Answer:
c. the tilt of the axis of rotation with respect to the Plane of the Ecliptic
Explanation:
The inclination of the ecliptic (or known only as obliqueness) refers to the angle of the axis of rotation with respect to a perpendicular to the plane of the eclipse. He is responsible for the seasons of the year that the planet Earth lends. It is not constant but changes through the movement of nutation. The terrestrial plane of Ecuador and the ecliptic intersect in a line that has an end at the point of Aries and at the diametrically opposite point of Libra.
When the Sun crosses the Aries, the spring equation occurs (between March 20 and 21, the beginning of spring in the northern hemisphere and the early autumn of the southern hemisphere), and from which the Sun is in the North Hemisphere; Pound until you reach the point of the autumn equinox (around September 22-23, beginning fall in the northern hemisphere and spring in the southern hemisphere).
Answer:
5.01×10⁴ J.
Explanation:
Applying,
q = Cm....................... Equation 1
Where q = amount of heat needed to melt the ice, m = mass of the ice, C = specific latent heat of ice.
From the question,
Given: m = 150 g = (150/1000) kg = 0.15 kg, C = 3.34×10⁵ J/kg
Substitute these values into equation 1
q = (0.15×3.34×10⁵)
q = 0.501×10⁵ J
q = 5.01×10⁴ J.
Answer:
Explanation:
Given
![B_z=(2.4\mu T)\sin (1.05\times 10^7x-\omega t)](https://tex.z-dn.net/?f=B_z%3D%282.4%5Cmu%20T%29%5Csin%20%281.05%5Ctimes%2010%5E7x-%5Comega%20t%29)
Em wave is in the form of
![B=B_0\sin (kx-\omega t)](https://tex.z-dn.net/?f=B%3DB_0%5Csin%20%28kx-%5Comega%20t%29)
where ![\omega =frequency\ of\ oscillation](https://tex.z-dn.net/?f=%5Comega%20%3Dfrequency%5C%20of%5C%20oscillation)
![k=wave\ constant](https://tex.z-dn.net/?f=k%3Dwave%5C%20constant)
![B_0=Maximum\ value\ of\ Magnetic\ Field](https://tex.z-dn.net/?f=B_0%3DMaximum%5C%20value%5C%20of%5C%20Magnetic%5C%20Field)
Wave constant for EM wave k is
![k=1.05\times 10^7 m^{-1}](https://tex.z-dn.net/?f=k%3D1.05%5Ctimes%2010%5E7%20m%5E%7B-1%7D)
Wavelength of wave ![\lambda =\frac{2\pi }{k}](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cfrac%7B2%5Cpi%20%7D%7Bk%7D)
![\lambda =\frac{2\pi }{1.05\times 10^7}](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cfrac%7B2%5Cpi%20%7D%7B1.05%5Ctimes%2010%5E7%7D)
![\lambda =5.98\times 10^{-7} m](https://tex.z-dn.net/?f=%5Clambda%20%3D5.98%5Ctimes%2010%5E%7B-7%7D%20m)
(a) The free body of all the forces include, frictional force, weight of the box acting perpendicular and another acting parallel to the plane.
(b) When the box is sliding down, the frictional force acts towards the right.
(c) When the box slides up, the direction of the frictional force changes, it acts towards the left.
<h3>
Free body diagram</h3>
The free body diagram of all the forces on the box is obtained by noting the upward force and downward forces on the box as shown below;
/ W2
Ф → Ff
↓W1
where;
- Ff is the frictional force resisting the down motion of the box
- W1 is the perpendicular component of the box weight = Wcos(33)
- W2 is the parallel component of the box weight = Wsin(33)
(b) When the box is sliding down, the frictional force acts towards the right.
(c) When the box slides up, the direction of the frictional force changes, it acts towards the left.
Learn more about free body diagram of inclined objects here: brainly.com/question/4176810