HEY CAN YALL PLS HELP ME IN DIS!!!!!!!!!!!!!

During a rockslide, a 670 kg rock slides from rest down a hillside that is 740 m along the slope and 240 m high. The coefficient

ElenaW [278]

a) $1.58\cdot 10^6 J$

b) $1.15\cdot 10^6 J$

c) $0.43\cdot 10^6 J$

d) 35.8 m/s

Explanation:

a)

The gravitational potential energy of an object is the energy possessed by the object due to its location with respect to the ground.

It is given by:

$U=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height of the object, relative to a reference level

Here, the reference level is taken at the bottom of the hill (where the potential energy is zero).

So, we have:

m = 670 kg is the mass of the rock

$g=9.8 m/s^2$

h = 240 m is the initial height of the rock

So, the potential energy of the rock just before the slide is

$U=(670)(9.8)(240)=1.58\cdot 10^6 J$

b)

The energy transferred to thermal energy during the slide is equal to the work done by friction, which is:

$W=F_f d$

where

$F_f$ is the force of friction

d = 740 m is the displacement of the rock along the ramp

The force of friction is given by:

$F_f=-\mu mg cos \theta$

where

$\mu=0.25$ is the coefficient of friction

m = 670 kg is the mass of the rock

$\theta$ is the angle of the ramp

Since we know the lenght of the ramp (d = 740 m) and the height (h = 240 m), we can find the angle:

$\theta=sin^{-1}(\frac{h}{d})=sin^{-1}(\frac{240}{740})=18.9^{\circ}$

Therefore, the work done by friction is:

$W=-\mu m g cos \theta d =-(0.25)(670)(9.8)(cos 18.9^{\circ})(740)=-1.15\cdot 10^6 J$

So, the energy transferred to thermal energy is $1.15\cdot 10^6 J$ .

c)

According to the law of conservation of energy, the kinetic energy of the rock as it reaches the bottom of the hill will be equal to the initial potential energy (at the top) minus the energy transformed into thermal energy.

Therefore, we have:

$K_f = U_i -E_t$

where here we have:

$U_i=1.58\cdot 10^6 J$ is the potential energy of the rock at the top of the hill

$E_t=1.15\cdot 10^6 J$ is the energy converted into thermal energy

Substituting, we find

$K_f=1.58\cdot 10^6-1.15\cdot 10^6=0.43\cdot 10^6 J$

So, this is the kinetic energy of the rock at the bottom of the hill.

d)

The kinetic energy of the rock at the bottom of the hill can be rewritten as

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

where

m is the mass of the rock

v is its final speed

In this problem, we have:

$K_f=0.43\cdot 10^6 J$ is the final kinetic energy of the hill

m = 670 kg is the mass of the rock

Therefore, the final speed of the rock is:

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.43\cdot 10^6)}{670}}=35.8 m/s$

7 0

3 years ago

A rocket moves through outer space at 11,000 m/s. At this rate, how much time would be required to travel the distance from Eart

Charra [1.4K]

11,000 m = 11 km

11 km/s over 380,000km

380,000 / 11 = 34545.4 seconds

34545.4 / 60 = 575.7 minutes

5 0

3 years ago

Unpolarized light of intensity Io is incident on a stack of 7 polarizing filters, each with its axis rotated 17Â°cw with respect

mash [69]

The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925

To answer the question, we need to know what polarization of light is.

<h3>What is polarization of light?</h3>

This is when the electric field vector of light is oscillating in one plane.

Now for light of intensity I' which is initially unpolarized, its intensity after polarization is I = 1/2I'.
Also, for light initially polarized, its intensity after polarization is I"' = I"cos²Ф where Ф is the angle between the initial direction and the direction of polarization.

<h3>Intensity of light through each polarized filter</h3>

Given that we have 7 polarizing filters, each rotated 17° cw with respect to the previous filter.

So, since the light is initially unpolarized,

The intensity through the first polarizing filter is I₁ = 1/2I₀ where I₀ is the initial intensity.

The intensity through the second polarizing filter is I₂ = I₁cos²17°= 1/2I₀cos²17°
The intensity through the third polarizing filter is I₃ = I₂cos²17° = 1/2I₀cos⁴17°
The intensity through the fourth polarizing filter is I₄ = I₃cos²17° = 1/2I₀cos⁶17°
The intensity through the fifth polarizing filter is I₅ = I₄cos²17° = 1/2I₀cos⁸17°
The intensity through the sixth polarizing filter is I₆ = I₅cos²17° = 1/2I₀cos¹⁰17°
The intensity through the seventh polarizing filter is I₇ = I₆cos²17° = 1/2I₀cos¹²17°.

<h3>The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity</h3>

Since I₇ is the last intensity I₇ = It = 1/2I₀cos¹²17°.

So, It/I₀ = 1/2cos¹²17°

= 1/2(0.9563)¹²

= 1/2 × 0.5850

= 0.2925

So, the ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925

Learn more about intensity of polarized light here:

brainly.com/question/25402491

5 0

2 years ago

una onda longitudinal tiene una frecuencia de 200 hz y una longitud de onda de 4.2m ¿cual es la rapidez de la onda?

swat32

Answer:

v = 8.4 m/s

Explanation:

The question ays, "A longitudinal wave has a frequency of 200 Hz and a wavelength of 4.2m. What is the speed of the wave?".

Frequency of a wave, f = 200 Hz

Wavelength = 4.2 cm = 0.042 m

We need to find the speed of the wave. The formula for the speed of a wave is given by :

$v=f\lambda\\\\v=200\times 0.042\\\\=8.4\ m/s$

So, the speed of the wave is equal to 8.4 m/s.

4 0

3 years ago

PLEASE help!!

sergeinik [125]

Answer:

in which standard you are, i am typing your answer till please reply.

5 0

3 years ago