As per given conditions there are two directions along which forces are acting
1. Net force along left direction is given as
2. Net force towards right direction is given as
now since the two forces here in opposite direction so here we will have net force given as
so here net forces must be 440 N towards right
The acceleration is given as:
a = g sin(30°) where g is the gravitational acceleration
For g = 10 m/s^2, we get
a = 10 sin(30°) = 10 * 1/2 = 5 m/s^2
Answer:
Explanation:
From the data it appears that A is the middle point between two charges.
First of all we shall calculate the field at point A .
Field due to charge -Q ( 6e⁻ ) at A
= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴
= 13.82 x 10⁻⁶ N/C
Its direction will be towards Q⁻
Same field will be produced by Q⁺ charge . The direction will be away
from Q⁺ towards Q⁻ .
We shall add the field to get the resultant field .
= 2 x 13.82 x 10⁻⁶
= 27.64 x 10⁻⁶ N/C
Force on electron put at A
= charge x field
= 1.6 x 10⁻¹⁹ x 27.64 x 10⁻⁶
= 44.22 x 10⁻²⁵ N
Answer:
static friction acting opposite to the direction of travel
Explanation:
Because the Frictional force of the front wheels act to oppose the spinning, so, For the front wheels to roll without slipping, the friction must be static friction pointing in the direction of travel of the car.
Explanation:
Answer:
magnitude = 7.446 km, direction = 75.22° north of east
Explanation:
From the questions,
To get the the magnitude of the resultant vector we use Pythagoras theorem
a² = b²+c²
From the diagram,
y² = 1.9²+7.2²
y² = 55.45
y = √(55.45)
y = 7.446 km.
The direction of the dolphin is given as,
θ = tan⁻¹(7.2/1.9)
θ = tan⁻¹(3.7895)
θ = 75.22° north of east
Hence the magnitude of the resultant vector = 7.446 km, and it direction is 75.22° north of east
