Answer:
Explanation:
Velocity at the bottom of height h
= √2gh
deceleration on rough horizontal surface
= μg , μ is coefficient of friction
= .27 x 9.8
= 2.646 m / s²
v² = u² - 2as
0 = 2gh - 2 x 2.646 x 19
h = 2 x 2.646 x 19 / 2 x 9.8
= 5.13 m
The initial speed of the bolt is not 58.86 m/s.
Let a be the acceleration of the rocket.
During the 4 sec lift off, the rocket has reached a height of
h = (1/2)*a*t^2
with t=4,
h = (1/2)*a^16
h = 8*a
Its velocity at 4 sec is
v = t*a
v = 4*a
The initial velocity of the bolt is thus 4*a.
During the 6 sec fall, the bolt has the initial velocity V0=-4*a and it drops a total height of h=8*a. From the equation of motion,
h = (1/2)*g*t^2 + V0*t
Substituting h0=8*a, t=6 and V0=-4*a into it,
8*a = (1/2)*g*36 - 4*a*6
Solving for a
a = 5.52 m/s^2
Answer:
3.88m/s
Explanation:
Using the law of conservation of momentum
m1u1+m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and 2 are the initial velocities
v is the final velocity
Given
m1 = 64kg
u1 = 4.2m/s
m2 = 25kg
u2 = 3.2m/s
Required
Final velocity v
Substitute the given values into the formula
64(4.2)+25(3.2) = (65+25)v
268.8+80 = 90v
348.8 = 90v
v = 348.8/90
v = 3.88m/s
Hence the velocity of the kayak after the swimmer jumps off is 3.88m/s
Answer:
W = M g weight of ball
T cos θ = W balancing vertical forces
T sin θ = F balancing horizontal forces
tan θ = F / W dividing equations
F = W tan θ when θ equals zero F equals zero
Depends on what type of gass
