That depends on what "objects" refer to.
If "objects" refer to the ones on Earth, then it is TRUE. These objects are in motion relative to the sun same as with Earth itself.
<span>If "objects" refer to the ones outside earth, then it may be a TRUE or FALSE depending on how far or near they are from the sun.</span>
Answer:
<h2>
<u>Joule</u><u>:</u></h2>
1 Joule of work is said to be done when a force of 1 Newton is applied to move/displace a body by 1 metre.
1 Joule= 1 Newton × 1 metre
1 Newton is the amount of force required to accelerate body of mass 1 kg by 1m/s²
So units of N is kgm/s²
So,
1 Joule
=1kgm/s² × m
=1kgm²/s²
<h2><u>Erg</u><u>:</u></h2>
1 erg is the amount of work done by a force of 1 dyne exerted for a distance of one centimetre.
1 Erg =1 Dyne × 1 cm
1 dyne is the force required to cause a mass of 1 gram to accelerate at a rate of 1cm/s².
1 Erg=1 gmcm/s² × cm
1 Erg=1 gmcm/s² × cm=1gmcm²/s²
this is what you need to convert 1gmcm²/s² to 1kgm²/s²
<h3><u>
what you need to know for conversion</u></h3>
[1gm=0.001kg
1cm²
=1cm ×1cm
=0.01 m × 0.01 m
=0.0001m²
second remains constant
]
So,
1gmcm²/s²
=0.001kg×0.0001m²/s²
=0.001kg×0.0001m²/s² =0.0000001kgm²/s²
Hence,
<h3>
<u>1 Erg</u><u>=</u><u>0.0000001</u><u> </u><u>Joule</u></h3><h3>
<u>1</u><u> </u><u>Joule</u><u>=</u><u>1</u><u>0</u><u>,</u><u>0</u><u>0</u><u>0</u><u>,</u><u>0</u><u>0</u><u>0</u><u> </u><u>Erg</u></h3>
<h2>⇒15 J=15×10000000 Erg</h2><h2> =150000000 Erg</h2><h2>
=1.5×10⁶ Erg</h2>
Answer:25.61 m/s
Explanation:
Given
truck is moving eastbound with a velocity of 16 m/s
Velocity of truck 
SUV is moving south with a velocity of 20 m/s
Velocity of SUV in vector form 
Velocity of truck relative to the SUV
Magnitude of relative velocity is
The total gauge pressure at the bottom of the cylinder would
simply be the sum of the pressure exerted by water and pressure exerted by the
oil.
The formula for calculating pressure in a column is:
P = ρ g h
Where,
P = gauge pressure
ρ = density of the liquid
g = gravitational acceleration
h = height of liquid
Adding the two pressures will give the total:
P total = (ρ g h)_water + (ρ g h)_oil
P total = (1000 kg / m^3) (9.8 m / s^2) (0.30 m) + (900 kg /
m^3) (9.8 m / s^2) (0.4 - 0.30 m)
P total = 2940 Pa + 882 Pa
P total = 3,822 Pa
Answer:
The total gauge
pressure at the bottom is 3,822 Pa.
