Answer:
The work done by a particle from x = 0 to x = 2 m is 20 J.
Explanation:
A force on a particle depends on position constrained to move along the x-axis, is given by,
We need to find the work done on a particle that moves from x = 0.00 m to x = 2.00 m.
We know that the work done by a particle is given by the formula as follows :
So, the work done by a particle from x = 0 to x = 2 m is 20 J. Hence, this is the required solution.
Answer:
45 degrees
Explanation:
The textbooks say that the maximum range for projectile motion (with no air resistance) is 45 degrees.
Explanation:
so sorry
don't know but please mark me as brainliest please
Our values can be defined like this,
The problem can be solved for part A, through the Work Theorem that says the following,
Where
KE = Kinetic energy,
Given things like that and replacing we have that the work is given by
W = Fd
and kinetic energy by
So,
Clearing F,
Replacing the values
B) The work done by the wall is zero since there was no displacement of the wall, that is d = 0.
For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s
