<span>The question is 'a centre seeking force related to acceleration is ............... force. The answer is centripetal force. Motion in a curved path is an accelerated motion and it requires a force that will direct the moving object towards the centre of curvature of the path of motion. This centre seeking force is known as centripetal force.</span>
Use Newton's second law and the free body diagram to determine the net force and acceleration of an object. In this unit, the forces acting on the object were always directed in one dimension.
The object may have been subjected to both horizontal and vertical forces but there was no single force directed both horizontally and vertically. Moreover, when free-body diagram analysis was performed, the net force was either horizontal or vertical, never both horizontal and vertical.
Times have changed and we are ready for situations involving two-dimensional forces. In this unit, we explore the effects of forces acting at an angle to the horizontal. This makes the force act in two dimensions, horizontal and vertical. In such situations, as always in situations involving one-dimensional network forces, Newton's second law applies.
Learn more about Newton's second law here:-brainly.com/question/25545050
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The frequency of the light source is 1.5 x 10¹⁵ Hz.
<h3>
Frequency of the light source</h3>
The frequency of the light source is determined using the following equations;
c = fλ
where;
c is speed of light
f is the frequency
λ is the wavelength
f = (3 x 10⁸) / (2 x 10⁻⁷)
f = 1.5 x 10¹⁵ Hz
Thus, the frequency of the light source is 1.5 x 10¹⁵ Hz.
Learn more about frequency of light here: brainly.com/question/10728818
Here is the full information about the question. <span>Ilya and Anya each can run at a speed of 8.50mph and walk at a speed of 3.50 mph . They set off together on a route of length 5.00 miles . Anya walks half of the distance and runs the other half, while Ilya walks half of the time and runs the other half.
the calculation would be:
</span><span>
t = d / s </span>
<span>t = 2.5 (half of the total distance) / 8.5 (speed of running) </span>
<span>This is .294 hours which is about 1058s... </span>
<span>for the walking part... </span>
<span>t = d / s </span>
<span>t = 2.5 / 3.5 </span>
<span>t = 5/7hours = 2571 s. </span>
