Density = (mass) / (volume)
4,000 kg/m³ = (mass) / (0.09 m³)
Multiply each side
by 0.09 m³ : (4,000 kg/m³) x (0.09 m³) = mass
mass = 360 kg .
Force of gravity = (mass) x (acceleration of gravity)
= (360 kg) x (9.8 m/s²)
= (360 x 9.8) kg-m/s²
= 3,528 newtons .
That's the force of gravity on this block, and it doesn't matter
what else is around it. It could be in a box on the shelf or at
the bottom of a swimming pool . . . it's weight is 3,528 newtons
(about 793.7 pounds).
Now, it won't seem that heavy when it's in the water, because
there's another force acting on it in the upward direction, against
gravity. That's the buoyant force due to the displaced water.
The block is displacing 0.09 m³ of water. Water has 1,000 kg of
mass in a m³, so the block displaces 90 kg of water. The weight
of that water is (90) x (9.8) = 882 newtons (about 198.4 pounds),
and that force tries to hold the block up, against gravity.
So while it's in the water, the block seems to weigh
(3,528 - 882) = 2,646 newtons (about 595.2 pounds) .
But again ... it's not correct to call that the "force of gravity acting
on the block in water". The force of gravity doesn't change, but
there's another force, working against gravity, in the water.
Answer:
9] V = D ÷ T
Take any distance value from the graph and its relevant time.
V = 4 ÷ 2
V = 2 m/s
[You will notice that any distance values with its time will give you 2 m/s as its speed. This means that speed is constant throughout.]
10] Take the distance value and its time for the highest peak of B.
V = 20 ÷ 2
V = 10 m/s
The concepts of force<span>, mass, and weight play critical roles. Newton's Laws of. Motion ... the person stops </span>pushing<span>? ... F </span>net<span> =10 N </span>2<span> N. = 8 N (to the right) a = F </span>net<span> m. = 8 N. 5 kg. =1.6 m s. </span>2<span> ... </span>Two equal forces<span> act on an </span>object<span> in the directions shown. </span>If<span> these ... </span>Two<span> connected carts </span>being accelerated by a force<span> F applied by.</span>
Answer:
i) 21 cm
ii) At infinity behind the lens.
iii) A virtual, upright, enlarged image behind the object
Explanation:
First identify,
object distance (u) = 42 cm (distance between object and lens, 50 cm - 8 cm)
image distance (v) = 42 cm (distance between image and lens, 92 cm - 50 cm)
The lens formula,
Then applying the new Cartesian sign convention to it,
Where f is (-), u is (+) and v is (-) in all 3 cases. (If not values with signs have to considered, this method that need will not arise)
Substituting values you get,
i) 
f = 21 cm
ii) u =21 cm, f = 21 cm v = ?
Substituting in same equation
v ⇒ ∞ and image will form behind the lens
iii) Now the object will be within the focal length of the lens. So like in the attachment, a virtual, upright, enlarged image behind the object.
