Answer:
he can explore other types of physical activity
Explanation:
lifting weights and paddling will help but running could also help
Answer:
Vb = k Q / r r <R
Vb = k q / R³ (R² - r²) r >R
Explanation:
The electic potential is defined by
ΔV = - ∫ E .ds
We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product
VB - VA = - ∫ E dr
Let's substitute every equation they give us and we find out
r> R
Va = - ∫ (k Q / r²) dr
-Va = - k Q (- 1 / r)
We evaluate with it Va = 0 for r = infinity
Vb = k Q / r r <R
We perform the calculation of the power with the expression of the electric field that they give us
Vb = - int (kQ / R3 r) dr
We integrate and evaluate from the starting point r = R to the final point r <R
Vb = ∫kq / R³ r dr
Vb = k q / R³ (R² - r²)
This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity
Answer:
a) 42 m/s, positive direction (to the east), b) 42 m/s, negative direction (to the west).
Explanation:
a) Let consider that Car A is moving at positive direction. Then, the relative velocity of Car A as seen by the driver of Car B is:
42 m/s, positive direction (to the east).
b) The relative velocity of Car B as seen by the drive of Car A is:
42 m/s, negative direction (to the west).
Answer:
discrete lines are observed by the spectroscope, the emission of the lamp is of the ATOMIC source
Explanation:
Bulbs can emit light in several ways:
* When the emission is carried out by the heating of its filament, the bulb is called incandescent, in general its spectrum is similar to that of a black body, this is a continuous spectrum with a maximum dependent on the fourth power of the temperature of the filament.
* The emission can be by atomic transitions, in this case there is a discrete spectrum formed by the spectral lines of the material that forms the gas of the lamp, in general for the yellow emission the most used materials are mercury and sodium or a mixture of they.
Consequently, as discrete lines are observed by the spectroscope, the emission of the lamp is of the ATOMIC type
It's hard to tell exactly what's happening in that 110 cm that you marked over the wave. What is under the ends of the long arrow ? How many complete waves ? I counted 4.5 complete waves ... maybe ?
If there are 4.5 complete waves in 110cm, then the length of 1 wave is (110/4.5)=24.44cm.
Frequency = speed/wavelength
Frequency = 2m/s /0.2444m
Frequency = 8.18 Hz
