please answer 1-3 and use a separate piece of paper thank you i've been trying those problems for 3 day.... ikr Sad

A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of

Veseljchak [2.6K]

Answer:

Explanation:

Given that,

Mass attached to spring

M = 0.52kg

Force constant K = 8N/m

Amplitude A = 11.6 cm

a. Maximum speed?

Angular velocity is calculated using

w = √k/m

w = √8/0.52

w = √15.385

w = 3.922rad/s

Then, the relation ship between angular velocity and linear velocity is given as

v = - w•A

v = - 3.922 × 11.6

v = - 45.5 cm/s

Then, the maximum velocity is

vmax = |v|= 45.5cm/s

b. Acceleration a?

Acceleration can be determine using the formula

a = -w²• A

a = -3.922² × 11.6

a = -178.46 cm/s²

Magnitude of the acceleration is 178.46cm/s²

c. Speed when the object is at 9.6cm from equilibrium position?

Generally,

The position of the object at equilibrium is

x(t) = A•Cos(wt)

x(t) = 11.6 Cos (3.922t)

Then, when x(t) = 9.6cm

9.6 = 11.6 Cos(3.92t)

Cos(3.922t) = 9.6/11.6

Cos(3.922t) = 0.8276

3.922t = ArcCos(0.8276)

Note: the angle is in radiant

3.922t = 0.596

t = 0.596/3.922

t = 0.152 second

Then, v(t) at that time is

v(t) = x'(t) = -11.6×3.92Sin(3.922t)

v(t) = -45.5Sin(3.922t)

Now, when t =0.152

v(t) = -45.5 Sin(3.922×0.152)

v(t) = -45.5Sin(0.596)

v(t) = -25.5 cm/s

Then, it's magnitude is 25.5cm/s

d. Acceleration at same position

t = 0.152s

a(t) = v'(t) = - 45.5×3.922Cos(3.922t)

a(t) = -178.46Cos(3.92t)

a(t) = -178.46 Cos(3.92×0.152)

a(t) = -178.46 Cos(0.596)

a(t) = -147.68 cm/s²

Magnitude of the acceleration is 147.68 cm/s²

5 0

3 years ago

Four point masses of 3.0 kg each are arranged in a square on masslessrods. The length of a side of the square is 0.50m. What is

Zigmanuir [339]

Answer:

Part a)

$I = 1.5 kg m^2$

Part b)

$I = 0.75 kg m^2$

Part c)

$I = 1.5 kg m^2$

Explanation:

Part a)

Moment of inertia of the system about an axis passing through B and C is given as

$I = mL^2 + mL^2 + m(0) + m(0)$

$I = 2mL^2$

$I = 2(3 kg)(0.50^2)$

$I = 1.5 kg m^2$

Part b)

Moment of inertia of the system about an axis passing through A and C is given as

$I = m(0^2) + m(\frac{L}{\sqrt2})^2 + m(0) + m(\frac{L}{\sqrt2})^2$

$I = 2m\frac{L^2}{2}$

$I = (3 kg)(0.50^2)$

$I = 0.75 kg m^2$

Part c)

Moment of inertia of the system about an axis passing through the center of the square and perpendicular to the plane of the square

$I = m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2$

$I = 4m\frac{L^2}{2}$

$I = 2(3 kg)(0.50^2)$

$I = 1.5 kg m^2$

8 0

3 years ago

How is it possible for a person watching a baseball game on TV to hear sounds before someone who is actually watching the same g

Andreyy89

Answer:

It is possible because, the TV broadcast audio and video signals in radio frequency which travels at the speed of light while the audio signals travel to those present in the stadium at the speed of sound which is over eight hundred thousand times slower than the speed of light

Explanation:

It is possible because of the following;

1. TV signals from the camera (including the captured sound) very close to the field of play are transmitted through the radio frequency bands and as such are a form of electromagnetic radiation that travels at the speed of light which is about 300,000 km/second

It will therefore, take 1 second for a sound of the game to reach someone located at 300,000,000 meters watching a live televised game

2. The speed of sound is about 343 m/second and it therefore takes up to 2 seconds for a sound to reach someone 686 meters away from the ball in the stadium.

5 0

3 years ago

An us bomber is flying horizontally at 300 mph at an altitude of 610 m. its target is an iraqi oil tanker crusing 25kph in the s

Pachacha [2.7K]

Answer:

=1419.19 meters.

Explanation:

The time it takes for the shell to drop to the tanker from the height, H =1/2gt²

610m=1/2×9.8×t²

t²=(610m×2)/9.8m/s²

t²=124.49s²

t=11.16 s

Therefore, it takes 11.16 seconds for a free fall from a height of 610m

Range= Initial velocity×time taken to hit the tanker.

R=v₁t

Lets change 300 mph to kph.

=300×1.60934 =482.802 kph

Relative velocity=482.802 kph-25 kph

=457.802 kph

Lets change 11.16 seconds to hours.

=11.16/(3600)

=0.0031 hours.

R=v₁t

=457.802 kph × 0.0031 hours.

=1.41918 km

=1.41919 km × 1000m/km

=1419.19 meters.

3 0

4 years ago

He starter motor of a car engine draws a current of 170 AA from the battery. The copper wire to the motor is 6.00 mmmm in diamet

sweet-ann [11.9K]

Answer:

129.2 C

0.33758239177 mm

Explanation:

n = Number density = $8.46\times 10^{28}\ electrons/m^3$

i = Current = 170 A

t = Time taken = 0.76 s

d = Diameter = 6 mm

Charge is given by

$q=it\\\Rightarrow q=170\times 0.76\\\Rightarrow q=129.2\ C$

The charge passing throught the motor is 129.2 C

Current density

$J=\dfrac{i}{A}\\\Rightarrow J=\dfrac{170}{\dfrac{\pi}{4}\times (6\times 10^{-3})^2}\\\Rightarrow J=6012520.07236\ A/m^2$

Drift velocity is given by

$v_d=\dfrac{J}{ne}\\\Rightarrow v_d=\dfrac{6012520.07236}{8.46\times 10^{28}\times 1.6\times 10^{-19}}\\\Rightarrow v_d=0.000444187357592\ m/s$

Distance traveled

$s=v_dt\\\Rightarrow s=0.000444187357592\times 0.76\\\Rightarrow s=0.00033758239177\ m=0.33758239177\ mm$

The electron traveled 0.33758239177 mm

8 0

3 years ago