Consider the mass and velocity values of Objects A and B below.

How long would it take for a sound impulse to travel through a copper Rod 20km

Nana76 [90]

Not sure.can you give me a clue?

8 0

3 years ago

If you were told an atom was an ion, you would know the atom must have a ______?

andreyandreev [35.5K]

Answer:

charge

Explanation:

7r0I and its etc. ,"!×_/;

3 0

3 years ago

A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The

USPshnik [31]

Answer:

a) see attached, a = g sin θ

b)

c) v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

Wₓ = m a

W sin θ = m a

a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

θ' = 9/2 θ

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

Em₀ = mg h = mg L (1-cos tea)

Lowest point

Emf = K = ½ m v²

Em₀ = Emf

g L (1-cos θ) = v² / 2

v = √(2gL (1-cos θ))

4 0

3 years ago

Whenever two apollo astronauts were on the surface of the moon, a third astronaut orbited the moon. assume the orbit to be circu

almond37 [142]

Missing question:
"Determine (a) the astronaut’s orbital speed v and (b) the period of the orbit"

Solution

part a) The center of the orbit of the third astronaut is located at the center of the moon. This means that the radius of the orbit is the sum of the Moon's radius r0 and the altitude ( $h=430 km=4.3 \cdot 10^5 m$ ) of the orbit:
$r= r_0 + h=1.7 \cdot 10^6 m + 4.3 \cdot 10^5 m=2.13 \cdot 10^6 m$
This is a circular motion, where the centripetal acceleration is equal to the gravitational acceleration g at this altitude. The problem says that at this altitude, $g=1.08 m/s^2$ . So we can write
$g=a_c= \frac{v^2}{r}$
where $a_c$ is the centripetal acceleration and v is the speed of the astronaut. Re-arranging it we can find v:
$v= \sqrt{g r}= \sqrt{(1.08 m/s^2)(2.13 \cdot 10^6 m)}=1517 m/s = 1.52 km/s$

part b) The orbit has a circumference of $2 \pi r$ , and the astronaut is covering it at a speed equal to v. Therefore, the period of the orbit is
$T= \frac{2 \pi r}{v} = \frac{2\pi (2.13 \cdot 10^6 m)}{1517 m/s} =8818 s = 2.45 h$
So, the period of the orbit is 2.45 hours.

6 0

3 years ago

A sled is pulled with a horizontal force of 18 n along a level trail, and the acceleration is found to be 0.39 m/s2. an extra ma

Natali5045456 [20]

From Newton's second law of motion, it is identified that the net force applied to the object with mass m, will make it move with an acceleration of a. This can be mathematically translated as,
F = m x a
To solve for the mass of the sled, we derive the equation above such that,
m = F / a
Substituting,
m = (18 N) / (0.39 m/s²)
m = 46.15 kg

Then, we add to the calculated mass the mass of the extra material.
total mass = 46.15kg + 4.5 kg
total mass = 50.65 kg
We solve for the normal force of the surface to the object by calculating its weight.
F₂ = (50.65 kg)(9.8 m/s²)
F₂ = 496.41 N

The force that would allow barely a movement for the object is equal to the product of the normal force and the coefficient of kinetic friction.
F = (F₂)(c)
c = F/F₂

Substituting,
c = 18 N/496.41 N
c = 0.0362

<em>ANSWER: c = 0.0362</em>

5 0

3 years ago