Answer:
In SI units, its value is approximately 6.674×10−11 m3⋅kg−1⋅s−2. The modern notation of Newton's law involving G was introduced in the 1890s by C. V. Boys. The first implicit measurement with an accuracy within about 1% is attributed to Henry Cavendish in a 1798 experiment.
Explanation:
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Answer:
271.862 N/m
Explanation:
From Hook's Law,
mgh = 1/2ke²............... Equation 1
Where
m = mass of the ball, g = acceleration due to gravity, k = spring constant, e = extension, h = height fro which the ball was dropped.
Making k the subject of the equation,
k =2mgh/k²....................... Equation 2
Note: The potential energy of the ball is equal to the elastic potential energy of the spring.
Given: m = 60.3 g = 0.0603 kg, g = 9.8 m/s², e = 4.68317 cm = 0.0468317 m, h = 53.7 cm = 0.537 m
Substitute into equation 2
k = 2(0.0603)(9.8)(0.537)/0.048317²
k = 0.6346696/0.0023345
k = 271.862 N/m
The frequency of a wave is the number of complete oscillations passing a given point per second.
In this case, assuming the duck is stationary, we have 4 complete waves passing the duck in one second: therefore, the frequency of the wave is
The answer of this question is B.
1)
Answer:
Part 1)
H = 30.6 m
Part 2)
t = 2.5 s
Part 3)
t = 2.5 s
Part 4)

Explanation:
Part 1)
initial speed of the ball upwards

so maximum height of the ball is given by



Part 2)
As we know that final speed will be zero at maximum height
so we will have



Part 3)
Since the time of ascent of ball is same as time of decent of the ball
so here ball will same time to hit the ground back
so here it is given as
t = 2.5 s
Part 4)
since the acceleration due to earth will be same during its return path as well as the time of the motion is also same
so here its final speed will be same as that of initial speed
so we have

2)
Answer:
a = 9.76 m/s/s
Explanation:
As we know that the object is released from rest
so the displacement of the object in vertical direction is given as



3)
Answer:
v = 29.7 m/s
Explanation:
acceleration of the rocket is given as

time taken by the rocket
t = 0.33 min
final speed of the rocket is given as



4)
Answer:
Part 1)
y = 25.95 m
Part 2)
d = 6.72 m
Explanation:
Part 1)
As it took t = 2.3 s to hit the water surface
so here we will have



Part 2)
Distance traveled by it in horizontal direction is given as


