What are the three components of the equation for the second law of motion?

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.360 mm wide. The diffraction pattern is observed

Bumek [7]

Answer:

a) 0.0130 m

b') w' = =6.46*10^{-3] m

Explanation:

given data:

\lambda of light = 633 nm

width of siit a =0.360 mm

distance from screen = 3.75 m

a) the first minima is located at

$sin\theta = \frac{\lambda}{a}$

= $= \frac{633 *10^{-9}}{.360*10^{-3}}$

$\theta = 0.100$

$y_1 = dtan\theta_1 = 3.75*tan(0.100) = 6.54 *10^{-3} m$

with of central fringe = 2y_1 = 2*6.54 *10^{-3} = 0.0130 m

b)

width of the first bright fringe on either side of the central one = $w' = y_2 -y_1$

calculation for y_2

$sin\theta = 2\frac{\lambda}{a}$

= $= 2*\frac{633 *10^{-9}}{.360*10^{-3}}$

$\theta = 2*0.100 = 0.200$

$y_2 = dtan\theta_1 = 3.75*tan(0.200) =0.0130 m$

$w' = 0.0130 -6.54 *10^{-3}$

w' = =6.46*10^{-3] m

6 0

3 years ago

To model this process, assume two charged spherical conductors are connected by a long conducting wire and a 1.20-mC charge is p

il63 [147K]

Answer:

Part a: The electric potential of each sphere is 1.35x10⁸V

Part b: The electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively

Explanation:

As the complete question is not given, the similar question is attached herewith. The values are used as indicated in the given question

Let r_1 = 6 cm=0.06 m

r2 = 2 cm = 0.02 m

Q = 1.2 mC

Let q1 and q2 are the charges on each sphere.

q1 + q2 = 1.2 mC -------(1)

In the equilibrium, V1 = V2

k*q1/r1 = k*q2/r2

q1/0.06 = q2/0.02

q1/q2 = 0.06/0.02

q1/q2 = 3 ---------(2)

On solving equation 1 and 2

we get

q1 = 0.9 mC

q2 = 0.3 mC

So

V1 = k*q1/r1 = (9*10^9*0.9*10^-3)/0.06 = 1.35*10^8 Volts

V2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02 = 1.35*10^8 Volts

So the electric potential of each sphere is 1.35x10⁸V

Part b

Now the electric potential is given as

E1 = k*q1/r1^2 = 9*10^9*0.9*10^-3/0.06^2 = 2.25*10^9 N/C

E2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02^2 = 6.75*10^9 N/C

So the electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively

7 0

3 years ago

Elena (60.0 kg) and Madison (65.0 kg) are ice-skating at the Rockefeller ice rink in New Yok city. Their friend Tanner sees Elen

Bas_tet [7]

1. +72.0 kg m/s

The momentum of an object is given by:

p = mv

where

m is the mass of the object

v is its velocity

Taking "to the right" as positive direction, for Elena we have

m = 60.0 kg is the mass

v = +1.20 m/s is the velocity

So, Elena's momentum is

$p_e=(60.0 kg)(+1.20 m/s)=+72.0 kg m/s$

2. -162.5 kg m/s

Here Madison is moving in the opposite direction of Elena (to the left), so her velocity is

v = -2.50 m/s

while her mass is

m = 65.0 kg

Therefore, her momentum is

$p_m= (65.0 kg)(-2.50 m/s)=-162.5 kg m/s$

3. -90.5 kg m/s

The total momentum of Elena and Madison is equal to the algebraic sum of their momenta; taking into account the correct signs, we have:

$p=p_e + p_m = +72.0 kg m/s - 162.5 kg m/s =-90.5 kg m/s$

4. 0.72 m/s to the left

We can find the final speed of Elena and Madison by using the law of conservation of momentum. In fact, the final momentum must be equal to the initial momentum (before the collision).

The initial momentum is the one calculated at the previous step:

$p_i = -90.5 kg m/s$

while the final momentum (after the collision) is given by

$p_f = (m_e + m_m) v$

where

$m_e$ is Elena's mass

$m_m$ is Madison's mass

v is their final velocity

According to the law of conservation of momentum,

$p_i = p_f\\p_i = (m_e + m_m) v$

So we can find v:

$v=\frac{p_i}{m_e + m_m}=\frac{-90.5 kg m/s}{60.0 kg+65.0 kg}=-0.72 m/s$

and the direction is to the left, since the sign is negative.

8 0

3 years ago

A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res

Natalija [7]

Answer:

The time constant is $\tau = 17.5 \ s$

Explanation:

From the question we are told that

The spring constant is $k = 11.5 \ N/m$

The mass of the ball is $m_b = 490 \ g = 0.49 \ kg$

The amplitude of the oscillation t the beginning is $x = 6.70 cm = 0.067 \ m$

The amplitude after time t is $x_t = 2.20 cm = 0.022 \ m$

The number of oscillation is $N = 30$

Generally the time taken to attain the second amplitude is mathematically represented as

$t = N * T$ Here T is the period of oscillation

$t = N * 2\pi \sqrt{\frac{m}{k} }$

=> $t = 30 * 2 * 3.142 * \sqrt{\frac{ 0.490}{11.5} }$

=> $t = 38.88 \ s$

Generally the amplitude at time t is mathematically represented as

$x(t) = x e^{-\frac{at}{2m} }$

Here a is the damping constant so

at $t = 38.88 \ s$ , $x_t = 2.20 cm = 0.022 \ m$

So

$0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }$

=> $0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }$

taking natural log of both sides

=> $ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }$

=> $a = 0.028$

Generally the time constant is mathematically represented as

$\tau = \frac{m}{a}$

=> $\tau = \frac{0.490}{ 0.028}$

=> $\tau = 17.5 \ s$

4 0

3 years ago

A copper block rests 30.0 cm from the center of a steel turntable. The coefficient of static friction between the block and the

PIT_PIT [208]

Answer:

refer to the above attachment

3 0

1 year ago