When a chemical reaction occurs,

A 190 g air-track glider is attached to a spring. The glider is pushed in 9.20 cm and released. A student with a stopwatch finds

Sholpan [36]

Answer:

$k = 12.136\,\frac{N}{m}$

Explanation:

The angular frequency of the system is:

$\omega = \sqrt{\frac{k}{m} }$

The frequency is:

$f = \frac{14\,osc}{11\,s}$

$f = 1.272\,hz$

The angular frequency is:

$\omega = 2\pi\cdot (1.272\,hz)$

$\omega = 7.992\,\frac{rad}{s}$

The spring constant is:

$k = \left(7.992\,\frac{rad}{s} \right)^{2}\cdot (0.190\,kg)$

$k = 12.136\,\frac{N}{m}$

6 0

3 years ago

A body with mass of 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time tequals0 the bod

11Alexandr11 [23.1K]

Answer:

X(t) = 13/13 cos(12t+α)

C =13/13

π/6 s

Explanation:

(A) A body with mass 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time t = 0 the body is pulled 1 m in to the right, stretching the spring, 3 set in motion with an initial velocity of 5 m/s to the left.

(a) Find X(t) in the form c • cos(w_o*t— α)

(b) Find the amplitude 3 Period of motion of the body 1

mass: m = 200g = 0.200 kg

displacement: ΔX = 20 cm = 0.20 m

Spring Constant: K = 9/0.20 = 45 N/m

IV: X(0) = 1m V(0) = -5 m/s

Simple Harmonic Motion: c•cos(cosw_t— α) = X(t)

Circular Frequency: w_o = √k/m= √36/(0.20) = 13 rad/s

X(0) = 1m =c_1

X'(0) = V(0) = c_2*w_o/w_o

= -5/12 = c_2

"radians Technically Unitless"

Amplitude: c = √ci^2 + c^2 ==> √1^2 + (-5/12)^2 = 1 m =13/13 = c

X(t) = 13/13 cos(12t+α)

since, C>0 : damped forced vibration c_1>0, c_2>0

phase angle 2π+tan^-1(c_2/c_1)

=2π+tan^-1(-5/12/1)= 5.884

period: T =2π/w_o

=π/6 s

6 0

4 years ago

A soap bubble, when illuminated with light of frequency 5.27 Hz × 1014 Hz, appears to be especially reflective. If it is surroun

denis23 [38]

Answer:

$1.07004\times 10^{-7}\ m$

Explanation:

$n_s$ = Refractive index of bubble = 1.33

f = Frequency of light = $5.27\times 10^{14}\ Hz$

c = Speed of light = $3\times 10^8\ m/s$

The wavelength of light is given by

$\lambda=\dfrac{2n_st}{m-\dfrac{1}{2}}$

Wavelength is also given by

$\lambda=\dfrac{c}{f}$

m = 1 for minimum thickness

$\dfrac{c}{f}=\dfrac{2n_st}{m-\dfrac{1}{2}}\\\Rightarrow t=\dfrac{m-\dfrac{1}{2}c}{2n_sf}\\\Rightarrow t=\dfrac{(1-\dfrac{1}{2})\times 3\times 10^8}{2\times 1.33\times 5.27\times 10^{14}}\\\Rightarrow t=1.07004\times 10^{-7}\ m$

The minimum thickness is $1.07004\times 10^{-7}\ m$

4 0

3 years ago

A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou

Hoochie [10]

Answer:

e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.

Explanation:

This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.

The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.

Mathematical expression for the Newton's second law of motion is given as:

$F=\frac{dp}{dt}$ ............................................(1)