Answer:
The solution is attached in the pictures below
Explanation:
Answer:
time to fall is 3.914 seconds
Explanation:
given data
half distance time = 1.50 s
to find out
find the total time of its fall
solution
we consider here s is total distance
so equation of motion for distance
s = ut + 0.5 × at² .........1
here s is distance and u is initial speed that is 0 and a is acceleration due to gravity = 9.8 and t is time
so for last 1.5 sec distance is 0.5 of its distance so equation will be
0.5 s = 0 + 0.5 × (9.8) × ( t - 1.5)² ........................1
and
velocity will be
v = u + at
so velocity v = 0+ 9.8(t-1.5) ..................2
so first we find time
0.5 × (9.8) × ( t - 1.5)² = 9.8(t-1.5) + 0.5 ( 9.8)
solve and we get t
t = 3.37 s
so time to fall is 3.914 seconds
Answer:
202.8m
Explanation:
Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.
First calculate the total time travelled by using the second equation of motion
h = Ut + 1/2gt^2
Let assume that u = 0
And h = 3.5
Substitute all the parameters into the formula
3.5 = 1/2 × 9.8 × t^2
3.5 = 4.9t^2
t^2 = 3.5/4.9
t^2 = 0.7
t = 0.845s
To know how far the cannonball travel, let's use the equation
S = UT + 1/2at^2
But acceleration a = 0
T = 2t
T = 1.69s
S = 120 × 1.69
S = 202.834 m
Therefore, the distance travelled by the cannon ball is approximately 202.8m.
Answer:
power emitted is 1.75 W
Explanation:
given data
length l = 5 cm = 5 ×
m
diameter d = 0.074 cm = 74 ×
m
total filament emissivity = 0.300
temperature = 3068 K
to find out
power emitted
solution
we find first area that is π×d×L
area = π×d×L
area = π×74 ×
×5 ×
area = 1162.3892 ×
m²
so here power emitted is express as
power emitted = E × σ × area × (temperature)^4
put here all value
power emitted = 0.300× 5.67 × 1162.3892 ×
× (3068)^4
power emitted = 1.75 W
Answer:

Explanation:
From the question we are told that:
Crane Length 
Crane Mass 
Arm extension at lifting end 
Arm extension at counter weight end 
Load 
Generally the equation for Torque Balance is mathematically given by


