Answer:
F = 41,954 N
Explanation:
given,
mass of bucket = 580 Kg
length of the cable = 20 m
velocity = 40 m/s
angle made = 38.0°
T cos 38° = m g..............(1)
T sin 38^0 = \dfrac{mv^2}{l} + F......(2)
dividing equation (2) by (1)
F = -46400 + 4445.36
F = -41,954 N
hence, the force is acting in the opposite direction as assumed.
F = 41,954 N
As per energy conservation we know that
Energy enter into the bulb = Light energy + Thermal energy
so now we have
energy enter into the bulb = 100 J
Light energy = 5 J
now from above equation we have
True this is true bc yes as you said you're at the most direct point of sunlight
Answer:
a. 13.7 s b. 6913.5 m
Explanation:
a. How much time before being directly overhead should the box be dropped?
Since the box falls under gravity we use the equation
y = ut - 1/2gt² where y = height of plane above ocean = 919 m, u = initial vertical velocity of airplane = 0 m/s, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the airplane to be directly overhead.
So,
y = ut - 1/2gt²
y = 0 × t - 1/2gt²
y = 0 - 1/2gt²
y = - 1/2gt²
t² = -2y/g
t = √(-2y/g)
So, t = √(-2 × 919 m/-9.8 m/s²)
t = √(-1838 m/-9.8 m/s²)
t = √(187.551 m²/s²)
t = 13.69 s
t ≅ 13.7 s
So, the box should be dropped 13.69 s before being directly overhead.
b. What is the horizontal distance between the plane and the victims when the box is dropped?
The horizontal distance x between plane and victims, x = speed of plane × time it takes for box to drop = 505 m/s × 13.69 s = 6913.45 m ≅ 6913.5 m
An interesting problem, and thanks to the precise heading you put for the question.
We will assume zero air resistance.
We further assume that the angle with vertical is t=53.13 degrees, corresponding to sin(t)=0.8, and therefore cos(t)=0.6.
Given:
angle with vertical, t = 53.13 degrees
sin(t)=0.8; cos(t)=0.6;
air-borne time, T = 20 seconds
initial height, y0 = 800 m
Assume g = -9.81 m/s^2
initial velocity, v m/s (to be determined)
Solution:
(i) Determine initial velocity, v.
initial vertical velocity, vy = vsin(t)=0.8v
Using kinematics equation,
S(T)=800+(vy)T+(1/2)aT^2 ....(1)
Where S is height measured from ground.
substitute values in (1): S(20)=800+(0.8v)T+(-9.81)T^2 =>
v=((1/2)9.81(20^2)-800)/(0.8(20))=72.625 m/s for T=20 s
(ii) maximum height attained by the bomb
Differentiate (1) with respect to T, and equate to zero to find maximum
dS/dt=(vy)+aT=0 =>
Tmax=-(vy)/a = -0.8*72.625/(-9.81)= 5.9225 s
Maximum height,
Smax
=S(5.9225)
=800+(0.8*122.625)*(5.9225)+(1/2)(-9.81)(5.9225^2)
= 972.0494 m
(iii) Horizontal distance travelled by the bomb while air-borne
Horizontal velocity = vx = vcos(t) = 0.6v = 43.575 m/s
Horizontal distace travelled, Sx = (vx)T = 43.575*20 = 871.5 m
(iv) Velocity of the bomb when it strikes ground
vertical velocity with respect to time
V(T) =vy+aT...................(2)
Substitute values, vy=58.1 m/s, a=-9.81 m/s^2
V(T) = 58.130 + (-9.81)T =>
V(20)=58.130-(9.81)(20) = -138.1 m/s (vertical velocity at strike)
vx = 43.575 m/s (horizontal at strike)
resultant velocity = sqrt(43.575^2+(-138.1)^2) = 144.812 m/s (magnitude)
in direction theta = atan(43.575,138.1)
= 17.5 degrees with the vertical, downward and forward. (direction)
