The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.
<h3>What is the time after being ejected is the boulder moving at a speed 20.7 m/s upward?</h3>
The motion of the boulder is a uniformly accelerated motion, with constant acceleration
a = g = -9.8 ![$$m / s^2](https://tex.z-dn.net/?f=%24%24m%20%2F%20s%5E2)
downward (acceleration due to gravity).
By using Suvat equation:
v = u + at
where: v is the velocity at time t
u = 40.0 m/s is the initial velocity
a = g = -9.8
is the acceleration
To find the time t at which the velocity is v = 20.7 m/s
Therefore,
![$t=\frac{v-u}{a}=\frac{20.7-40}{-9.8}=2.0204 \mathrm{~s}](https://tex.z-dn.net/?f=%24t%3D%5Cfrac%7Bv-u%7D%7Ba%7D%3D%5Cfrac%7B20.7-40%7D%7B-9.8%7D%3D2.0204%20%5Cmathrm%7B~s%7D)
The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.
The complete question is:
A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. At what time after being ejected is the boulder moving at 20.7 m/s upward?
To learn more about uniformly accelerated motion refer to:
brainly.com/question/14669575
#SPJ4
Answer:
Doing science could be defined as carrying out scientific processes, like the scientific method, to add to science's body of knowledge.
<h2>The distance between students is 2.46 m</h2>
Explanation:
The force of attraction due to Newton's gravitation law is
F = ![\frac{Gm_1m_2}{r^2}](https://tex.z-dn.net/?f=%5Cfrac%7BGm_1m_2%7D%7Br%5E2%7D)
Here G is the gravitational constant
m₁ is the mass of one student
m₂ is the mass of second student .
and r is the distance between them
Thus r = ![\sqrt{\frac{Gm_1m_2}{F} }](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7BGm_1m_2%7D%7BF%7D%20%7D)
If we substitute the values in the above equation
r = ![\sqrt{\frac{6.673x10^-^1^1x31.9x30.0}{2.59x10^-^8} }](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B6.673x10%5E-%5E1%5E1x31.9x30.0%7D%7B2.59x10%5E-%5E8%7D%20%7D)
= 2.46 m
Answer:
2 m/s^2, west
Explanation:
Vf=final velcoity
Vi=initial velocity
t=timw
![a = \frac{vf - vi}{t}](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7Bvf%20-%20vi%7D%7Bt%7D%20)
=
![\frac{15 - 25}{5}](https://tex.z-dn.net/?f=%20%5Cfrac%7B15%20-%2025%7D%7B5%7D%20)
= - 2 m/s^2
The - changes direction and makes it opposite
2 m/s, west