The legend is that he discovered gravity when an apple feel on his head. I don’t know what the true story is, but that’s what I’ve heard so maybe A??
Although, I’m pretty sure it could also be C
So... between A and C, however, I don’t want you to get it wrong so I would recommend getting another opinion
Hope this helps!
"60 kg" is not a weight. It's a mass, and it's always the same
no matter where the object goes.
The weight of the object is
(mass) x (gravity in the place where the object is) .
On the surface of the Earth,
Weight = (60 kg) x (9.8 m/s²)
= 588 Newtons.
Now, the force of gravity varies as the inverse of the square of the distance from the center of the Earth.
On the surface, the distance from the center of the Earth is 1R.
So if you move out to 5R from the center, the gravity out there is
(1R/5R)² = (1/5)² = 1/25 = 0.04 of its value on the surface.
The object's weight would also be 0.04 of its weight on the surface.
(0.04) x (588 Newtons) = 23.52 Newtons.
Again, the object's mass is still 60 kg out there.
___________________________________________
If you have a textbook, or handout material, or a lesson DVD,
or a teacher, or an on-line unit, that says the object "weighs"
60 kilograms, then you should be raising a holy stink.
You are being planted with sloppy, inaccurate, misleading
information, and it's going to be YOUR problem to UN-learn it later.
They owe you better material.
It depends because it’s might be lolilolololol 21212132
Answer:
Explanation:
This does not violate Newton's 1st law because the net force would still be 0 in order to produce uniform motion (aka constant velocity). The other forces acting on the vehicles is air resistance which is non-zero. So we need car internal force to counter balance this force, which require extra gas for the car.
Use VFR1 = VFR2 to discover the velocity at in the hose VFR =
A * V
D hose =10 * D nozzle, R hose = 5 * D nozzle
Area of a circle = πR^2
Area h=3.14*25*D^2 = 75.5D^2
(Radius=Diameter/2) area n = 3.14*(D^2/4) = .785D^2
Use VFR = VFR v2 = 0.4m/s
0.4*.785D^2 = 75.5*D^2* v1 D^2
= .314 =75.5*V1
v1 = 0.004m/s
Now we have the velocity, we can use Bernoulli's equation.
P1+ρgh1+ρV1^2 /2 = constant
There is no atmospheric pressure before so the P1= the gauge
pressure at the pump, let’s call the height of the hose 0m and the height of
the nozzle 1m so the is no ρgh1 Likewise, there is only atmospheric pressure at
the nozzle which is 100000 PA, and lastly the density ρ of water is 1000 KG/M^3
Pg + 1000*.004^2/2 = 100000+1000*9.8*1+ 1000*0.4^2/2
Pg + .008= 100000+9800+80
Pg+.008= 109880
Pg=109880.008 PA
