Answer:
When the ball hits the ground, its velocity is -128 ft/s.
Explanation:
Hi there!
First, let's find the time it takes the ball to reach the ground (the value of t for which s(t) = 0):
s(t) = -16t² + 32t + 240
0 = -16t² + 32t + 240
Solving the quadratic equation with the quadratic formula:
t = 5.0 s (the other solution of the equation is rejected because it is negative).
Now, we have to find the velocity of the ball at t = 5.0 s.
The velocity of the ball is the change of height over time (the derivative of s(t)):
v = ds/dt = s'(t) = -32t + 32
at t = 5.0 s:
s'(5.0) = -32(5.0) + 32 = -128 ft/s
When the ball hits the ground, its velocity is -128 ft/s.
Answer:
0.51
Explanation:
m = mass of the book = 3.5 kg
F = force applied by the broom on the book = 21 N
a = acceleration of the book
v₀ = initial speed of the book = 0 m/s
v = final speed of the book = 1.2 m/s
d = distance traveled = 0.74 m
Using the equation
v² = v₀² + 2 a d
1.2² = 0² + 2 a (0.74)
a = 0.973 m/s²
f = kinetic frictional force
Force equation for the motion of the book is given as
F - f = ma
21 - f = (3.5) (0.973)
f = 17.6 N
μ = Coefficient of kinetic friction
Kinetic frictional force is given as
f = μ mg
17.6 = μ (3.5 x 9.8)
μ = 0.51
Answer:
F = 2553.6 N
Explanation:
v² = u² + 2as
0² = 16² + 2a4.0
a = -32 m/s²
F = ma = 79.8(32) = 2553.6 N
Grav. Potential at surface of the asteroid:
V = - G.Ma./ R
V = (-) 6.67^-11 x 4.0^20kg / 5.7^5m .. .. V = (-) 4.681 *10 ^5 J/kg
The GPE of the package on the asteroid = 9.0kg x (-) 4.681*10^5J/kg = (-) 4.21 ^5
J
This is the amount of energy required to come back the
package to infinity.
The total energy that needs to be transported to the package:
GPE + KE(for 187m/s)
Total energy required E = 4.21*10^5 + (½x 9.0kg x 168²) = 5.48 * 10^5 J
When the required energy is to be complete by releasing a compressed spring,
Elastic PE stored in spring = ½.ke² = 5.48 * 10^5 J where e = compression
distance
e = √ (2 x 5.48*10^5 / 2.1*10^5)
e = 2.28 m
