A Honda Civic Hybrid weighs about 3200 lb .

A 12-kg projectile is launched with an initial vertical speed of 20 m/s. It rises to a maximum height of 18 m above the launch p

kirill115 [55]

Answer:

Change in mechanical energy, $\Delta E=283.2\ J$

Explanation:

It is given that,

Mass of the projectile, m = 12 kg

Speed of the projectile, v = 20 m/s

Maximum height, h = 18 m

Initially, the projectile have only kinetic energy. it is given by :

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

$K=\dfrac{1}{2}\times 12\ kg\times (20\ m/s)^2$

K = 2400 J

Finally, it have only potential energy. it is given by :

P = mgh

$P=12\ kg\times 9.8\ m/s^2\times 18\ m$

P =2116.8 J

The change in mechanical energy is given by :

$\Delta E=K-P$

$\Delta E=2400-2116.8$

$\Delta E=283.2\ J$

So, the change in mechanical energy is 283.2 J. Hence, this is the required solution.

8 0

2 years ago

Each of the two coils has 320 turns. The average radius of the coil is 6 cm. The distance from the center of one coil to the ele

Temka [501]

Answer:

Magnetic field, $B=2.39\times 10^{-3}\ T$

Explanation:

It is given that,

Number of turns, N = 320

Radius of the coil, r = 6 cm = 0.06 m

The distance from the center of one coil to the electron beam is 3 cm, x = 3 cm = 0.03 m

Current flowing through the coils, I = 0.5 A

We need to find the magnitude of the magnetic field at a location on the axis of the coils, midway between the coils. The magnetic field midway between the coils is given by :

$B=\dfrac{\mu_oINr^2}{(x^2+r^2)^{3/2}}$

$B=\dfrac{4\pi \times 10^{-7}\times 0.5\times 320\times (0.06)^2}{(0.03^2+0.06^2)^{3/2}}$

B = 0.00239 T

or

$B=2.39\times 10^{-3}\ T$

So, the magnitude of the magnetic field at a location on the axis of the coils, midway between the coils is $2.39\times 10^{-3}\ T$ . Hence, this is the required solution.

7 0

3 years ago

The astronomical unit (AU) is defined as the mean center-to-center distance from Earth to the Sun, namely 1.496x10^(11) m. The p

Rudiy27

Answer:

a) How many parsecs are there in one astronomical unit?

$4.85x10^{-6}pc$

(b) How many meters are in a parsec?

$3.081x10^{16}m$

(c) How many meters in a light-year?

$9.46x10^{15}m$

(d) How many astronomical units in a light-year?

$63325AU$

(e) How many light-years in a parsec?

3.26ly

Explanation:

The parallax angle can be used to find out the distance using triangulation. Making a triangle between the nearby star, the Sun and the Earth, knowing that the distance between the Earth and the Sun ( $1.496x10^{11} m$ ) is defined as 1 astronomical unit:

$\tan{p} = \frac{1AU}{d}$

Where d is the distance to the star.

Since p is small it can be represent as:

$p(rad) = \frac{1AU}{d}$ (1)

Where p(rad) is the value of in radians

However, it is better to express small angles in arcseconds

$p('') = p(rad)\frac{180^\circ}{\pi rad}.\frac{60'}{1^\circ}.\frac{60''}{1'}$

$p('') = 2.06x10^5 p(rad)$

$p(rad) = \frac{p('')}{2.06x10^5}$ (2)

Then, equation 2 can be replace in equation 1:

$\frac{p('')}{2.06x10^5} = \frac{1AU}{d}$

$\frac{d}{1AU} = \frac{2.06x10^5}{p('')}$ (3)

From equation 3 it can be see that $1pc = 2.06x10^5 AU$

a) How many parsecs are there in one astronomical unit?

$1AU . \frac{1pc}{2.06x10^5AU}$ ⇒ $4.85x10^{-6}pc$

(b) How many meters are in a parsec?

$2.06x10^{5}AU . \frac{1.496x10^{11}m}{1AU}$ ⇒ $3.081x10^{16}m$

(c) How many meters in a light-year?

To determine the number of meters in a light-year it is necessary to use the next equation:

$x = c.t$

Where c is the speed of light ( $c = 3x10^{8}m/s$ ) and x is the distance that light travels in 1 year.

In 1 year they are 31536000 seconds

$x = (3x10^{8}m/s)(31536000s)$

$x = 9.46x10^{15}m$

(d) How many astronomical units in a light-year?

$9.46x10^{15}m . \frac{1AU}{1.496x10^{11}m}$ ⇒ $63325AU$

(e) How many light-years in a parsec?

$2.06x10^{5}AU . \frac{1ly}{63235AU}$ ⇒ $3.26ly$

5 0

3 years ago

Alicia está a punto de perder su bus. En un desesperado intento, corre a una velocidad constante de 5 m/s. Cuando está a 15 m de

pochemuha

Answer:

Si logra alcanzar el bus.

Explanation:

Para poder solucionar este problema debemos de tener en cuenta que Alicia corre a velocidad constante para poder alcanzar el bus. La formula de la cinematica que tiene en cuenta la velocidad constante es la siguiente:

$x_{f} = x_{o}+(v*t)$

donde:

Xf = Ubicacion del punto donde se encuentra el bus [m]

Xo = Ubicacion desde donde esta Alicia [m]

v = velocidad constante = 5 [m/s]

t = tiempo [s]

Xf - Xo = 15 [m]

15 = 5*t

t = 3 [s]

Ahora con el tiempo podemos encontrar la velocidad del bus por medio de la siguiente ecuacion de cinematica para la aceleracion constante:

$v_{f} = v_{i}+(a*t)$

donde:

Vf = velocidad del bus despues de los 3 [s]

Vi = velocidad inicial = 0

a = aceleracion = 0.5 [m/s^2]

Vf = 0 + (0.5*3)

Vf = 1.5 [m/s]

La velocidad del bus es menor que la velocidad de Alicia, por ende Alicia alcanzara el bus.

7 0

3 years ago

Explain how to determine the volume of an irregularly shaped object such as a rock

alexira [117]

If you have no way to accurately measure all of the object's bumps and dimples, then the only way to measure its volume is by means of fluid displacement.

-- Put some water into a graduated (marked) container, read the amount of water, drop the object into the container, and read the new volume in the container. The volume of the object is the difference between the two readings.

-- Alternatively, stand an unmarked container in a large pan, and fill it to the brim. Slowly slowly lower the object into the unmarked container, while the pan catches the water that overflows from it. When the object is completely down in the container, carefully remove the container from the pan, and measure the volume of the water in the pan. It's equal to the volume of the object.

6 0

3 years ago