Answer:
<h2>154.73N</h2>
Explanation:
The question is incomplete. Here is the complete question.
Using the strap at an angle of 31° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15 kg school bag across the floor at a constant velocity. (a) If the force of tension in the strap is 51 N, what is the normal force.
Check the diagram related to the question in the attachment below for better understanding.
The normal force is the reaction acting perpendicular to the force of tension in the strap and opposite the weight of the bag. They are the forces acting along the vertical.
The normal force N will be the sum of the force of tension acting along the vertical (Ty) and the weight of the bag (W).
Ty = 15sin31°
Ty = 7.73N
W = mass * acceleration due to gravity
W = 15.0*9.8
W = 147N
The normal force is therefore expressed as;
N = Ty + W
N = 7.73 + 147
N = 154.73N
<u><em>Developed countries will see a decrease in natural resources, because their population will decrease.</em></u>
Answer:
Explanation:
energy emitted by source per second = .5 J
Eg = 1.43 eV .
Energy converted into radiation = .5 x .12 = .06 J
energy of one photon = 1.43 eV
= 1.43 x 1.6 x 10⁻¹⁹ J
= 2.288 x 10⁻¹⁹ J .
no of photons generated = .06 / 2.288 x 10⁻¹⁹
= 2.6223 x 10¹⁷
wavelength of photon λ = 1275 / 1.43 nm
= 891.6 nm .
momentum of photon = h / λ ; h is plank's constant
= 6.6 x 10⁻³⁴ / 891.6 x 10⁻⁹
= .0074 x 10⁻²⁵ J.s
Total momentum of all the photons generated
= .0074 x 10⁻²⁵ x 2.6223 x 10¹⁷
= .0194 x 10⁻⁸ Js
b ) spectral width in terms of wavelength = 30 nm
frequency width = ?
n = c / λ , n is frequency , c is velocity of light and λ is wavelength
differentiating both sides
dn = c x dλ / λ²
given dλ = 30 nm
λ = 891.6 nm
dn = 3 x 10⁸ x 30 x 10⁻⁹ / ( 891.6 x 10⁻⁹ )²
= 11.3 x 10¹² Hz .
c )
10 nW = 10 x 10⁻⁹ W
= 10⁻⁸ W .
energy of 50 dB
50 dB = 5 B
I / I₀ = 10⁵ ; decibel scale is logarithmic , I is energy of sound having dB = 50 and I₀ = 10⁻¹² W /s
I = I₀ x 10⁵
= 10⁻¹² x 10⁵
= 10⁻⁷ W
= 10 x 10⁻⁸ W
power required
= 10⁻⁸ + 10 x 10⁻⁸ W
= 11 x 10⁻⁸ W.
Answer:
No, it is not proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinder.
Explanation:
A cylinder is said to be infinitely long when is of a sufficient length. Also, when the diameter of the cylinder is relatively small compared to the length, it is called infinitely long cylinder.
Cylindrical rods can also be treated as infinitely long when dealing with heat transfers at locations far from the top or bottom surfaces. However, it not proper to treat the cylinder as being infinitely long when:
* When the diameter and length are comparable (i.e have the same measurement)
When finding the temperatures near the bottom or top of a cylinder, it is NOT PROPER TO USE AN INFINITELY LONG CYLINDER because heat transfer at those locations can be two-dimensional.
Therefore, the answer to the question is NO, since it is not proper to use an infinitely long cylinder when finding temperatures near the bottom or top of a cylinder.
