Answer:
Explanation:
The x-component is found in the magnitude of the vector times the cosine of the angle.
and, to 3 sig dig,
<u>Increase the thickness of the wire</u> would decrease the resistance in a wire
Explanation:
Thicker wires have a larger cross-section that increases the surface area with which electrons can flow unimpeded. The thicker the wire, therefore, the lower the resistance.
Thin wires have very high resistance the reason the thin tungsten in a bulb glows because it is heated from the high resistance of many electrons trying to pass through a very small cross-section.
Answer:
The heat transferred through the wall that day is 13728 BTUs
Explanation:
Here, we have the area of the wall given as
Area of wall = 2 × Length × Height + 2 × Width × Height
Length = 15 feet
Width = 11 Feet and
Height = 9 feet
Therefore, the area = 2×15×9 + 2×11×9 = 468 ft²
Temperature difference is given by
Average outside temperature - Wall temperature = 40 - 18 = 22 °F
Therefore the heat transferred through the wall that day (24 hours) at 18 sq.ft. hr/BTU is given by;
468 × 22 × 24/18 = 13728 = 13728 BTUs.
Given conditions:
height of object = 7.5cmdistance of object from mirror = 14 cmfocus length = -7 cmimage distance = ?
Using mirror formula:
1/(focus length) = 1/(object distance) + 1/(image distance)
or, -1/7 = 1/14 + 1/(image distance)
or, image distance = -4.66cm (the image formed is a virtual image)
Also, magnification of image is:
image height /height of object = - image distance /object distance
or, image height = - image distance / object distance * height of object
or, image height = -(-4.66) / 14 * 7.5 = 2.49 = 3(nearest whole number)
Answer:
6
Explanation:
We are given that
Slid width,a=0.110 mm=
Wavelength,
m
We have to find the number of diffraction maxima are contained in a region of the Fraunhofer single-slit pattern.
Using the formula
Hence, 6 diffraction maxima are contained in a region of the Fraunhofer single-slit pattern
