I believe it's A. Sorry if I'm incorrect.
Answer:
A. 1.71 m
B. 2.66 m
Explanation:
A. Determination of the height of the pier.
We'll begin by calculating the time taken for the ball to get to the water
This can be obtained as follow:
Horizontal velocity (u) = 1.27 m/s,
Horizontal distance (s) = 0.75 m
Time (t) =?
s = ut
0.75 = 1.27 × t
Divide both side by 1.27
t = 0.75 / 1.27
t = 0.59 s
Finally, we shall determine the height of the pier as follow:
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) = 0.59 s
Height of pier (h) =?
h = ½gt²
h = ½ × 9.8 × 0.59²
h = 4.9 × 0.3481
h = 1.71 m
Thus, the height of the pier is 1.71 m
B. Determination of the horizontal distance.
Horizontal velocity (u) = 4.50 m/s
Time (t) = 0.59 s
Horizontal distance (s) =?
s = ut
s = 4.5 × 0.59
s = 2.66 m
Thus, if the ball moved at a velocity of 4.50 m/s off the pier, it will land at a distance of 2.66 m from the end of the pier.
Answer:
Explanation:
For acceleration due to gravity g , the expression is
g = GM / R² , where G is gravitational constant M is mass of the earth and R is radius of the earth .
At height h , let the value of it becomes g / 4 , so
g / 4 = GM / ( R + h )²
dividing
4 = [( R+ h)² / R² ]
2 = (R + h) / R
2 = 1 + h / R
h / R = 1
h = R
So at height equal to radius of the earth , acceleration due to gravity becomes 1 /4 of value on the surface of the earth .
Answer:
The answer is C, a thermometer reads 0.1 degree higher than it should.
Answer:
The Tension T is 42120N
The Horizontal force component is 18322.2N
The Vertical force component is - 4729N
Explanation:
First, you have to find the angle between the drawbridge and the cable using sine and cosine rule. This will result in angle 44.2°. Hence, the angle between the horizontal axis and the cable will be 64.2° (44.2° + 20°).
Having done that, you apply two conditions of equilibrium.
1. THE VECTOR SUM OF ALL FORCES EQUAL ZERO.
∑Fx = 0
∑Fx = Rx - Tcos64.2 = 0
Rx = 0.435T
∑Fy = 0
∑Fy = Ry + Tsin64.2 - W - w = 0
W = 2000kg × 9.8 = 19600N
w =1000kg × 9.8 = 9800N
Ry + 0.9T = 29400N
Ry = 29400 - 0.9T
2. THE SUM TOTAL OF TORQUES EQUALS ZERO
Rx: τ = 0
Ry: τ = 0
T: τ = 5 × Tsin44.2
= 3.49T m
W: τ = 4 × 19600sin90
= 78400Nm
w: τ = 7 × 9800sin9
= 68600Nm
Note:
Rx = x component of Reaction force
Ry = y component of Reaction force.
T = Tension
W = weight of bridge
w = weight of Sir Lance a Lost and his steed
τ = torque
Note: The torque of Tension is counter clockwise while that of the weights is clockwise.
Hence,
∑τccw = ∑τcw
3.49T = 78400 + 68600
3.49T = 14700Nm
T = 147000/3.49
T = 42120N
Rx = 0.435 × 42120
Rx = 18322.2N
Ry = 29400N - (0.9×42120)N
Ry = 29400 - 34129
Ry = -4729N
Note: Ry being negative means that the hinge of the drawbridge exerts a downward force.