Answer:
<h2>154.73N</h2>
Explanation:
The question is incomplete. Here is the complete question.
Using the strap at an angle of 31° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15 kg school bag across the floor at a constant velocity. (a) If the force of tension in the strap is 51 N, what is the normal force.
Check the diagram related to the question in the attachment below for better understanding.
The normal force is the reaction acting perpendicular to the force of tension in the strap and opposite the weight of the bag. They are the forces acting along the vertical.
The normal force N will be the sum of the force of tension acting along the vertical (Ty) and the weight of the bag (W).
Ty = 15sin31°
Ty = 7.73N
W = mass * acceleration due to gravity
W = 15.0*9.8
W = 147N
The normal force is therefore expressed as;
N = Ty + W
N = 7.73 + 147
N = 154.73N
Answer:
Explanation:
Impulse of a force is measured by force x time or F X t
Impulse also equals change in momentum or
F x t = m v₂ - m v₁
The given case is as follows
in the first case
F x t = mv - o = mv
F = mv / t
in the second case
F₁ x 4 t = mv
F₁ = 1/4 x mv /t
F₁ = F / 4
option a) is correct .
iii )
In the last case
F₂ X t = m v/2 -0
F₂ = 1/2 x mv / t
= 1/2 x F
F₂ = F/2
Option e ) is correct.
Answer:
change in entropy is 1.44 kJ/ K
Explanation:
from steam tables
At 150 kPa
specific volume
Vf = 0.001053 m^3/kg
vg = 1.1594 m^3/kg
specific entropy values are
Sf = 1.4337 kJ/kg K
Sfg = 5.789 kJ/kg
initial specific volume is calculated as





FROM STEAM Table
at 200 kPa
specific volume
Vf = 0.001061 m^3/kg
vg = 0.88578 m^3/kg
specific entropy values are
Sf = 1.5302 kJ/kg K
Sfg = 5.5698 kJ/kg
constant volume so




Change in entropy 
=3( 3.36035 - 2.88) = 1.44 kJ/kg
Answer:
Weight
Explanation:
The spring balance is used to measure weight of an object.
Answer:
d=360 miles
Donna lives 360 miles from the mountains.
Explanation:
Conceptual analysis
We apply the formula to calculate uniform moving distance[
d=v*t Formula (1)
d: distance in miles
t: time in hours
v: speed in miles/hour
Development of problem
The distance Donna traveled to the mountains is equal to the distance back home, equal to d,then,we pose the kinematic equations for d, applying formula 1:
travel data to the mountains: t₁= 8 hours , v=v₁
d= v₁*t₁=8*v₁ Equation (1)
data back home : t₂=4hours , v=v₂=v₁+45
d=v₂*t₂=(v₁+45)*4=4v₁+180 Equation (2)
Equation (1)=Equation (2)
8*v₁=4v₁+180
8*v₁-4v₁=180
4v₁=180
v₁=180÷4=45 miles/hour
we replace v₁=45 miles/hour in equation (1)
d=8hour*45miles/hour
d=360 miles