Answer:
12 cm/s
Explanation:
Quite simply, you are looking for cm/s
so 60 cm / 5s = 12 cm/s
Answer:19.5 J
Explanation:
Given
mass of block=3 kg
angular frequency=20 rad/sec
spring constant 
we know total energy remain conserved
Where 
=kinetic energy
=potential Energy
When mass reaches amplitude its velocity becomes zero
there is only potential energy which is equal to Total energy
Answer:
Explanation:
A lot. That's why he was one of Boston's star hitters.
m = 0.15 kg
a = 3.0 * 10^4
F = ?
F = 0.15 * 3*10^4
F = 0.45 * 10^4
F = 4.5 * 10^3 Newtons
You should always express scientific notation as a number between 1 and 10 or 1 < x < 10. The power will determine the exact value of the number expressed as a scientific notation.
Given what we know, we can confirm that doubling the distance between you and a source of radiation decreases your exposure by 75%.
<h3>How is distance related to radiation exposure?</h3>
- As expected, increasing the distance from the source of the radiation will reduce its negative effects.
- Counter-intuitively however, doubling the distance does not reduce by half, but rather reduces its effects by 3/4th.
- This is due to the fact that the radiation effects from the source are inversely proportional to the square of the distance.
- This causes the changes to be far greater than expected.
Therefore, given that the radiation is proportional to the square of the distance, instead of being of a more direct relation, we can confirm that when doubling the distance between yourself and the source of the radiation, you can reduce its effects by 3/4 or 75%.
To learn more about radiation visit:
brainly.com/question/9815840?referrer=searchResults
Define
g = 9.8 m/s², acceleration due to gravity, positive downward.
Assume that wind resistance may be neglected.
Frog A:
u = 0.551 m/s, launch velocity, upward.
When the frog lands back on the pad, its vertical position is zero, and its vertical velocity will be 0.551 m/s downward.
If the time of flight is t, then
(0.551 m/s)*(t s) - 0.5*(9.8 m/s²)*(t s)² = 0
0.551t - 4.9t² = 0
t = 0, or t = 0.1124 s
t = 0 corresponds to launch, and t = 0.1124 s corresponds to landing.
Frog B:
Launch velocity is 1.75 m/s
When t = 0.1124 s, the position of the frog is
s = (1.75 m/s)(0.1124 s) - 0.5*(9.8 m/s²)*(0.1124 s)²
= 0.135 m
The velocity of frog B is
v = (1.75 m/s) - (9.8 m/s²)*(0.1124 s)
= 0.6485 m/s
Answer:
When frog A lands on the ground,
Frog B is 0.135 m above ground and its velocity is 0.649 m/s upward.
