Answer:
60km/hr due east
Explanation:
Given parameters:
Displacement = 60km East
Start time = 10am
End time = 11am
Unknown:
Average velocity = ?
Solution:
The average velocity is the displacement divided by the time taken.
For this journey, the time taken is;
11am - 10am = 1hr
So;
Average velocity = = 60km/hr due east
Answer:
vertical distance, s = 0.16 meters
Explanation:
It is given that,
An amateur player is about to throw a dart with an initial velocity of 15 meters/second onto a dartboard.
Dashboard is placed at a distance of 2.7 meters.
Calculating time from velocity and distance i.e.
Using second equation of motion for finding vertical distance :
Here, u = 0 (for vertical velocity )
s = 0.162 meters
or s = 0.16 meters
Hence, the vertical distance by which the player will miss the target if he throws the dart horizontally, in line with the dartboard is 0.16 meters.
Answer:
1. a. 3,000 N
b. Repulsión
2. 46.875 × 10⁶ N/C
3. a. 81,000 J
b. 6.75 × 10⁹ V
Explanation:
1. Los parámetros dados son;
Q₁ = +12 μC, Q₂ = +16 μC
La distancia entre las cargas, r = 0.024
La magnitud de la fuerza electrostática, F, entre cargas se da como sigue;
Donde, k = constante de Coulomb = 9.0 × 10⁹ N · m² / C²
Por lo tanto, obtenemos;
F = 9.0 × 10⁹ × 12 × 10⁻⁶ × 16 × 10⁻⁶ / 0.024² = 3.000
La magnitud de la fuerza electrostática, entre las cargas, F = 3000 N
(b) Dado que tanto Q₁ como Q₂ son cargas positivas, y las cargas iguales se repelen entre sí, la fuerza es la repulsión.
2) La intensidad de un campo eléctrico, E, se da como sigue;
La magnitud de la carga, Q = 24 μC
La distancia donde se mide el campo, r = 48 mm = 0.048 m
Por lo tanto, E = 9.0 × 10⁹ × 12 × 10⁻⁶ / 0.048² = 46,875,000 N / C
La intensidad de un campo eléctrico, E = 46,875,000 N / C = 46.875 × 10⁶ N / C
3. La magnitud de las cargas son;
Q₁ = 24 mC
Q₂ = -12 μC
La distancia entre las cargas, r = 0.032 m
un. El potencial eléctrico de una carga, , se da de la siguiente manera;
Por lo tanto;
= 9.0×10⁹ × 24 × 10⁻³ × (-12) × 10⁻⁶ /0.032 = -81,000
La energía potencial eléctrica entre la carga, Q₁ y Q₂= -81,000 J
b. El potencial eléctrico de Q₁ en Q₂, V₁ =
Por lo tanto, V₁ = 9.0×10⁹ × 24 × 10⁻³/0.032 = 6.75 × 10⁹
El potencial eléctrico de Q₁ en Q₂, V₁ = 6.75 × 10⁹ V
Answer:
(A) Capacitance per unit length =
(B) The magnitude of charge on both conductor is C and the sign of charge on inner conductor is
and the sign on outer conductor is
Explanation:
Given :
Radius of inner part of conductor =
m
Radius of outer part of conductor =
m
The length of the capacitor =
m
(A)
Capacitance is purely geometrical property. It depends only on length, radius of conductor.
From the formula of cylindrical capacitor,
Where,
But we need capacitance per unit length so,
capacitance per unit length =
(B)
The charge on both conductors is given by,
Where, C = capacitance of cylindrical capacitor and value of F,
V
∴ C
The magnitude of charge on both conductor is same as above but the sign of charge is different.
Charge on inner conductor is and Charge on outer conductor is
.