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2 years ago

A thin rod of length d on a frictionless surface is pivoted about one end

1 answer:

3 0

The magnitude of the angular momentum of the rod immediately after the collision is given as (pf + pi) * d. (Option A).

<h3>What is Angular Momentum?</h3>

Angular Momentum may be defined or described as the vector quantity of the rotation of a body, which is arrived at by multiplying its moment of inertia by its angular velocity.

The formula for Angular Momentum is given as:

L = mvr; Where

L = Angular Momentum

m = mass

v = velocity; and

r = radius.

Learn more about Angular Momentum at:
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A projectile is launched from the ground with an initial velocity of 12ms at an angle of 30° above the horizontal. The projectil

netineya [11]

$vi^{2}sin2thita/g =12^{2}sin2[30]/9.8=12.7$ Answer:

Explanation:

range is given as

6 0

3 years ago

Read 2 more answers

HELP ASAP!! WILL MARK BRAINLIEST!! WRITE IN YOUR OWN WORDS!! What is something that you would like to see a physicist develop in

Schach [20]

Answer:

Phones as sunglasses with a mic. I put on my glasses and I say what's the weather today, The sunglasses will tell me the weather and can be charged just like phones

Explanation:

8 0

3 years ago

3.7. A dog searching for a bone walks 3.50 m south, then 8.20 m at an angle 23.1 degrees north of east, and finally 15.0 m west.

andriy [413]

Answer:

Explanation:

We shall represent displacement of dog in vector form , in terms of i , j , i representing east and j representing north .

Dog travels 3.5 m south .

Displacement D₁ = - 3.5 j

then dog travels 8.2 m , 23.1 degree north of east

Displacement D₂ = 8.2 cos23.1 i + 8.2 sin23 j

D₂ = 8.2 cos23.1 i + 8.2 sin23.1 j

= 7.54 i + 3.22 j

Third displacement

D₃ = - 15i

Total displacement = D₁ + D₂ + D₃

= - 3.5 j + 7.54 i + 3.22 j -15i

= - 7.46 i - 0.28 j

Magnitude of displacement = √ ( 7.46² + .28²)

= √(55.65 + .08 )

= 7.46 m

b ) Direction of displacement

If Ф be angle , displacement makes with west direction

TanФ = .08 / 55.65 = .00143

Ф = .082 degree south of west or almost west .

From east , this angle = 180 + .082 = 180.082 , counterclockwise .

5 0

2 years ago

A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

5 0

3 years ago

(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50

Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, $p=m_Av_A=m_Bv_B=m_Cv_C$ , were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as $m_h$ , and the mass of the first and second coals as $m_1$ and $m_2$ respectively