Answer:
293.7 degrees
Explanation:
A = - 8 sin (37) i + 8 cos (37) j
A + B = -12 j
B = a i+ b j , where and a and b are constants to be found
A + B = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j
- 12 j = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j
Comparing coefficients of i and j:
a = 8 sin (37) = 4.81452 m
b = -12 - 8cos(37) = -18.38908
Hence,
B = 4.81452 i - 18.38908 j ..... 4 th quadrant
Hence,
cos ( Q ) = 4.81452 / 12
Q = 66.346 degrees
360 - Q = 293.65 degrees from + x-axis in CCW direction
Complete Question
In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
Answer:
The speed of the helicopter is 
Explanation:
From the question we are told that
The height at which he let go of the brief case is h = 130 m
The time taken before the the brief case hits the water is t = 6 s
Generally the initial speed of the briefcase (Which also the speed of the helicopter )before the man let go of it is mathematically evaluated using kinematic equation as
Here s is the distance covered by the bag at sea level which is zero
=>
=> 
=> 
Answer:
d = 3.54 x 10⁴ Km
Explanation:
Given,
The distance between the two objects, r = 2.5 x 10⁴ Km
The gravitational force between them, F = 580 N
The gravitational force between the two objects is given by the formula
F = GMm/r² newton
When the gravitational force becomes half, then the distance between them becomes
Let us multiply the above equation by 1/2 on both sides
( 1/2) F = (1/2) GMm/r²
= GMm/2r²
= GMm/(√2r)²
Therefore, the distance becomes √2d, when the gravitational force between them becomes half
d = √2r = √2 x 2.5 x 10⁴ Km
= 3.54 x 10⁴ Km
Hence, the two objects should be kept at a distance, d = 3.54 x 10⁴ Km so that the gravitational force becomes half.
If no frictional work is considered, then the energy of the system (the driver at all positions is conserved.
Let
position 1 = initial height of the diver (h₁), together with the initial velocity (v₁).
position 2 = final height of the diver (h₂) and the final velocity (v₂).
The initial PE = mgh₁ and the initial KE = (1/2)mv₁²
where g = acceleration due to gravity,
m = mass of the diver.
Similarly, the final PE and KE are respectively mgh₂ and (1/2)mv₂².
PE in position 1 is converted into KE due to the loss in height from position 1 to position 2.
Therefore
(KE + PE) ₁ = (KE + PE)₂
Evaluate the given answers.
A) The total mechanical energy of the system increases.
FALSE
B) Potential energy can be converted into kinetic energy but not vice versa.
TRUE
C) (KE + PE)beginning = (KE + PE) end.
TRUE
D) All of the above.
FALSE
Answer:
air
Explanation:
The car is being slowed down by air.