The coefficient of expansion is 13 * 10^-6 m per meter length.per oK
The temperature difference = 42 - - 8 = 50 oC
delta T = (42 + 273) - (-8 + 273) = 50 oK
delta L = L * 13* 10^6 m/oK
oK = 50 oK delta L = 19.5 cm = 19.5 cm [1m / 100 cm] = 0.195m
So we need to find the length and it is computed by:
0.195= L * 13 * 10^-6 * 50 L = 0.195 / (13*10^-6*50) L = 300 m
Answer:
a) Therefore 2.6km is greater than 2.57km.
Statement A is greater than statement B.
b) Therefore 5.7km is equal to 5.7km
Statement A is equal to statement B
Explanation:
a) Statement A : 2.567km to two significant figures.
2.567km 2. S.F = 2.6km
Statement B : 2.567km to three significant figures.
2.567km 3 S.F = 2.57km
Therefore 2.6km is greater than 2.57km.
Statement A is greater than statement B.
b) statement A: (2.567 km + 3.146km) to 2 S.F
(2.567km + 3.146km) = 5.713km to 2 S.F = 5.7km
Statement B : (2.567 km, to two significant figures) + (3.146 km, to two significant figures).
2.567km to 2 S.F = 2.6km
3.146km to 2 S.F = 3.1km
2.6km + 3.1km = 5.7km
Therefore 5.7km is equal to 5.7km
Statement A is equal to statement B
Answer:
Option (A) , (b) and (d) are correct option
Explanation:
According to Coulomb's law electric force between two charges is given by
From the relation we can say that force is directly proportional to magnitude of charges and inversely proportional to distance between them '
So if we increase the distance then force will decrease
Increase if any of the charge get larger
If force is attractive then both the charge will be of different sign and is force is repulsive then both the charges of same sign
From above conclusion we can say that (a), (b) and (d) are correct option
Answer:
1/6 del peso en la tierra.
Answer:
The temperature and gravity both affects the density
