I really need help with this problem :)

The suspension system of a 2100 kg automobile "sags" 8.5 cm when the chassis is placed on it. Also, the oscillation amplitude de

Gennadij [26K]

Answer:

Part a)

$k = 6.06 \times 10^4 N/m$

Part b)

$b = 1795.4 kg/s$

Explanation:

Part a)

as the mass of the suspension system is given as

$m = 2100 kg$

also we have

$x = 8.5 cm$

so now for force balance we have

$mg = kx$

$(525)(9.81) = k(0.085)$

$k = 6.06 \times 10^4 N/m$

Part b)

Now we know that amplitude decreases by 63% in each cycle

so after one cycle the amplitude will become 37% of initial amplitude

so it is given as

$A = 0.37 A_o$

also we know

$A = A_o e^{-bt/2m}$

$0.37 A_o = A_o e^{-bt/2m}$

$\frac{bt}{2m} = 1$

$b = \frac{2m}{t}$

here t = time period of one oscillation

so it is

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

$t = 2\pi\sqrt{\frac{525}{6.06 \times 10^4}}$

$t = 0.58 s$

now damping constant is

$b = \frac{2(525)}{0.58}$

$b = 1795.4 kg/s$

7 0

4 years ago

When illuminated with monochromatic light, a double slit produces a pattern that is a combination of single-slit diffraction and

d1i1m1o1n [39]

Answer:

The ratio is $k:d = 1 : 5$

Explanation:

From the question we are told that

The first minimum of the single slit pattern falls on the fifth maximum of the double slit pattern.

Generally the condition for constructive interference for as single slit is

$ksin(\theta) = n\lambda$

Here k is the width of the slit and n is the order of the fringe and for single slit n = 1 (cause we are considering the first maxima)

Generally the condition for constructive interference for as double slit is

$dsin\theta = m\lambda$

Here d is the separation between the slit and m is the order of the fringe and for double slit m = 5 (cause we are considering the first maxima)

=> $dsin\theta = 5\lambda$

So

$\frac{ksin(\theta)}{dsin(\theta)} = \frac{\lambda}{5\lambda}$

=> $\frac{k }{d} = \frac{1}{5}$

So

$k:d = 1 : 5$

5 0

3 years ago

Fig. 2.1 shows the extension-load graph for a spring.

trapecia [35]

Answer:

(1) Hooke's law

(2) a) Extension is directly proportional to the applied load

b) The starting point of the graph is the origin (0, 0) or absence of load, no extension

Explanation:

(1) The law obeyed by the spring is known as Hooke's law which states that the extension or compression, x, of a spring proportional to the applied force, F

F = -k × x

Where;

k = The spring constant

(2) Given that the law mathematically is F = -k × x

The two features of the graph that show that the law is obeyed are;

a) The extension increases as the load is increased

b) The extension is zero when the there is no applied load.

3 0

3 years ago

According to ohms law if you don't change the value of the resistor and you double the voltage in a circuit the amount of curren

padilas [110]

Answer: D (doubled)

According to ohm's law

I = V/R ,

V = IR ; R = constant and V is doubled

From the equation

V is directly proportional to the current, and it is given that R is constant ;

2V = 2I. R since R = constant

Hence I is doubled.

5 0

4 years ago

If, in

STatiana [176]

Answer:

a) λ = 121.5 nm , b) 102.6, 97, 91.1 nm

Explanation:

Bohr's model describes the energy of the hydrogen atom

$E_{n}$ = k² e² / 2m (1 / n²)

A transition occurs when the electron passes from n level to a lower one

$E_{i}$ - $E_{n}$ = k² e² / 2m (1 / $n_{i}$ ² - 1 / $n_{f}$ ²)

Planck's relationship is

E = h f = h c / lam

hc /λ = k² e²/ 2m(1 / $n_{i}$ ² - 1 / $n_{f}$ ²)

1 / λ = [k² e² / 2m h c] (1 / $n_{i}$ ² - 1 / $n_{f}$ ²)

1 /λ = Ry] (1 / $n_{i}$ ² - 1 / $n_{f}$ ²)

a) the first element of the series occurs for $n_{f}$ = 2

1 / λ = 1.097 10⁷ (1- 1/2²)

1 / λ = 1.097 10⁷ (1- 0.25)

1 / λ = 0.82275 10⁷

λ = 1.215 10⁻⁷ m

λ = 1,215 10⁻⁷ m (10⁹nm / m)

λ = 121.5 nm

b) the next elements of the series occur to

$n_{f}$ $n_{i}$ 1 /λ λ (10-7m) λ (nm)

3 1 1,097 10⁷ (1-1 / 9) 1,0255 102.6

4 1 1,097 10⁷ (1-1 / 16) 0.9723 97.2

∞ 1 1,097 10⁷ (1 - 0) 0.91158 91.1

5 0

3 years ago