Answer:
a. Quadruped arm and opposite leg raise
Explanation:
Quadruped arm and opposite leg lift
- Kneel on the floor, lean forward and place your hands down.
- Keep your knees in line with your hips and hands directly under your shoulders.
- Simultaneously raise one arm and extend the opposite leg, so that they are in line with the spine.
- Go back to the starting position.
This method is usually used as an alternative to iso-abs exercise or also known as a bridge, which allows you to exercise the abdominal and spinal area at the same time.
It is also used together with other exercises for the treatment of hyperlordosis.
Answer:
The difference between frictionless ramp and a regular ramp is that on a frictionless ramp the ball cannot roll it can only slide, but on a regular ramp the ball can roll without slipping.
We will use conversation of energy.

Note that initial potential energy is zero because the ball is on the bottom, and the final kinetic energy is zero because the ball reaches its maximum vertical distance and stops.
For the ball B;


The initial velocities of the balls are equal. Their maximum climbing point will be proportional to their final potential energy. Since their initial kinetic energies are equal, their final potential energies must be equal as well.
Hence, both balls climb the same point.
Explanation:
Answer:
A isolate the DNA From a sample of cells
Answer:
2000 kg m/s
Explanation:
The momentum of an object is a vector quantity whose magnitude is given by

where
m is the mass of the object
v is the velocity of the object
and its direction is the same as the velocity.
In this problem, we have:
- Spaceship 1 has
m = 200 kg (mass)
v = 0 m/s (zero velocity)
So its momentum is

- Spaceship 2 has
m = 200 kg (mass)
v = 10 m/s (velocity)
So its momentum is

Therefore, the combined momentum of the two spaceships is

Answer:
90 m
Explanation:
Acceleration,
where v and u are final and initial velocities respectively, t is the time taken
Substituting
for a, 4 m/s for u and 10 s for t then
1*10=v-4
v=14 m/s
From kinematic equations

Making s the subject then
