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:
A. 3,000,000 m
B. 0.25 km
C. 10 m
D. 1,000 cm
Explanation:
no hablo español, así que solo ingrese esto en el traductor de G*ogle
A. One kilometer equals 1000 meters, so
3,000*1,000 = 3,000,000 m
B. One meter equals 0.001 kilometer, so
250*0.001 = 0.25 km
C. One centimeter equals 0.01 meter
1,000*0.01 = 10 m
D. One milimeter equals 0.1 centimer, so
10,000*0.1 = 1,000
Answer:
≈933.3kg/m^3
Explanation:
Density=Mass/Volume
11200kg/12.0= 933.3333kg/m^3
Answer:
I think u have to do 580N times 40KG