Answer:
a)Work done by fireman= 2.15 Btu
b) Time t= 0.86 sec
Explanation:
Given that
Weight = 280 lbf
We know that 1 lbf = 4.44 N
so 280 lbf = 1245.5 N
Weight =1245.5 N
Height h = 60 ft
We know that
1 ft = 0.3048 m
So 60 ft = 18.28 m
h =18.28 m
Power = 3.5 hp
We know that
1 hp =0.74 KW
So 3.5 hp = 2.61 KW
Power = 2.61 KJ/s
So the work done by fireman = Weight x h
Now by putting the values
Work done by fireman= 1245.5 x 18.28 J
Work done by fireman= 2267.74 J
Work done by fireman= 2.26774 KJ
We know that 1 Btu= 1.05 KJ
So 2.266 KJ = 2.15 Btu
Work done by fireman= 2.15 Btu
We know that ,rate of work is called power.
Power x time = work
2.61 x t = 2.26
So t= 0.86 sec
Answer:
Explanation:
mass of object, m = 3 kg
spring constant, K = 750 n/m
compression, x = 8 cm = 0.08 m
angle of gun, θ = 30°
(a) As the ball is launched, it has some velocity due to the compression in the spring, so it has some kinetic energy.
(b) Let v be th evelocity of ball at the tim eof launch.
by using the conservation of energy
1/2 Kx² = 1/2 mv²
750 x 0.08 x 0.08 = 3 x v²
v = 1.265 m/s
By use of the formula of maximum height
h = 0.02 m
h = 2 cm
Answer:
She should drop two balls with different masses from the same height
Explanation:
Answer:
the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15
Explanation:
Given that;
speed of car V = 120 km/h = 33.3333 m/s
Reaction time of an alert driver = 0.8 sec
Reaction time of an alert driver = 3 sec
extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec
now, extra distance that car will travel in case of sleepy driver will be'
S_d = V × 2.2 sec
S_d = 33.3333 m/s × 2.2 sec
S_d = 73.3333 m
hence, number of car of additional car length n will be;
n = S_n / car length
n = 73.3333 m / 5m
n = 14.666 ≈ 15
Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15
