"on a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. wh
en max pulls on the sled with 12 n of force, directed at an angle of 15° above the ground, how much work does max do on the sled as he pulls his sister 25 m in the snow?"
1 answer:
Refer to the diagram shown below.
The angle θ = 15° relative to the direction of motion.
Therefore the force acting in the direction of motion is
(12 N) cos(15°) = 11.59 N
By definition,
Work = Force x Distance.
Therefore work done in moving the sled by 25 m is
Work = (11.59 N)*(25 m) = 289.8 J
Answer: 289.8 J
The angle θ = 15° relative to the direction of motion.
Therefore the force acting in the direction of motion is
(12 N) cos(15°) = 11.59 N
By definition,
Work = Force x Distance.
Therefore work done in moving the sled by 25 m is
Work = (11.59 N)*(25 m) = 289.8 J
Answer: 289.8 J
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Answer:
4.75 m/s
Explanation:
The computation of the velocity of the existing water is shown below:
Data provided in the question
Tall = 2 m
Inside diameter tank = 2m
Hole opened = 10 cm
Bottom of the tank = 0.75 m
Based on the above information, first we have to determine the height which is
= 2 - 0.75 - 0.10
= 2 - 0.85
= 1.15 m
We assume the following things
1. Compressible flow
2. Stream line followed
Now applied the Bernoulli equation to section 1 and 2
So we get
where,
P_1 = P_2 = hydrostatic
z_1 = 0
z_2 = h
Now
= 4.7476 m/sec
= 4.75 m/s
