Answer:
0.71 m/s
Explanation:
We find the time it takes the stone to hit the water.
Using y = ut - 1/2gt² where y = height of bridge, u = initial speed of stone = 0 m/s, g = acceleration due to gravity = -9.8 m/s² (negative since it is directed downwards)and t = time it takes the stone to hit the water surface.
So, substituting the values of the variables into the equation, we have
y = ut - 1/2gt²
82.2 m = (0m/s)t - 1/2( -9.8 m/s²)t²
82.2 m = 0 + (4.9 m/s²)t²
82.2 m = (4.9 m/s²)t²
t² = 82.2 m/4.9 m/s²
t² = 16.78 s²
t = √16.78 s²
t = 4.1 s
This is also the time it takes the raft to move from 5.04 m before the bridge to 2.13 m before the bridge. So, the distance moved by the raft in time t = 4.1 s is 5.04 m - 2.13 m = 2.91 m.
Since speed = distance/time, the raft's speed v = 2.91 m/4.1 s = 0.71 m/s
Answer:
(a) A = 0.700m
(b) k = 80.6N/m
(c) x = -0.699m
(d) x = -0.350m
(e) t = 0.168s
Explanation:
Given the equation of motion for the spring
X = 0.700cos(12.0t), m = 0.56kg
(a) A = amplitude = 0.700m
(b) The angular velocity ω = 12rad/s
ω = √(k/m)
ω² = k/m
k = m×ω² = 0.56×12² = 80.6N/m
Spring constant k = 80.6N/m
(c) T = 2π/ω = 2π/12
T = 0.524s
At t = T/2 = 0.524/2 = 0.262s
So x = 0.700cos(12×0.262) = –0.699m
(d) At t = 2/3×T = 2×0.524/3 = 0. 349s
x = 0.700cos(12×0.349) = –0.350m
(e) to find t at x = -0.300m
–0.300 = 0.700cos(12t)
–0.300/0.700 = cos(12t)
cos(12t) = –0.429
12t = cos-¹(-0.429)
12t = 2.01
t = 2.01/12
t = 0.168s
Answer:c>a>b
Explanation:
Given
All the engine extracts same amount of heat(Q) from High-temperature reservoir
For a) 400 and 500 K
For b)500 K and 600K
For c) 400 K and 600 K
So c will give the highest amount of work
c>a>b
Answer:
v=10.75m/s
Explanation:
It is given that,
Pete drives 215 Meters in 20 seconds. He is driving down 7th floor. We need to find his speed.
Speed = distance/time
⇒
Hence, the speed of Pete is 10.75 m/s.
