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3 years ago

10

Suppose that the mirror described in Part A is initially at rest a distance R away from the sun. What is the critical value of a

rea density for the mirror at which the radiation pressure exactly cancels out the gravitational attraction from the sun?

1 answer:

Sati [7]3 years ago

5 0

Ans;

6.25m²/Kd

Explanation: see attached file

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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

4 years ago

How much heat is absorbed by a 75g iron skillet when it’s temperature rises from 5c to 20c?

alexira [117]

Answer: $Q=499.5 J$

Explanation:

The heat (thermal energy) absorbed by the iron skillet can be found using the following equation:

$Q=m.C.\Delta T$ (1)

Where:

$Q$ is the heat (absorbed)

$m=75 g$ is the mass of the element (iron in this case)

$C$ is the specific heat capacity of the material. In the case of iron is $C=0.444\frac{J}{g\°C}$

$\Delta T=T_{f}-T_{i}=20\°C - 5\°C= 15\°C$ is the variation in temperature

Knowing this, lets rewrite (1) with these values:

$Q=(75 g)(0.444\frac{J}{g\°C})(15\°C)$ (2)

Finally:

$Q=499.5 J$

6 0

3 years ago

In the equation for Ohm’s law what does I stand for?

Ket [755]

A) current

(I is always current in electricity)

3 0

3 years ago

Sara is riding her bike down the road at 35 km/hr. She starts to peddle faster as she

steposvetlana [31]

Answer:

Distance covered to top of the hill was : 1.755 km

Explanation:

Initial velocity = 35 km/hr

Acceleration = 2.0 km/hr²

Time taken to accelerate = 3 minutes = 3/60 hours = 1/20 hours

Formula for acceleration : a = Δv /t

v-u/t ---where u is initial velocity , v is final velocity and t is time taken for acceleration

v- 35 / 0.05 = 2

v = 35.10 km/h

Formula for distance is product of speed and time

Distance covered = 35.10 * 0.05 = 1.755 km

6 0

3 years ago

Explain three ways you can get home safely, without getting behind the wheel, if there are drugs or alcohol in your system.

Ad libitum [116K]

Call your parent, call a local taxi ,or a sober friend

4 0

3 years ago