What do the length of the cord and gravity determine for a pendulum?

A thread holds a 1.5-kg and a 4.50-kg cart together. After the thread is burned, a compressed spring pushes the carts apart, giv

maksim [4K]

Answer:

c. 8.67 cm/s

Explanation:

From the law of conservation of momentum,

Total momentum before the thread was burned = Total momentum after was burned

mu + m'u' = mv + m'v'...................... Equation

Where m = mass of the heavier cart, m' = mass of the lighter cart, u = initial velocity of the bigger cart, u' = initial velocity of the smaller cart, v = final velocity of the bigger cart, v' = final velocity of the smaller cart.

Note: Both cart where momentarily at rest, as such u = u' = 0. i.e the total momentum before the thread was burn = 0

And assuming the left is positive,

We can rewrite equation 1 as

mv + m'v' = 0............................................ Equation 2

Given: m = 4.5 kg, m' = 1.5 kg, v' = 26 cm/s

Substitute into equation 2,

4.5v + 1.5(26) = 0

4.5v + 39 = 0

4.5v = -39

v = -39/4.5

v = -8.67 cm/s.

Note: v is negative because it moves to right.

Hence the velocity of the 4.5 kg cart = 8.67 cm/s.

The right option is c. 8.67 cm/s

7 0

3 years ago

What must be known to determine the direction of the magnetic force on a charge? Check all that apply.

sergeinik [125]

The strength of the magnetic field

8 0

3 years ago

Read 2 more answers

To get an accurate data reading, scientists must be sure to

joja [24]

D.
Always use the right tool to get accurate measurements

8 0

4 years ago

What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

Wittaler [7]

Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

$\cos \theta = \frac{adjacent}{hypotenuse}$

$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

$r_{71.9^{\circ}}$ can be isolated from equation 3:

$r_{71.9^{\circ}} = r_{e} \cos \theta$ (4)

$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

$v = \frac{2 \pi (1.97x10^{6} m)}{24h}$

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

$T = 24h . \frac{3600s}{1h}$ ⇒ $84600s$