The answer to this question is:
C-"That moving clocks run slower"
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Answer:
acceleration, a = 9.8 m/s²
Explanation:
'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.
u = 0 m/s
After 2 seconds, velocity of the ball is 19.6 m/s.
t = 2s, v = 19.6 m/s
Using
v = u + at
19.6 = 0 + 2a
a = 9.8 m/s²
Answer:
3.5 N
Explanation:
Let the 0-cm end be the moment point. We know that for the system to be balanced, the total moment about this point must be 0. Let's calculate the moment at each point, in order from 0 to 100cm
- Tension of the string attached at the 0cm end is 0 as moment arm is 0
- 2 N weight suspended from the 10 cm position: 2*10 = 20 Ncm clockwise
- 2 N weight suspended from the 50 cm position: 2*50 = 100 Ncm clockwise
- 1 N stick weight at its center of mass, which is 50 cm position, since the stick is uniform: 1*50 = 50 Ncm clockwise
- 3 N weight suspended from the 60 cm position: 3*60 = 180 Ncm clockwise
- Tension T (N) of the string attached at the 100-cm end: T*100 = 100T Ncm counter-clockwise.
Total Clockwise moment = 20 + 100 + 50 + 180 = 350Ncm
Total counter-clockwise moment = 100T
For this to balance, 100 T = 350
so T = 350 / 100 = 3.5 N
<h3><u>Answer;</u></h3>
= 1.256 m
<h3><u>Explanation;</u></h3>
We can start by finding the spring constant
F = k*y
Therefore; k = F/y = m*g/y
= 0.40kg*9.8m/s^2/(0.76 - 0.41)
= 11.2 N/m
Energy is conserved
Let A be the maximum displacement
Therefore; 1/2*k*A^2 = 1/2*k*(1.20 - 0.41)^2 + 1/2*m*v^2
Thus; A = sqrt((1.20 - 0.55)^2 + m/k*v^2)
= sqrt((1.20 -0.55)^2 + 0.40/9.8*1.6^2)
= 0.846 m
Thus; the length will be 0.41 + 0.846 = 1.256 m
Answer:
5308.34 N/C
Explanation:
Given:
Surface density of each plate (σ) = 47.0 nC/m² = 
Separation between the plates (d) = 2.20 cm
We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:
Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,
Now, plug in for 'σ' and 
for 
and solve for the electric field. This gives,
Therefore, the electric field between the plates has a magnitude of 5308.34 N/C
