Akito pushes a wheelbarrow with 800 W of power. How much work is required to get the wheelbarrow across the yard in 12 s? Round
2 answers:
Result:
9600 J
Clarification:
Power can be expressed as the quotient of work performed and time utilized:
where W represents work done and t denotes time taken.
In this scenario, we know the power:
P = 800 W
and the time allocated:
t = 12 s
Thus, we can rearrange the equation to determine the required work:
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#1
The volume of lead measures 100 cm^3
with a density of lead at 11.34 g/cm^3
. Thus, the mass of the lead block equals density multiplied by volume
Therefore, its weight in air is noted as
Next, the buoyant force acting on the lead is defined as
We know that
After solving, we find
V = 11.22 cm^3
(ii) This corresponding volume of water exerts the same weight as the buoyant force, resulting in 0.11 N
(iii) The buoyant force measures 0.11 N
(iv) The lead block sinks in water due to its density being greater than that of water.
#2
The buoyant force acting on the lead block counterbalances its weight
(ii) This volume of mercury corresponds to the buoyant force weight, confirming that the block floats within mercury, resulting in 11.11 N as its weight.
(iii) The buoyant force is recorded as 11.11 N
(iv) Given that lead's density is less than mercury's, the lead will float in the mercury medium.
#3
Indeed, an object that has lesser density than a liquid will float; otherwise, it will sink in the liquid.
Answer:
0.0984
Explanation:
The first diagram below illustrates a free body diagram that will aid in resolving this problem.
According to the diagram, the force's horizontal component can be expressed as:
Substituting 42° for θ and 87.0° for
Meanwhile, the vertical component is:
Again substituting 42° for θ and 87.0° for
In resolving the vector, let A denote the components in mutually perpendicular directions.
The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ
The frictional force can be expressed as:
Where;
is the coefficient of friction
N = the normal force
Also, the normal reaction (N) is calculated as mg - F sin θ
Substituting . Normal reaction becomes:
By balancing the forces, the horizontal component of the force equals the frictional force.
The horizontal component is described as follows:
Rearranging the equation above to isolate leads to:
Substituting in the following values:
m = 73 kg
g = 9.8 m/s²
Thus:
Therefore, the coefficient of friction is = 0.0984
Response:
Clarification:
Provided
weight of disk
diameter of disc
weight of ring
Force
Overall moment of inertia
=Disc's moment of inertia +Ring's Moment of Inertia
At this point, Torque is
Utilizing
in this scenario