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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

3 years ago

How much energy (in kW-h) does a 900 Watt stove use in a week if it is used for 1.5 hours each day?

ioda

Answer:

9.45 kWh

Explanation:

Energy = Power × time

E = 900 W × (1.5 h/day × 7 day)

E = 9450 Wh

E = 9.45 kWh

3 0

3 years ago

Human reaction times are worsened by alcohol. How much further (in feet) would a drunk driver's car travel before he hits the br

sweet-ann [11.9K]

Answer:

A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes

Explanation:

Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:

2as = Vf² - Vi²

s = (Vf² - Vi²)/2a

where,

Vf = Final Velocity of Car = 0 mi/h

Vi = Initial Velocity of Car = 50 mi/h

a = deceleration of car

s = distance covered

Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.

So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:

s = vt

FOR SOBER DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 0.33 s

s = s₁

Therefore,

s₁ = (73.33 ft/s)(0.33 s)

s₁ = 24.2 ft

FOR DRUNK DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 1 s

s = s₂

Therefore,

s₂ = (73.33 ft/s)(1 s)

s₂ = 73.33 ft

Now, the distance traveled by drunk driver's car further than sober driver's car is given by:

ΔS = s₂ - s₁

ΔS = 73.33 ft - 24.2 ft

6 0

3 years ago

The speed of sound through diamond is about 12,000 m/s. The speed of sound through wood is about 3,300 m/s. Which statement expl

Damm [24]

Answer: They have different rigidities.

Explanation:

7 0

2 years ago

You, Archimedes, suspect that the king’s crown is not solid gold but is instead gold-plated lead. To test your theory, you weigh

hjlf

Answer:

a) 16675.75 Kg/m³ b) 77.6%

Explanation:

the weight of the crown = 60 N, density of gold = 19300 Kg/m^3, density of lead = 11340 kg/m^3, density of water = 1000kg/m^3 and acceleration due to gravity = 9.8 m/s^2

upthrust on the crown = weight in air - weight when fully submerged in water = 60 - 56.4 = 3.6 N

mass of water displaced = 3.6 / 9.8 since weight = mass × g

mass of water displaced = 0.367 Kg

density of water = mass / volume

1000 = 0.367 / volume

cross multiply and find volume

volume of the crown = 0.367 / 1000 = 0.000367 m³ since the crown will displace water of equal volume according to Archimedes principle

Let V1 represent the volume of Gold and let V2 represent the volume of lead

Total volume of the crown = V1 + V2

also

density of gold = mass of gold / V1 and density of lead = mass of lead / V2

19300 = mass of gold in the crown / V1 and 11340 = mass of lead in the crown / V2

19300 V1 = mass of gold and 11340 V2 = mass of lead

add the two together

19300 V1 + 11340 V2 = weigth of the crown / 9.8

19300 V1 + 11340 V2 = 6.12 also

V1 + V2 = 0.000367

make V1 subject of the formula in equation 2

V1 = 0.000367 - V2

substitute for V1 in equation 1

19300 (0.000367 - V2) + 11340 V2 = 6.12

open the bracket

7.083 - 19300 V2 + 11340 V2 = 6.12

rearrange the equation

-7960 V2 = 6.12 - 7.083

-7960 V2 = -0.963

V2 = -0.963 / -7960 = 0.000121 (volume of lead in the crown)

substitute V2 into equation 2

V1 + 0.000121 = 0.000367m³

V1 = 0.000367 - 0.000121 = 0.000246m³ (volume of gold in the crown)

so mass of gold in the crown = 19300 × 0.000246 = 4.748 kg

and mass of lead = 11340 × 0.000121 = 1.372 kg

average density of the crown = (mass of gold + mass of lead) / total volume = 6.12 / 0.000367 = 16675.75 kg/ m³

b) percentage make of gold = mass of gold / total mass × 100 = 77.6 % approx

4 0

3 years ago