Answer:
a) V_f = 25.514 m/s
b) Q =53.46 degrees CCW from + x-axis
Explanation:
Given:
- Initial speed V_i = 20.5 j m/s
- Acceleration a = 0.31 i m/s^2
- Time duration for acceleration t = 49.0 s
Find:
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Solution:
- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:
V_f = V_i + a*t
V_f = 20.5 j + 0.31 i *49
V_f = 20.5 j + 15.19 i
- The magnitude of the velocity vector is given by:
V_f = sqrt ( 20.5^2 + 15.19^2)
V_f = sqrt(650.9861)
V_f = 25.514 m/s
- The direction of the velocity vector can be computed by using x and y components of velocity found above:
tan(Q) = (V_y / V_x)
Q = arctan (20.5 / 15.19)
Q =53.46 degrees
- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.
Answer: 161.3
I have a acellus too and got this question correct, so I hope this helps y’all out
Answer:
9 meters
Explanation:
Given:
Mass of Avi is, 
Spring constant is, 
Compression in the spring is, 
Let the maximum height reached be 'h' m.
Now, as the spring is compressed, there is elastic potential energy stored in the spring. This elastic potential energy is transferred to Avi in the form of gravitational potential energy.
So, by law of conservation of energy, decrease in elastic potential energy is equal to increase in gravitational potential energy.
Decrease in elastic potential energy is given as:

Now, increase in gravitational potential energy is given as:

Now, increase in gravitational potential energy is equal decrease in elastic potential energy. Therefore,

Therefore, Avi will reach a maximum height of 9 meters.
Answer:
The fraction of the protons would have no electrons 
Explanation:
We are given that
Amoeba has total number of protons=
Net charge, Q=0.300pC
Electrons are fewer than protons=
We have to find the fraction of protons would have no electrons.
The fraction of the protons would have no electrons
=
The fraction of the protons would have no electrons
=

Hence, the fraction of the protons would have no electrons 