I'm going to assume this is over a horizontal distance. You know from Newton's Laws that F=ma --> a = F/m. You also know from your equations of linear motion that v^2=v0^2+2ad. Combining these two equations gives you v^2=v0^2+2(F/m)d. We can plug in the given values to get v^2=0^2+2(20/3)0.25. Solving for v we get v=1.82 m/s!
Answer:
Explanation:
Let the velocity be v
Total energy at the bottom
= rotational + linear kinetic energy
= 1/2 Iω² + 1/2 mv² ( I moment of inertia of shell = mr² )
= 1/2 mr²ω² + 1/2 mv² ( v = ω r )
= 1/2 mv² +1/2 mv²
= mv²
mv² = mgh ( conservation of energy )
v² = gh
v = √gh
= √9.8 x 1.8
= 4.2 m /s
Answer:
<h3>The answer is 15 N</h3>
Explanation:
The force acting on an object can be found by using the formula
<h3>Force = mass × acceleration</h3>
From the question
mass = 50 g = 0.05 kg
acceleration = 300 m/s²
We have
force = 0.05 × 300
We have the final answer as
<h3>15 N</h3>
Hope this helps you
If the solution is treated as an ideal solution, the extent of freezing
point depression depends only on the solute concentration that can be
estimated by a simple linear relationship with the cryoscopic constant:
ΔTF = KF · m · i
ΔTF, the freezing point depression, is defined as TF (pure solvent) - TF
(solution).
KF, the cryoscopic constant, which is dependent on the properties of the
solvent, not the solute. Note: When conducting experiments, a higher KF
value makes it easier to observe larger drops in the freezing point.
For water, KF = 1.853 K·kg/mol.[1]
m is the molality (mol solute per kg of solvent)
i is the van 't Hoff factor (number of solute particles per mol, e.g. i =
2 for NaCl).
Vf = Vo + at
Vf = 20 m/s
Vo = 50 m/s
a = ?
t = 15
Therefore
20 = 50 + 15a
20 - 50 = 15a
-30 = 15a
a = -30 / 15
a = -2 m/s²
