Answer:
To achieve the velocity of 40 m/sec height will become 4 times
Explanation:
We have given initially truck is at rest and attains a speed of 20 m/sec
Let the mass of the truck is m
At the top of the hill potential energy is mgh and kinetic energy is 
So total energy at the top of the hill 
At the bottom of the hill kinetic energy is equal to
and potential energy will be 0
So total energy at the bottom of the hill is equal to 
Form energy conservation 
, for v = 20 m/sec

Squaring both side

h = 20.408 m
Now if velocity is 0 m/sec


h = 81.63 m
So we can see that to achieve the velocity of 40 m/sec height will become 4 times
Answer:
Yes.
Explanation:
A negative power would just represent a loss of power. So in your case it lost -1252.16 W
Answer:
<h2>
15m/s</h2>
Explanation:
The equation for a traveling wave as expressed as y(x, t) = A cos(kx −
t) where An is the amplitude f oscillation,
is the angular velocity and x is the horizontal displacement and y is the vertical displacement.
From the formula;
where;

Before we can get the transverse speed, we need to get the frequency and the wavelength.
frequency = 1/period
Given period = 2/15 s
Frequency = 
frequency = 1 * 15/2
frequency f = 15/2 Hertz
Given wavelength
= 2m
Transverse speed 

Hence, the transverse speed at that point is 15m/s
10.00 °C this is the right answer need more question feel free to post
Coulomb's Law
Given:
F = 3.0 x 10^-3 Newton
d = 6.0 x 10^2 meters
Q1 = 3.3x 10^-8 Coulombs
k = 9.0 x 10^9 Newton*m^2/Coulombs^2
Required:
Q2 =?
Formula:
F = k • Q1 • Q2 / d²
Solution:
So, to solve for Q2
Q2 = F • d²/ k • Q1
Q2 = (3.0 x 10^-3 Newton) • (6.0 x 10^2 m)² / (9.0 x 10^9
Newton*m²/Coulombs²) • (3.3x 10^-8 Coulombs)
Q2 = (3.0 x 10^-3 Newton) • (360 000 m²) / (297 Newton*m²/Coulombs)
Q2 = 1080 Newton*m²/ (297 Newton*m²/Coulombs)
Then, take the reciprocal of the denominator and start
multiplying
Q2 = 1080 • 1 Coulombs/297
Q2 = 1080 Coulombs / 297
Q2 = 3.63636363636 Coulombs
Q2 = 3.64 Coulumbs