Answer:
A) s = 796.38 m
B) t = 12.742 s
C) T = 25.484 s
Explanation:
A) First of all let's find the time it takes to get to maximum height using Newton's first equation of motion.
v = u + gt
u = 125 m/s
v = 0 m/s
g = 9.81 m/s²
Thus;
0 = 125 - 9.81(t)
g is negative because motion is against gravity. Thus;
9.81t = 125
t = 125/9.81
t = 12.742 s
Max height will be gotten from Newton's 2nd equation of motion;
s = ut + ½gt²
s = (125 × 12.742) + (½ × -9.81 × 12.742²)
s = 1592.75 - 796.37
s = 796.38 m
B) time to reach maximum height is;
t = u/g
t = 125/9.81
t = 12.742 s
C) Total time elapsed is;
T = 2u/g
T = 2 × 125/9.81
T = 25.484 s
Answer:
1 × 10⁶ N/C
Explanation:
The magnitude of the electric field between the membrane = surface density / permittivity of free space = 10 ⁻⁵C/ m² / (8.85 × 10⁻¹²N⁻¹m⁻²C²) = 1.13 × 10⁶ N/C approx 1 × 10⁶ N/C
Answer:
w = 0.943 rad / s
Explanation:
For this problem we can use the law of conservation of angular momentum
Starting point. With the mouse in the center
L₀ = I w₀
Where The moment of inertia (I) of a rod that rotates at one end is
I = 1/3 M L²
Final point. When the mouse is at the end of the rod
= I w + m L² w
As the system is formed by the rod and the mouse, the forces during the movement are internal, therefore the angular momentum is conserved
L₀ = L_{f}
I w₀ = (I + m L²) w
w = I / I + m L²) w₀
We substitute the moment of inertia
w = 1/3 M L² / (1/3 M + m) L² w₀
w = 1 / 3M / (M / 3 + m) w₀
We substitute the values
w = 1/3 / (1/3 + 0.02) w₀
w = 0.943 w₀
To finish the calculation the initial angular velocity value is needed, if we assume that this value is w₀ = 1 rad / s
w = 0.943 rad / s
Answer:
A
Explanation:
sound waves hit the canyon walls, then bounce back towards you. That's why you hear an echo
To find the relative frequencies, divide each frequency by the total number of throws - in this case, 100.
Red: 10/100=1/10=0.1
Yellow:35/100=0.35
Blue:48/100= 0.48
Misses: 7/100=0.07
The event of landing in the red region has a relative frequency of 0.1 which means the dart landed in the red region about 10% (0.1 x 100%) of the time.