Answer:
Explanation:
(a) The applied force has two components Fx and Fy. The Fx component is the only one that does work

(b) There in no net force in the vertical component

(c)

(d)

I attached an scheme of the force diagram
Answer:
The resistance that will provide this potential drop is 388.89 ohms.
Explanation:
Given;
Voltage source, E = 12 V
Voltage rating of the lamp, V = 5 V
Current through the lamp, I = 18 mA
Extra voltage or potential drop = 12 V - 5 V = 7 V
The resistance that will provide this potential drop (7 V) is calculated as follows:

Therefore, the resistance that will provide this potential drop is 388.89 ohms.
Answer:
a) A = 4.0 m
, b) w = 3.0 rad / s
, c) f = 0.477 Hz
, d) T = 20.94 s
Explanation:
The equation that describes the oscillatory motion is
x = A cos (wt + fi)
In the exercise we are told that the expression is
x = 4.0 cos (3.0 t + 0.10)
let's answer the different questions
a) the amplitude is
A = 4.0 m
b) the frequency or angular velocity
w = 3.0 rad / s
c) angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 3 / 2π
f = 0.477 Hz
d) the period
frequency and period are related
T = 1 / f
T = 1 / 0.477
T = 20.94 s
e) the phase constant
Ф = 0.10 rad
f) velocity is defined by
v = dx / dt
v = - A w sin (wt + Ф)
speed is maximum when sine is + -1
v = A w
v = 4 3
v = 12 m / s
g) the angular velocity is
w² = k / m
k = m w²
k = 1.2 3²
k = 10.8 N / m
h) the total energy of the oscillator is
Em = ½ k A²
Em = ½ 10.8 4²
Em = 43.2 J
i) the potential energy is
Ke = ½ k x²
for t = 0 x = 4 cos (0 + 0.1)
x = 3.98 m
j) kinetic energy
K = ½ m v²
for t = 00.1
²
v = A w sin 0.10
v = 4 3 sin 0.10
v = 1.98 m / s
Answer:78 N
Explanation:Impulse=Force*time
=39*2 =78 Ns
Answer:
3.25 seconds
Explanation:
It is given that,
A person throws a baseball from height of 7 feet with an initial vertical velocity of 50 feet per second. The equation for his motion is as follows :

Where
s is the height in feet
For the given condition, the equation becomes:

When it hits the ground, h = 0
i.e.

It is a quadratic equation, we find the value of t,
t = 3.25 seconds and t = -0.134 s
Neglecting negative value
Hence, for 3.25 seconds the baseball is in the air before it hits the ground.