<span>Acceleration is the change in velocity divided by time taken. It has both magnitude and direction. In this problem, the change in velocity would first have to be calculated. Velocity is distance divided by time. Therefore, the velocity here would be 300 m divided by 22.4 seconds. This gives a velocity of 13.3928 m/s. Since acceleration is velocity divided by time, it would be 13.3928 divided by 22.4, giving a final solution of 0.598 m/s^2.</span>
Answer:
18 W
Explanation:
Applying,
P = V²/R.................. Equation 1
Where P = Power of both glowing bulbs, V = Voltage, R = Combined Resistance of both bulbs
Since: It is a series circuit,
Then,
R = R1+R2............. Equation 2
Where R1= Resistance of the first bulb, R2 = Resistance of the second bulb
Given: R1 = R2 = 8 Ω
Substitute into equation 1
R = 8+8
R = 16 Ω
Also Given: V = 12 V
Substitute into equation 1
P = 12²/8
P = 144/8
P = 18 W
Answer:
v = 38.73 m/s
Explanation:
Given
Extension of the bow, x = 50 cm = 0.5 m
Force of the arrow, F = 150 N
Mass of the arrow, m = 50 g = 0.05 kg
speed of arrow, v = ? m/s
We start by finding the spring constant
Remember, F = kx, so
k = F/x
k = 150 / 0.5
k = 300 N/m
the potential energy if the bow when pulled back is
E = 1/2kx²
E = 1/2 * 300 * 0.5²
E = 0.5 * 300 * 0.25
E = 37.5 J
The speed of the arrow will now be found by using the law of conservation of energy
1/2kx² = 1/2mv²
kx² = mv²
v² = kx²/m, on substituting, we have
v² = (300 * 0.5²) / 0.05
v² = 75 / 0.05
v² = 1500
v = √1500
v = 38.73 m/s
Net flux through the cylindrical surface is given as

here q = enclosed charge in the surface
so here in order to find the value of q

so now we have

so this is the total flux
now by Gauss's law we can find the electric field




<em>by above expression we can find the electric field at required position</em>
Answer:
a S orbital
Explanation:
Atomic orbitals is the place where we are most likely to find at least one electron, this definition is based on the equation posed by Erwin Schrödinger.
It is said that each electron occupies an atomic orbital that is defined by a series of quantum numbers s, n, ml, ms. In any atom each orbital can contain two electrons. It is possible that thanks to the function of the orbitals, the appearance that atoms can have is that of a diffuse cloud.
The orbitals s (l = 0) have a spherical shape. The extent of this orbital depends on the value of the main quantum number, so a 3s orbital has the same shape but is larger than a 2s orbital.
The orbitals p (l = 1) are formed by two identical lobes that project along an axis. The junction zone of both lobes coincides with the atomic nucleus. There are three orbitals p (m = -1, m = 0 and m = + 1) in the same way, which differ only in their orientation along the x, y or z axes.
The orbitals d (l = 2) are also formed by lobes. There are five types of d orbitals (corresponding to m = -2, -1, 0, 1, 2)