Answer:
-2.5m/s²
Explanation:
The acceleration of a body is giving by the rate of change of the body's velocity. It is given by
a = Δv / t ----------------(i)
Where;
a = acceleration (measured in m/s²)
Δv = change in velocity = final velocity - initial velocity (measure in m/s)
t = time taken for the change (measured in seconds(s))
From the question;
i. initial velocity = 5m/s
final velocity = 0 [since the body (ball) comes to rest]
Δv = 0 - 5 = -5m/s
ii. time taken = t = 2s
<em>Substitute these values into equation (i) as follows;</em>
a = (-5m/s) / (2s)
a = -2.5m/s²
Therefore, the acceleration of the ball is -2.5m/s²
NB: The negative sign shows that the ball was actually decelerating.
Answer
Hi,
An increase in amplitude from 3m to 6 m increases the energy it transports. The frequency of the wave is not affected
Explanation
Amplitude is the height of a wave where as frequency is the number of waves that pass by each second. A wave with bigger amplitude has more energy than a wave with smaller amplitude. A point where more waves pass contains more energy that is transferred every second. The change in the amplitude of a wave does not change its frequency. However, frequency is inversely related to the wavelength of a wave.
Best Wishes!
<span>an apparatus for producing gas, steam, or another product. here is what it means sorry if i couldn't help. </span>
Answer:
As the temperature increases, the kinetic energy of the particles increases.
Explanation:
When the temperature of the substance increases, the velocity increases which makes the movement of the particles to speed up. This causes the particles to increase. Therefore, as the temperature increases, the kinetic energy of the particles also increases.
Answer:
Explanation:
Let the plastic rod extends from - L to + L .
consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .
It will create a field at point P on y -axis . Distance of point P
= √ x² + .15²
electric field at P due to small charged length
dE = k λ dx x / (x² + .15² )
Its component along Y - axis
= dE cosθ where θ is angle between direction of field dE and y axis
= dE x .15 / √ x² + .15²
= k λ dx .15 / (x² + .15² )³/²
If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L
E = ∫ k λ .15 / (x² + .15² )³/² dx
= k λ x L / .15 √( L² / 4 + .15² )
