4. Table 2.4 shows how the displacement of a runner changed
during a sprint race. Draw a displacement–time graph to show
this data, and use it to deduce the runner’s speed in the middle
of the race.
Table 2.4 Data for a sprinter during a race
Displacement
(m)
0 4 10 20 50 80 105
Time (s) 1 2 3 6 9 12
Answer:
Explanation:
Electric field due to charge at origin
= k Q / r²
k is a constant , Q is charge and r is distance
= 9 x 10⁹ x 5 x 10⁻⁶ / .5²
= 180 x 10³ N /C
In vector form
E₁ = 180 x 10³ j
Electric field due to q₂ charge
= 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²
= 30.33 x 10³ N / C
It will have negative slope θ with x axis
Tan θ = .5 / √.5² + .8²
= .5 / .94
θ = 28°
E₂ = 30.33 x 10³ cos 28 i - 30.33 x 10³ sin28j
= 26.78 x 10³ i - 14.24 x 10³ j
Total electric field
E = E₁ + E₂
= 180 x 10³ j +26.78 x 10³ i - 14.24 x 10³ j
= 26.78 x 10³ i + 165.76 X 10³ j
magnitude
= √(26.78² + 165.76² ) x 10³ N /C
= 167.8 x 10³ N / C .
La respuesta es "Un avion que vuela al norte con rapidez constante y altitud constante".
Para que la fuerza neta sea 0, la aceleracion debe ser 0, para esto la velocidad debe ser constante.
Para que la velocidad sea constante el objecto debe estar moviendo con rapidez (magnitud de la velocidad) constante y sin cambiar direccion; ya que la velocidad es un vector asi es que depende en magnitud y direccion.
En las demas opciones la magnitud de la velocidad (rapidez) cambia y/o la direccion.
Answer:
e*P_s = 11 W
Explanation:
Given:
- e*P = 1.0 KW
- r_s = 9.5*r_e
- e is the efficiency of the panels
Find:
What power would the solar cell produce if the spacecraft were in orbit around Saturn
Solution:
- We use the relation between the intensity I and distance of light:
I_1 / I_2 = ( r_2 / r_1 ) ^2
- The intensity of sun light at Saturn's orbit can be expressed as:
I_s = I_e * ( r_e / r_s ) ^2
I_s = ( 1.0 KW / e*a) * ( 1 / 9.5 )^2
I_s = 11 W / e*a
- We know that P = I*a, hence we have:
P_s = I_s*a
P_s = 11 W / e
Hence, e*P_s = 11 W
Answer:
- 1 m/s, 20 m
Explanation:
u = 9 m/s, a = - 2 m/s^2, t = 5 sec
Let s be the displacement and v be the velocity after 5 seconds
Use first equation of motion.
v = u + a t
v = 9 - 2 x 5 = 9 - 10 = - 1 m/s
Use second equation of motion
s = u t + 1/2 a t^2
s = 9 x 5 - 1/2 x 2 x 5 x 5
s = 45 - 25 = 20 m
