Perfect
symmetrical cone type volcanoes would have radial drainage pattern.
A stratovolcano, also known as a composite volcano, is a
conical volcano built up by many layers (strata) of hardened lava, tephra,
pumice, and volcanic ash.
If you have any further questions, please don’t hesitate to
ask again.
Use kinematic equations to solve:
1) yf = yo + vo*t + 1/2at²
yf = final height
yo = initial height
vo = initial velocity
a = acceleration
t = time
yf - yo = vo*t + 1/2at²
yf - yo = h
vo = 0
Thus,
h = 1/2at²
h = 1/2(9.8)(12)² = 705.6 m
2) vf = vo + at
vo = 0
Thus,
vf = at
vf = (9.8)(12) = 117.6 m/s
Complete question is;
A rocket ship starts from rest and turns on its forward booster rockets, causing it to have a constant acceleration of 4 m/s² rightward. After 3s, what will be the velocity of the rocket ship?
Answer:
v = 12 m/s
Explanation:
We are given;
Initial velocity; u = 0 m/s (because ship starts from rest)
Acceleration; a = 4 m/s²
Time; t = 3 s
To find velocity after 3 s, we will use Newton's first equation of motion;
v = u + at
v = 0 + (4 × 3)
v = 12 m/s
Answer:
B it decreases
Explanation:
the movement of a positive test charge in the direction of an electric field would be like a mass falling downward within Earth's gravitational field. Both movements would be like going with nature and would occur without the need of work by an external force. This motion would result in the loss of potential energy
Answer:
The friction force is 250 N
Explanation:
The desk is moving at constant velocity. This means that its acceleration is zero: a = 0. Newton's second law states that the resultant of the forces acting on the desk is equal to the product between mass (m) and acceleration (a):
In this case, we know that the acceleration is zero: a = 0, so also the resultant of the forces must be zero:
(1)
We are only interested in the forces acting along the horizontal direction, since it is the direction of motion. There are two forces acting in this direction:
- the pull, forward, F = 250 N
- the friction force, backward,
Given (1), we have
So the force of friction must be equal to the pull: