All categories

3 years ago

In general, how do you find the average velocity of any object falling in a vacuum?

1 answer:

5 0

In general, how do you find the average velocity of any object falling in a vacuum? (Assume you know the final velocity.) Multiply the final velocity by final time. 3. Calculate : Distance, average velocity, and time are related by the equation, d = v • t A

You might be interested in

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

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

You are looking down on a N = 9 turn coil in a magnetic field B = 0.5 T which points directly down into the screen. If the diame

Novay_Z [31]

Answer:

The magnitude of the voltage is $2.27\times10^{-4}\ V$ and the direction of the current is clockwise.

Explanation:

Given that,

Number of turns = 9

Magnetic field = 0.5 T

Diameter = 3 cm

Time t = 0.14 s

We need to calculate the flux

Using formula of flux

$\phi=NAB$

Put the value into the formula

$\phi=9\times\pi\times(1.5\times10^{-2})^2\times0.5$

$\phi=0.003180$

We need to calculate the emf

Using formula of emf

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{0.003180}{0.14}$

$\epsilon =-0.000227\ V$

$\epsilon=-2.27\times10^{-4}\ V$

Negative sign shows the direction of current.

Hence, The magnitude of the voltage is $2.27\times10^{-4}\ V$ and the direction of the current is clockwise.

8 0

3 years ago

A vessel that contains a gas has two pressure gauges attached to it. One contains liquid mercury, and the other an oil such as d

castortr0y [4]

Answer:

Pressure of the gas = 12669 (Pa) and height of the oil is 1,24 meters

Explanation:

First, we can use the following sketch for an easy understanding, in the attached image we can see the two pressure gauges the one with mercury to the right and the other one with oil to left. We have all the information needed in the mercury pressure gauge, so we can determine the pressure inside the vessel because the fluid is a gas it will have the same pressure distributed inside the vessel (P1).

Since P1 = Pgas, we can use the same formula, but this time we need to determine the height of the column of oil in the pressure gauge.

The result is that the height of the oil column is higher than the height of the one that uses mercury, this is due to the higher density of mercury compared to oil.

Note: the information given in the units of the fluids is not correct because the density is always expressed in units of (mass /volume)

4 0

3 years ago

Nobel gases share which characteristics?

Leto [7]

Full outer shell of electrons, they are colorless and odorless ,their melting and boiling points are close together which gives them very narrow liquid range

8 0

3 years ago

Read 2 more answers

Part A If the velocity of a pitched ball has a magnitude of 47.5 m/s and the batted ball's velocity is 51.5 m/s in the opposite

nalin [4]

Explanation:

Let us assume that the mass of a pitched ball is 0.145 kg.

Initial velocity of the pitched ball, u = 47.5 m/s

Final speed of the ball, v = -51.5 m/s (in opposite direction)

We need to find the magnitude of the change in momentum of the ball and the impulse applied to it by the bat. The change in momentum of the ball is given by :

$\Delta p=m(v-u)\\\\\Delta p=0.145\times ((-51.5)-47.5)\\\\\Delta p=-14.355\ kg-m/s$

So, the magnitude of the change in momentum of the ball is 14.355 kg-m/s.

Let the the ball remains in contact with the bat for 2.00 ms. The impulse is given by :

$J=\dfrac{\Delta p}{t}\\\\J=\dfrac{14.355}{2\times 10^{-3}}\\\\J=7177.5\ kg-m/s$

Hence, this is the required solution.

7 0

4 years ago