Answer: Option (b) is the correct answer.
Explanation:
The force of gravity acting on an object helps in determining the weight of an object. But a place where there will be no gravity or have zero gravitational pull then it means the person will be weightless.
For example, force of gravity on moon is zero which means any object or person on moon will be weightless.
On the other hand, when a child is in the air as she plays on a trampoline then it means gravitational pull form the earth is acting on it. So, it will definitely has some weight.
Similarly, a scuba diver exploring a deep-sea wreck is under the ground where there will be force of gravity. Hence, it will also have some weight.
Thus, we can conclude that an astronaut on the Moon is the person who is weightless.
Answer:
The last two bearings are
49.50° and 104.02°
Explanation:
Applying the Law of cosine (refer to the figure attached):
we have
x² = y² + z² - 2yz × cosX
here,
x, y and z represents the lengths of sides opposite to the angels X,Y and Z.
Thus we have,

or

substituting the values in the equation we get,

or

or
X = 26.47°
similarly,

or

or
Y = 49.50°
Consequently, the angel Z = 180° - 49.50 - 26.47 = 104.02°
The bearing of 2 last legs of race are angels Y and Z.
Answer:
1 × 10⁶ N/C
Explanation:
The magnitude of the electric field between the membrane = surface density / permittivity of free space = 10 ⁻⁵C/ m² / (8.85 × 10⁻¹²N⁻¹m⁻²C²) = 1.13 × 10⁶ N/C approx 1 × 10⁶ N/C
Answer:


Explanation:
Wave 1, 
Wave 2, 
Wave 3, 
Wave 4, 
The general equation of travelling wave is given by :

The value of
will remain the same if we take phase difference into account.
For first wave,



For second wave,



For the third wave,



For the fourth wave,



It is clear from above calculations that waves 1 and 3 have same time period. Also, wave 2 and 4 have same time period. Hence, this is the required solution.