Answer:
N = 177843 sheets
Explanation:
We are given;
Mass;m = 0.0035 kg
Pressure; p = 101325 pa = 101325 N/m²
L = 0.279m
W = 0.216m
The weight of N sheets is N(mg)
Where;
m is the mass of one sheet
N is number of sheets
g is the acceleration due to gravity.
The pressure equals weight divided by the area on which the weight presses:
Thus,
p= F/A = Nmg/(L•W)
Therefore, making N the subject;
N = pLW/(mg)
N = 101325 x 0.279 x 0.216/ (0.0035 x 9.81)
N = 177843
100N describes the weight of the sandbag, while 100kg is the mass of the sandbag.
To calculate acceleration, divide your weight by the mass, thus the accleration is:
Answer:
Missing numbers are 3,11,13, 15,17,19, 23,27,29 and wrong are 6,68
(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.
(b) The maximum height above the ground reached by the ball is 8.6 m.
(c) The distance off course the ball would be carried is 0.38 m.
(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
<h3>
Horizontal and vertical components of the ball's velocity</h3>
Vx = Vcosθ
Vx = 39.7 x cos(17.8)
Vx = 37.8 m/s
Vy = Vsin(θ)
Vy = 39.7 x sin(17.8)
Vy = 12.14 m/s
<h3>Maximum height reached by the ball</h3>

Maximum height above ground = 7.51 + 1.09 = 8.6 m
<h3>Distance off course after 2 second </h3>
Upward speed of the ball after 2 seconds, V = V₀y - gt
Vy = 12.14 - (2x 9.8)
Vy = - 7.46 m/s
Horizontal velocity will be constant = 37.8 m/s
Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

<h3>Resultant speed of the ball and crosswind</h3>

<h3>Distance off course the ball would be carried</h3>
d = Δvt = (38.72 - 38.53) x 2
d = 0.38 m
The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
Learn more about projectiles here: brainly.com/question/11049671