The correct answer would be 1.375 < t < 3 i hope this helps anyone
It's the second graph!
it's the only one with a negative gradient.
so the temperature of the ball will fall in water as it looses its heat.
activate windows,:-P
Does this help?
When an object is
immersed in a fluid (in this case water, but may include both liquids and
gases) the fluid exerts an upward force on the object which is called buoyancy
force or <span>up-thrust. Archimedes’ Principle states that the buoyant
force (upward push or force) applied to an object is equal to the weight of the fluid that the object takes the space of by
that object. Thus when an object is
placed in water the rise in the water level is dictated by the mass of that
object.</span>
<span>
</span>
<span>So for example if you fill a bucket with water and you drop a stone in that bucket, if you measure the weight of the water that overflows from the bucket due to the stone being dropped into the bucket is equivalent to the pushing force that the water has on the stone (as the stone drops to the bottom of the bucket the water is pushing it to stay afloat but the rock is more dense than water and as such its downthrust exceeds water's upthrust).</span>
Answer:
Explanation:
Given that,
At one instant,
Center of mass is at 2m
Xcm = 2m
And velocity =5•i m/s
One of the particle is at the origin
M1=? X1 =0
The other has a mass M2=0.1kg
And it is at rest at position X2= 8m
a. Center of mass is given as
Xcm = (M1•X1 + M2•X2) / (M1+M2)
2 = (M1×0 + 0.1×8) /(M1 + 0.1)
2 = (0+ 0.8) /(M1 + 0.1)
Cross multiply
2(M1+0.1) = 0.8
2M1 + 0.2 =0.8
2M1 = 0.8-0.2
2M1 = 0.6
M1 = 0.6/2
M1 = 0.3kg
b. Total momentum, this is an inelastic collision and it momentum after collision is given as
P= (M1+M2)V
P = (0.3+0.1)×5•i
P = 0.4 × 5•i
P = 2 •i kgm/s
c. Velocity of particle at origin
Using conversation of momentum
Momentum before collision is equal to momentum after collision
P(before) = M1 • V1 + M2 • V2
We are told that M2 is initially at rest, then, V2=0
So, P(before) = 0.3V1
We already got P(after) = 2 •i kgm/s in part b of the question
Then,
P(before) = P(after)
0.3V1 = 2 •i
V1 = 2/0.3 •i
V1 = 6 ⅔ •i m/s
V1 = 6.667 •i m/s
What kind of analogy is this?
A. synonyms
B. part to whole
C. degrees of intensity
<span>D. cause to effect
This is because a synonym refers to things like "normal" or "regular" like your calling someone something.
</span><span /><span>
</span>