The acceleration is the principal subordinate of the speed if the speed is steady the subsidiary is invalid if the speed is diminishing the subsidiary is negative. When discussing so much stuff we consider the momentary esteem.
<span>Note that when you back off, you back off by and large yet can locally in time quicken a tiny bit, suppose amid 1/tenth of a sec since you achieved a segment of the street which was slanting. In any case, this does not change the way that when the speed diminishes, the quickening is negative.</span>
Answer:
the reflected wave is inverted and the transmitted wave is up
Explanation:
To answer this question we must analyze the physical phenomenon, with an wave reaching a discontinuity, we can analyze it as a shock.
Let's start when the discontinuity is with a fixed, very heavy and rigid obstacle, in this case the reflected wave is inverted, since the contact point cannot move
In the event that it collides with an object that can move, the reflected wave is not inverted, this is because the point can rise, they form a maximum at this point.
In the proposed case the shock is when the thickness changes, in this case we have the above phenomena, a part of the wave is reflected by being inverted and a part of the wave is transmitted without inverting.
The amplitude sum of the amplitudes of the two waves is proportional to the lanería that is distributed between them.
When checking the answers the correct one is the reflected wave is inverted and the transmitted wave is up
Answer is B. According to the equation of motion s = vt + 1/2 at2 Where s is distance covered, v is velocity, a is acceleration and t is time taken. So, by putting all the values, we get s = (20)(5) + 1/2 (3)(5)2 s = 100 + 1/2 (3)(25) s = 100 + 1/2 75 s = 100 + 37.5 s = 137.5 meters
Answer:
11.7 m/s
Explanation:
To find its speed, we first find the acceleration of the center of mass of a rolling object is given by
a = gsinθ/(1 + I/MR²) where θ = angle of slope = 4, I = moment of inertia of basketball = 2/3MR²
a = 9.8 m/s²sin4(1 + 2/3MR²/MR²)
= 9.8 m/s²sin4(1 + 2/3)
= 9.8 m/s²sin4 × (5/3)
= 1.14 m/s²
To find its speed v after rolling for 60 m, we use
v² = u² + 2as where u = initial speed = 0 (since it starts from rest), s = 60 m
v = √(u² + 2as) = √(0² + 2 × 1.14 m/s × 60 m) = √136.8 = 11.7 m/s
Answer:
53.32°C
Explanation:
Length of the aluminium wing = 35 m
Change in length of aluminium wing = 0.03 m
The linear expansion coefficient of aluminium 
We know that change in length is given by 
So 
So final temperature 
