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
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<u>Answer</u>
0.00346 hL
<u>Explanation</u>
cL means Centilitre while hL means Hectolitre.
10,000 cL = 1 hL
∴ 34.6 cL = 34.6/10,000 hL
= <em>0.00346 hL</em>
Answer:
The angular velocity at the beginning of the interval is
.
Explanation:
Given that,
Angular acceleration 
Angular displacement 
Angular velocity 
We need to calculate the angular velocity at the beginning
Using formula of angular velocity


Where,
= angular acceleration
= angular velocity
Put the value into the formula



Hence, The angular velocity at the beginning of the interval is
.
Answer:
<h2>73 kg</h2>
Explanation:
The mass of the object can be found by using the formula

f is the force
a is the acceleration
From the question we have

We have the final answer as
<h3>73 kg</h3>
Hope this helps you