An aluminum rod and a nickel rod are both 5.00 m long at 20.0 degree Celsius. The temperature of each is raised to 70.0 degrees
1 answer:
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since airplane is thrown towards west with speed 6 m/s
while air is blowing with speed 8 m/s towards north
so here the net speed of air plane will be the resultant of airplane speed and wind speed always
SO here we can say it would be a combination of vector along west with must be of length 6 m/s and other vector is towards north with is of length 8 m/s
so correct answer must be 1st option
The acceleration of the wagon is found by applying Newton's Second Law of motion.
1. The responses for question 1 are;
- x-component of the tension in the rope on the right is approximately <u>91.93 N</u>
- y-component of the tension in the rope on the right is approximately <u>71.135 N</u>
- x-component of the tension in the rope on the left is -80.0 N
- y-component of the tension in the rope on the left is 0
2. The net force in the x-direction is approximately <u>11.93 N</u>
3. The net acceleration of the wagon in the horizontal direction is approximately <u>0.1193 m/s²</u>.
Reasons:
The given parameters are;
Mass of the wagon, m = 100.0 kg
Angle of inclination to the horizontal of the rope to the right, θ = 40.0°
Tension in the rope on the right = 120.0 N
Direction in which the rope on the left is pulled = To the west
Tension in the rope on the left = 80.0 N
1. The <em>x</em> and <em>y</em> component of the tension in the rope on the right are;
x-component = 120.0 N × cos(40.0°) ≈ <u>91.93 N</u>
y-component = 120.0 N × sin(40.0°) ≈ <u>77.135 N</u>
The <em>x</em> and <em>y</em> component of the tension in the rope on the left are;
x-component = 80.0 N × cos(180°) = <u>-80.0 N</u>
y-component = 80.0 N × sin(180°) = <u>0.0 N</u>
2. The net force in the horizontal direction, Fₓ, is found as follows;
Fₓ = The x-component of the rope on the left + The x-component of the rope on the right
Which gives;
Fₓ = 91.93 N - 80.0 N = <u>11.93 N</u>
3. The net acceleration of the block is given as follows;
According to Newton's Second Law of motion, we have;
Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ
Fₓ = m × aₓ
Therefore;
- The acceleration of the wagon in the horizontal direction, aₓ ≈ <u>0.1193 m/s²</u>.
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Answer: 313920
Explanation:First, we’re going to assume that the top of the circular plate surface is 2 meters under the water. Next, we will set up the axis system so that the origin of the axis system is at the center of the plate.
Finally, we will again split up the plate into n horizontal strips each of width Δy and we’ll choose a point y∗ from each strip. Attached to this is a sketch of the set up.
The water’s surface is shown at the top of the sketch. Below the water’s surface is the circular plate and a standard xy-axis system is superimposed on the circle with the center of the circle at the origin of the axis system. It is shown that the distance from the water’s surface and the top of the plate is 6 meters and the distance from the water’s surface to the x-axis (and hence the center of the plate) is 8 meters.
The depth below the water surface of each strip is,
di = 8 − yi
and that in turn gives us the pressure on the strip,
Pi =ρgdi = 9810 (8−yi)
The area of each strip is,
Ai = 2√4− (yi) 2Δy
The hydrostatic force on each strip is,
Fi = Pi Ai=9810 (8−yi) (2) √4−(yi)² Δy
The total force on the plate is found on the attached image.
