Answer:
True
Explanation:
Bohr proposed an atomic model in which;
- the electrons found in an atom can only occupy orbits for which the angular momentum is quantized, which results in discrete values of energy level.
- the electrons in such stationary state or ground state emits no light, but
- if an electron jumps to a lower state, it emits a photon whose energy equals the difference in energy between the two states.
<span>FIRST SECTION
You should use the formula for uniformly accelerated linear movement.
Initial speed is 0 because it starts from rest.
d=(1/2)*a*t^2+vo*t =(1/2)*(4.0 m/s^2)*(3s)^2+0*3s=(1/2)*(4.0 m/s^2)*3^2*s^2+0=2.0 m*9=18m
You can calculate the final speed with the other formula:
v=a*t+vo=(4.0 m/s^2)*(3s)+0=(4.0 m/s)*(3)=12m/s
SECOND SECTION
You should use the formula for uniform linear movement.
Velocity is a constant: it remains in 12m/s.
d=v*t=12m/s*2s=12m*2=24m
THIRD SECTION
We should use the same formulas as the first section, but with different numbers.
Initial velocity will be 12m/s, and then velocity will start to decrease until it gets to 0.
We don’t know what the time is for this section.
Acceleration is negative, because it’s slowing down.
v=a*t+vo
0=-3.0 m/s^2*t+12m/s
3.0 m/s^2*t=12m/s
t=(12m/s)/(3.0 m/s^2)=4(1/s)/(1/s^2)=4s^2/s=4s
Now let’s use that time in the other formula:
d=(1/2)*a*t^2+vo*t =(1/2)*(-3.0 m/s^2)*(4s)^2+(12m/s)*3s=(-1.5 m/s^2)*4^2*s^2+12*3m*s/s=-1.5 m*4^2+36m=-1.5*16m+36m=-24m+36m=12m
Now let’s add the 3 stages:
d=18m+24m+12m=54m
</span>
Answer:
the power (outside) of the train must be increased for the increase in resistance by the weight component.
Explanation:
We analyze the movement of the train at constant speed, on flat ground the force of the motor must counteract the friction force that opposes the movement, therefore to maintain the constant speed the two forces must be of the same magnitude.
When the train goes up a slope, it is subjected to two forces: the friction force that opposes the movement and a component of the weight in the opposite direction to the movement, therefore the power (outside) of the train must be increased for the increase in resistance by the weight component.
To solve this problem it is necessary to apply the concepts related to the Conservation of Energy, for which it is necessary that any decrease made through the potential energy, is equivalent to the gain given in the kinetic energy or vice versa.
Mathematically this can be expressed as
Since there is no final potential energy (the height is zero), and the initial potential energy is equivalent to the work done we have to
Therefore the non-conservative work was done on the boy is 1.4kJ