<h2>
Velocity at the end of 3.6 s is -8.64 m/s</h2>
Explanation:
We have equation of motion v = u + at
Initial velocity, u 0 m/s
Final velocity, v = ?
Time, t = 3.6 s
Acceleration, a = -2.40 m/s²
Substituting
v = u + at
v = 0 + -2.40 x 3.6
v = -8.64 m/s
Velocity at the end of 3.6 s is -8.64 m/s
Answer:
Explanation:
Explain how a projectile might be modified to decrease the air resistance impacting its trajectory.
Answer:
27.22 m/s
Explanation:
Let the speed of clay before impact is u.
the speed of clay and target is v after impact.
use conservation of momentum
momentum before impact momentum after impact
mass of clay x u = (mass of clay + mass of target) x v
100 x u = (100 + 500) x v
u = 6 v .....(1)
distance, s = 2.1 m
μ = 0.5
final velocity is zero. use third equation of motion
v'² = v² + 2as
0 = v² - 2 x μ x g x s
v² = 2 x 0.5 x 9.8 x 2.1 = 20.58
v = 4.54 m/s
so by equation (1)
u = 6 x 4.54 = 27.22 m/s
thus, the speed of clay before impact is 27.22 m/s.
Answer:
The hypotenuse is 3.68
3.682 = 2.852 + x2
14.8669 = 8.1225 + x2
6.7444 = x2
2.59699827 ≈ x
The monkey can climb about 2.6 feet up the pole
Explanation:
Explanation:
1) For a positive point charge, the lines radiate outwards
for a negative point charge, the lines converge inwards
2) F = 2.3 X 10^-26 N
k = 9 X 10^9 N.m²/C
q1 = q2 = e = 1.6 X 10^-19 C
r = ?
F = kq1q2/r²
r² = kq1q2/F
r = √[kq1q2/F ]
r = √0.0100
r = 0.10m
The two protons are 0.10 m apart
3) The unit if electric field intensity is Newton-per-coulomb N/C