Answer: A is Compression and B is Rarefaction.
Explanation:
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Answer:
Refractive index of a medium = Speed of light in vacuum/Speed of light in a medium
1.5 = 3 x 108 / Speed of light in medium
Speed of light in the medium = 3 x 108 /1.5
= 2 x 108 m/s.
Explanation:
Answer:
Explanation:
Given that,
Bathysphere radius
r = 1.5m
Mass of bathysphere
M = 1.2 × 10⁴ kg
Constant speed of descending.
v = 1.2m/s
Resistive force
Fr = 1100N upward direction
Density of water
ρ = 1.03 × 10³kg/m³
The volume of the bathysphere can be calculated using
V = 4πr³ / 3
V = 4π × 1.5³ / 3
V = 14.14 m³
The Bouyant force can be calculated using
Fb = ρgV
Fb = 1.03 × 10³ × 9.81 × 14.14
Fb = 142,846.18 N
Buoyant force is acting upward
Weight of the bathysphere
W = mg
W = 1.2 × 10⁴ × 9.81
W = 117,720 N
Weight is acting downward
The net positive buoyant using resolving
Fb+ = Fb — W
Fb+ = 142,846.18 — 117,720
Fb+ = 25,126.18 N
The force acting downward is the weight of the submarine and it is equal to the positive buoyant force and the resistive force
W = Fb+ + Fr
W = 25,126.18 + 1100
W = 26,226.18
mg = 26,226.18
m = 26,226.18 / 9.81
m = 2673.4kg
Mass of submarine is 2673.4kg
Answer:
the tension in the part of the cord attached to the textbook is 7.4989 N
Explanation:
Given the data in the question;
As illustrated in the image below;
first we determine the value of the acceleration,
along vertical direction; we use the second equation of motion;
y = ut + at²
we substitute;
0 m/s for u, 1.29 m for y, 0.850 s for t,
1.29 = 0×0.850 + ×a×(0.850)²
1.29 = 0.36125a
a = 1.29 / 0.36125
a = 3.5709 m/s²
Now when the text book is moving with acceleration , the dynamic equation will be;
T₁ = m₁a
where m₁ is the mass of the text book ( 2.10 kg )
a is the vertical acceleration ( 3.5709 m/s² )
so we substitute
T₁ = 2.10 × 3.5709
T₁ = 7.4989 N
Therefore, the tension in the part of the cord attached to the textbook is 7.4989 N
Answer:
Explanation:
In multiplications or divisions, the result must be written using a number of significant figures equal to the smallest number of significant figures contained in the original numbers.
For instance, in this case the two numbers are:
3278 --> 4 significant figures
42 --> 2 significant figures
So, the product must be written using 2 significant figures.
Calculating the product, we get:
Therefore, using 2 significant figures only, the result is