Answer:
Explanation:
A.
Given:
Vo = 21 m/s
Vf = 0 m/s
Using equation of Motion,
Vf^2 = Vo^2 - 2aS
S = (21^2)/2 × 9.8
= 22.5 m.
B.
Given:
S = 22.5 + 21 mm
= 22.521 m
Vo = 0 m/s
Using the equation of motion,
S = Vo × t + 1/2 × a × t^2
22.521 = 0 + 1/2 × 9.8 × t^2
t^2 = (2 × 22.521)/9.8
= 4.6
t = 2.14 s
if i am correct it should be that an atom is made up of three smaller particles and those particles i can see you have written down -Protons which are positive, Neutrons which are neutral and electrons which are negative the neutrons and protons are heavier and stay in the middle of the atom which is called the nucleus hope this helps!!
Answer:
The mass and gravity.
Explanation:
Depending on how much mass there is in an object, will determine how fast/slow and object moves. This is the same with gravity, if there is a larger slope, there will be more energy, and if there is less there will be less movement.
Answer:
The suitcase will land 976.447m from the dog.
Explanation:
The velocity in its component in the X and Y axis is decomposed:
Vx= 100m/s × cos(25°)= 90.63m/s
Vy= 100m/s × sen(25°)= 42.26m/s
Time it takes for the suitcase to reach maximum height, the final speed on the axis and at the point of maximum height is zero whereby:
VhmaxY= Voy- 9.81(m/s^2) × t ⇒ t= (42.26 m/s) / (9.81(m/s^2)) = 4.308s
The space traveled on the axis and from the moment the suitcase is thrown until it reaches its maximum height will be:
Dyhmax= Voy × t - (1/2) × 9.81(m/s^2) × (t^2) =
= 42.26m/s × 4.308s - 4.9 (m/s^2) × (4.308s)^2 =
=182.056m - 90.938m= 91.118m
The time from the maximum height to touching the ground is:
Dtotal y= 114m + 91.118m = (1/2) × 9.81(m/s^2) × (t^2) =
= 205.118m = 4.9 (m/s^2) × (t^2) ⇒ t= (41.818 s^2) ^ (1/2)= 6.466s
The total time of the bag in its rise and fall will be:
t= 4.308s + 6.466s = 10.774s
With this time and the initial velocity at x which is constant I can obtain the distance traveled by the suitcase on the x-axis:
Dx= 90.63 (m/s) × 10.774s = 976.447m
Answer:
The gravitational potential energy the barbells have at the maximum height, is P.E. = 3680·h J = 1.10 × 10⁴ units
Where;
h = The maximum height to which he lifts the barbells
Explanation:
The given parameters are;
The amount of work done by the championship lifter in lifting the weight, W = 1.10 × 10⁴ units
The weight of the barbells lifted by the championship lifter, N = 3680 N
The gravitational potential energy, P.E., the barbells had at their maximum height of lift is given as follows;
P.E. = m × g × h
Where;
m = The mass of the barbells;
g = The acceleration due to gravity = 9.8 m/s²
h = The maximum height to which the barbells are lifted by the championship weight lifter
m × g = The weight of the barbells = 3680 N
∴ P.E. = 3680 N × h = 3680·h J
By the conservation of energy principle, work done by the championship weight lifter = The maximum gravitational potential energy gained by the barbell = The gravitational potential energy at the maximum height, P.E.
∴ The gravitational potential energy the barbells have at the maximum height, P.E. = 3680·h J = W = 1.10 × 10⁴ units
