Answer:
Explanation:
Let v be the velocity acquired by electron in electric field
V q = 1/2 m v²
V is potential difference applied on charge q , m is mass of charge , v is velocity acquired
2400 x 1.6 x 10⁻¹⁹ = .5 x 9.1 x 10⁻³¹ x v²
v² = 844 x 10¹²
v = 29.05 x 10⁶ m /s
Maximum force will be exerted on moving electron when it moves perpendicular to magnetic field .
Maximum force = Bqv , where B is magnetic field , q is charge on electron and v is velocity of electron
= 1.7 x 1.6 x 10⁻¹⁹ x 29.05 x 10⁶
= 79.02 x 10⁻¹³ N .
Minimum force will be zero when electron moves along the direction of magnetic field .
The answer is d, gravity is the only force acting on the object
There's no air in space, so there's no air resistance there.
Answer:
a) a = 4.9 m / s², N = 16.97 N and b) F = 9.8 N
Explanation:
a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry
sin 30 = Wx / W
cos 30 = Wy / W
Wx = W sin30
Wy = W cos 30
Let's write the equations on each axis
X axis
Wx = ma
Y Axis
N- Wy = 0
N = Wy = mg cos 30
N = 2.0 9.8 cos 30
N = 16.97 N
We calculate the acceleration
a = Wx / m
a = mg sin 30 / m
a = g sin 30
a =9.8 sin 30
a = 4.9 m / s²
b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component
F -Wx = 0
F = Wx
F = m g sin 30
F = 2.0 9.8 sin 30
F = 9.8 N
The energy of the wave decreases gradually
