Answer:
Part a)
Part b)
Part c)
Part d)
Explanation:
Part a)
As we know that speed of package is same as that of helicopter in horizontal direction
So after time "t" the velocity in x direction will remain constant while in Y direction it will go free fall
So we have
Part b)
Distance from helicopter is same as the distance of free fall
so we will have
Part c)
If helicopter is rising upwards with uniform speed
then final speed of the package after time t is given as
Part d)
distance from helicopter
There is no picture given so I can't be really sure what color of the cable you're referring to. However, the only relationship I can think of when the power and the current is given would be: P=IV or P = I²R, where P is power, I is current, V is voltage and R is resistance. Solving both equations:
120 W = (24 A)(Voltage)
Voltage = 5 V
120 W = (24 A)²(R)
R = 0.2083 Ω
So, i think the cable would have specification of 5 Volts and 0.2083 ohms.
The roadway with the highest number of hazards is <span>city streets</span>
Answer:
The value is 
Explanation:
From the question we are told that
The mass of the block is 
The force constant of the spring is 
The amplitude is 
The time consider is 
Generally the angular velocity of this block is mathematically represented as
=> 
=> 
Given that the block undergoes simple harmonic motion the velocity is mathematically represented as
=> 
=>
Answer:
The gravitational potential energy of a system is -3/2 (GmE)(m)/RE
Explanation:
Given
mE = Mass of Earth
RE = Radius of Earth
G = Gravitational Constant
Let p = The mass density of the earth is
p = M/(4/3πRE³)
p = 3M/4πRE³
Taking for instance,a very thin spherical shell in the earth;
Let r = radius
dr = thickness
Its volume is given by;
dV = 4πr²dr
Since mass = density* volume;
It's mass would be
dm = p * 4πr²dr
The gravitational potential at the center due would equal;
dV = -Gdm/r
Substitute (p * 4πr²dr) for dm
dV = -G(p * 4πr²dr)/r
dV = -G(p * 4πrdr)
The gravitational potential at the center of the earth would equal;
V = ∫dV
V = ∫ -G(p * 4πrdr) {RE,0}
V = -4πGp∫rdr {RE,0}
V = -4πGp (r²/2) {RE,0}
V = -4πGp{RE²/2)
V = -4Gπ * 3M/4πRE³ * RE²/2
V = -3/2 GmE/RE
The gravitational potential energy of the system of the earth and the brick at the center equals
U = Vm
U = -3/2 GmE/RE * m
U = -3/2 (GmE)(m)/RE
