The work done to pull the object 7.0 m is the total area under the graph from 0.0 m to 7.0 m, determined as 245 J.
<h3>Work done by the applied force</h3>
The area under force versus displacement graph is work done.
The total work done by pulling the object 7 m, can be grouped into two areas;
- First area, A1 = area of triangle from 0 m to 2.0 m
- Second area, A2 = area of trapezium, from 2.0 m to 7.0 m
A1 = ¹/₂ bh
A1 = ¹/₂ x (2) x (20)
A1 = 20 J
A2 = ¹/₂(large base + small base) x height
A2 = ¹/₂[(7 - 2) + (7-3)] x 50
A2 = ¹/₂(5 + 4) x 50
A2 = 225 J
<h3>Total work done </h3>
W = A1 + A2
W = 20 J + 225 J
W = 245 J
Learn more about work done here: brainly.com/question/8119756
Answer:
Conduction, Convection and Radiation
Explanation:
Hope this helps!
You will need to turn this into scientific notation, so:
1.176 × 10^1 cm^2
To two significant figures means to two digits so since 7 is greater than 5, you will need to round up.
Final answer is:
1.2×10^1 cm^2
Answer:
15.106 N
Explanation:
From the given information,
The weight of the bucket can be calculated as:
The mass of the water accumulated in the bucket after 3.20s is:
To determine the weight of the water accumulated in the bucket, we have:
For the speed of the water before hitting the bucket; we have:
v = 8.4 m/s
Now, the force required to stop the water later when it already hit the bucket is:
F = 1.68 N
Finally, the reading scale is:
= 7.154 N + 6.272 N + 1.68 N
= 15.106 N
Answer:
∆PE = 749.7 J
At 0.9 m high, PE = 793.8 J
At 1.75 m high, PE = 1543.5 J
