Answer:
a) X = 17.64 m
b) X = 17.64 + 4∆t^2 + 16.8∆t
c) Velocity = lim(∆t→0)〖∆X/∆t〗 = 16.8 m/s
Explanation:
a) The position at t = 2.10s is:
X = 4t^2
X = 4(2.10)^2
X = 17.64 m
b) The position at t = 2.10 + ∆t s will be:
X = 4(2.10 + ∆t)^2
X = 17.64 + 4∆t^2 + 16.8∆t m
c) ∆X is the difference between position at t = 2.10s and t = 2.10 + ∆t so,
∆X= 4∆t^2 + 16.8∆t
Divide by ∆t on both sides:
∆X/∆t = 4∆t + 16.8
Taking the limit as ∆t approaches to zero we get:
Velocity =lim(∆t→0)〖∆X/∆t〗 = 4(0) + 16.8
Velocity = 16.8 m/s
Answer:
98 m √
Explanation:
How about s = Vo * t + ½at² ?
s = h = Vo * 2s - 4.9m/s² * (2s)² = 2Vo - 19.6
and
h = Vo * 10s - 4.9m/s² * (10s)² = 10Vo - 490
Subtract 2nd from first:
0 = -8Vo + 470.4
Vo = 58.8 m/s
h = 58.8m/s * 2s - 4.9m/s² * (2s)² = 98 m
r = radius of the circle traveled by the particle = 76 cm = 0.76 m
T = time period of revolution for the particle = 4.5 s
w = angular velocity of the particle
angular velocity of the particle is given as
w = 2π/T
inserting the values
w = 2 (3.14)/4.5
w = 1.4 rad/s
a = centripetal acceleration of the particle in the circle
centripetal acceleration is given as
a = r w²
inserting the values
a = (0.76) (1.4)²
a = 1.5 m/s²
The answer is: none of the above.
Explanation:
When light reflects from a surface, the frequency, wavelength, and speed do not change. They remain the same.
