Answer:
Explanation:
Deceleration of solid disk = g sin10/1 + k²/r² = g sin 10 / 1 + 1/2 = g sin 10 x 2/3
[ k is radius of gyration of disk which is equal to( 1/√2)x r ]
deceleration a = -1.1345 m/s²
v = u - at , t = u / a = 1.5 / 1.1345 = 1.322 s.
The magnetic field or force seems to be associated with the lineup of electrons withim the magnet
Distance of fall from rest,
without air resistance = (1/2) (gravity) (time)²
= (1/2) (9.8 m/s²) (95 sec)²
= (4.9 m/s²) (9,025 sec²)
= 44,222.5 meters .
The depth of the mine shaft is five times the height of Mt. Everest !
Answer:
<h2>Virtual image</h2>
Explanation:
<h3>
<em>Virtual</em><em> </em><em>image</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>caught</em><em> </em><em>on</em><em> </em><em>a</em><em> </em><em>screen</em></h3>
<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>you</em><em>.</em>
<em>will</em><em> </em><em>give</em><em> </em><em>the</em><em> </em><em>brainliest</em><em>!</em>
<em>follow</em><em> </em><em>~</em><em>H</em><em>i</em><em>1</em><em>3</em><em>1</em><em>5</em><em>~</em>
Answer:
D. Calculate the area under the graph.
Explanation:
The distance made during a particular period of time is calculated as (distance in m) = (velocity in m/s) * (time in s)
You can think of such a calculation as determining the area of a rectangle whose sides are velocity and time period. If you make the time period very very small, the rectangle will become a narrow "bar" - a bar with height determined by the average velocity during that corresponding short period of time. The area is, again, the distance made during that time. Now, you can cover the entire area under the curve using such narrow bars. Their areas adds up, approximately, to the total distance made over the entire span of motion. From this you can already see why the answer D is the correct one.
Going even further, one can make the rectangular bars arbitrarily narrow and cover the area under the curve with more and more of these. In fact, in the limit, this is something called a Riemann sum and leads to the definition of the Riemann integral. Using calculus, the area under a curve (hence the distance in this case) can be calculated precisely, under certain existence criteria.
