Answer:
Explanation:
Initial velocity , u = 30 m/s
final velocity , v = 10 m/s
time , t = 5 seconds
1. Acceleration = v - u / t
= 10 - 30 / 5
= -20 / 5
= <u><em>- 4 m/s</em></u>
Answer:
1. The precession of the equinoxes.
2. Changes in the tilt angle of Earth’s rotational axis relative to the plane of Earth’s orbit around the Sun.
3. Variations in the eccentricity
Explanation:
These variations listed above; the precession of the equinoxes (refers, changes in the timing of the seasons of summer and winter), this occurs on a roughly about 26,000-year interval; changes in the tilt angle of Earth’s rotational axis relative to the plane of Earth’s orbit around the Sun, this occurs roughly in a 41,000-year interval; and changes in the eccentricity (that is a departure from a perfect circle) of Earth’s orbit around the Sun, occurring on a roughly 100,000-year timescale. which influences the mean annual solar radiation at the top of Earth’s atmosphere.
Answer:
a) the elastic force of the pole directed upwards and the force of gravity with dissects downwards
Explanation:
The forces on the athlete are
a) at this moment the athlete presses the garrolla against the floor, therefore it acquires a lot of elastic energy, which is absorbed by the athlete to rise and gain potential energy,
therefore the forces are the elastic force of the pole directed upwards and the force of gravity with dissects downwards
b) when it falls, in this case the only force to act is batrachium by the planet, this is a projectile movement for very high angles
c) When it reaches the floor, it receives an impulse that opposes the movement created by the mat. The attractive force is the attraction of gravity.
Answer:
(A) -2940 J
(B) 392 J
(C) 212.33 N
Explanation:
mass of bear (m) = 25 kg
height of the pole (h) = 12 m
speed (v) = 5.6 m/s
acceleration due to gravity (g) = 9.8 m/s
(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)
height at the bottom = 0
= 25 x 9.8 x (0-12) = -2940 J
(B) kinetic energy of the Bear (KE) = 
= 
= 392 J
(C) average frictional force = 
- change in KE (ΔKE) = initial KE - final KE
- ΔKE = -
- when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes ΔKE =
- 0 = 392 J
\frac{-(ΔKE+ΔU)}{h}[/tex] = 
= 
= 212.33 N
Answer:
Your correct answer is option C. Displacement is equal to the total distance traveled by an object in motion.
Explanation:
This is something that is definitely not a displacement.
