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1 year ago

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Two masses, each having a value of M, are vibrating vertically on a spring with a Hooke's law constant, k. At the lowest point o

f the vibration, one of the masses falls off, so that now the total mass is M instead of 2M. Comparing the new vibrational motion to the original vibrational motion: 1) How is the period of vibration different, if at all

1 answer:

ELEN [110]1 year ago

7 0

What are you asking for they

You might be interested in

What is the change in internal energy (in J) of a system that does 4.50 ✕ 105 J of work while 3.20 ✕ 106 J of heat transfer occu

Dmitrij [34]

Answer:

$-3.25\times 10^6 J$

Explanation:

We are given that

Work done by the system= $4.5\times 10^5$ J

Heat transfer into the system= $U_1=3.2\times 10^6$ J

Heat transfer to the environment= $U_2=6\times 10^6$ J

We have to find the change in internal energy

By first law of thermodynamics

$\Delta Q=\Delta U+w$

$\Delta Q=U_1-U_2=3.2\times 10^6-6\times 10^6=-2.8\times 10^6J$

Substitute the values then we get

$-2.8\times 10^6=\Delta U+4.5\times 10^5$

$\Delta U=-2.8\times 10^6-4.5\times 10^5=-28\times 10^5-4.5\times 10^5=-32.5\times 10^5=-3.25\times 10^6 J$

Hence, the change in internal energy = $-3.25\times 10^6 J$

7 0

2 years ago

A car of mass 998 kilograms moving in the positive y–axis at a speed of 20 meters/second collides on ice with another car of mas

goldfiish [28.3K]

<span> Let’s determine the initial momentum of each car.
#1 = 998 * 20 = 19,960
#2 = 1200 * 17 = 20,400

This is this is total momentum in the x direction before the collision. B is the correct answer. Since momentum is conserved in both directions, this will be total momentum is the x direction after the collision. To prove that this is true, let’s determine the magnitude and direction of the total momentum after the collision.

Since the y axis and the x axis are perpendicular to each other, use the following equation to determine the magnitude of their final momentum.

Final = √(x^2 + y^2) = √(20,400^2 + 19,960^2) = √814,561,600

This is approximately 28,541. To determine the x component, we need to determine the angle of the final momentum. Use the following equation.

Tan θ = y/x = 19,960/20,400 = 499/510
θ = tan^-1 (499/510)

The angle is approximately 43.85˚ counter clockwise from the negative x axis. To determine the x component, multiply the final momentum by the cosine of the angle.

x = √814,561,600 * cos (tan^-1 (499/510) = 20,400</span>

3 0

2 years ago

A bicyclist is riding to the left with a velocity of 14 \,\dfrac{\text m}{\text s}14 s m 14, start fraction, start text, m, en

Gnesinka [82]

Answer:

-2.0 m/s²

Explanation:Acceleration is the rate of change of velocity.

\begin{aligned}a&=\dfrac{\text{Change in velocity}}{\text{Change in time}}\\ \\ &=\dfrac{v_f-v_i}{\Delta t} \end{aligned}

a

=

Change in time

Change in velocity

=

Δt

v

f

−v

i

Hint #22 / 3

We can calculate the bicyclist's acceleration from the final velocity v_fv

f

v, start subscript, f, end subscript, initial velocity v_iv

i

v, start subscript, i, end subscript, and time interval \Delta tΔtdelta, t.

\begin{aligned}a&=\dfrac{v_f-v_i}{\Delta t}\\ \\ &=\dfrac{-21\,\dfrac{\text m}{\text s}-(-14\,\dfrac{\text m}{\text s})}{3.5\,\text s}\\ \\ &=-2.0\,\dfrac{\text m}{\text s^2}\end{aligned}

a

=

Δt

v

f

−v

i

=

3.5s

−21

s

m

−(−14

s

m

)

=−2.0

s

2

m

Hint #33 / 3

The acceleration of the bicyclist is -2.0\,\dfrac{\text m}{\text s^2}−2.0

s

2

m

minus, 2, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction.

5 0

2 years ago

A projectile is launched at an angle of 45° from the horizontal and lands 21 s later at the same height from which it was launch

irinina [24]

Answer:

a) initial speed of projectile = 145.5 m/s

b) Maximum altitude = 540 m

c) Range = 2160.6 m

d) r = (1440î + 480j) m

Explanation:

The distance at any time for the projectile is given by the relation - r² = x² + y²

where x = horizontal distance covered covered by the projectile and y = vertical distance coveredby the projectile

Let the initial velocity be u = ?

angle of projection be θ with respect to the horizontal = 45°

u = (uₓî + uᵧj) m/s

T = total time of flight = 21 s

t = any time during the flight of the projectile

a) Total time of flight = 2 uᵧ/g = (2u sin θ)/g

21 = (2u sin 45°)/9.8

u = 145.5 m/s

b) maximum altitude of the projectile = H

H = (u² sin² θ)/2g

H = (145.5² sin² 45°)/(2 × 9.8)

H = 540 m

c) According to projectile motion the maximum horizontal displacement is given by

x = R = uₓT = u cos(θ) T (since uₓ = u cos θ)

R = (145.5 cos 45°) × 21 = 2160.6 m

d) At 14 s,

x = uₓt = u cos(θ) t (since uₓ = u cos θ)

x = (145.5 cos 45°) × 14 = 1440.1 m

y = uᵧ t - 0.5gt² = [u sin(θ)] t - 0.5gt² = (145.5 sin 45°) × 14 - 0.5(9.8)(14) = 480 m

r = (1440î + 480j) m

6 0

2 years ago

Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section),

salantis [7]

Answer:

The drag coefficient is $D_z = 1.30512$

Explanation:

From the question we are told that

The density of air is $\rho_a = 1.21 \ kg/m^3$

The diameter of bottom part is $d = 0.15 \ m$

The power trend-line equation is mathematically represented as

$F_{\alpha } = 0.9226 * v^{0.5737}$

let assume that the velocity is 20 m/s

Then

$F_{\alpha } = 0.9226 * 20^{0.5737}$

$F_{\alpha } = 5.1453 \ N$

The drag coefficient is mathematically represented as

$D_z = \frac{2 F_{\alpha } }{A \rho v^2 }$

Where

$F_{\alpha }$ is the drag force

$\rho$ is the density of the fluid

$v$ is the flow velocity

A is the area which mathematically evaluated as

$A = \pi r^2 = \pi \frac{d^2}{4}$

substituting values

$A = 3.142 * \frac{(0.15)^2}{4}$

$A = 0.0176 \ m^2$

Then

$D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }$

$D_z = 1.30512$

3 0

2 years ago