speed equal to v. Therefore, the period of the orbit is

So, the period of the orbit is 2.45 hours.