1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Likurg_2 [28]
2 years ago
15

Suppose a car approaches a hill and has an initial speed of 108 km/h at the bottom of the hill. The driver takes her foot off of

the gas pedal and allows the car to coast up the hill.a. If the car has the initial speed stated at a height of h = 0, how high, in meters, can the car coast up a hill if work done by friction is negligible?b. If, in actuality, a 710 kg car with an initial speed of 108 km/h is observed to coast up a hill and stops at a height 21 m above its starting point, how much thermal energy was generated by friction in J?c. What is the magnitude of the average force of friction, in newtons, if the hill has a slope 2.9 degrees above the horizontal?

Physics
1 answer:
Aleks04 [339]2 years ago
6 0

Answer:

a) The car will reach a height of 45.9 m.

b) The amount of thermal energy generated is 173382 J.

c) The magnitude of the force of friction is 417.8 N.  

Explanation:

Hi there!

a) In this problem, we have to use the conservation of energy. The energy conservation theorem states that the energy of a system remains constant. Energy can´t be created nor destroyed, only transformed. In the case of the car, the initial kinetic energy is transformed into potential energy as the car´s height increases while coasting up the hill.

Then, all the initial kinetic energy (KE) will be transformed into potential energy (PE) (only if there is no friction).

The equation of KE is the following:

KE = 1/2 · m · v²

Where:

m = mass of the car.

v = speed of the car.

The equation of PE is the following:

PE = m · g · h

Where:

m = mass of the car.

g = acceleration due to gravity.

h = height at which the car is located.

Since work done by friction is negligible, we can assume that all the initial kinetic energy will be transformed into potential energy. Then:

KE at the bottom of the hill = PE at the top of the hill

1/2 · m · v² = m · g · h

Solving for h:

1/2 · v² / g = h

Let´s convert the speed unit into m/s:

108 km/h · 1000 m/ 1 km · 1 h / 3600 s = 30 m/s

Now, let´s calculate h:

h = 1/2 · (30 m/s)² / 9.8 m/s²

h = 45.9 m

The car will reach a height of 45.9 m.

b) In this case, all the kinetic energy is not transformed into potential energy because some energy is transformed into thermal energy due to friction. The thermal energy generated is equal to the work done by friction. Then:

KE at the bottom of the hill = PE + work done by friction

KE = PE + Wfr  (where Wfr is the work done by friction).

1/2 · m · v² = m · g · h + Wfr

1/2 · m · v² - m · g · h = Wfr

1/2 · 710 kg · (30 m/s)² - 710 kg · 9.8 m/s² · 21 m = Wfr

Wfr = 173382 J

The amount of thermal energy generated is 173382 J.

c) The work done by friction is calculated as follows:

Wfr = Ffr · Δx

Where:

Ffr = friction force.

Δx = traveled distance

Please, see the attached figure to notice that the traveled distance can be calculated by trigonometry using this trigonometric rule of right triangles:

sin angle = opposite side / hypotenuse

In our case:

sin 2.9° = h / Δx

Δx = h / sin 2.9°

Δx = 21 m / sin 2.9° = 415 m

Then, solving for the friction force using the equation of the work done by friction:

Wfr = Ffr · Δx

Wfr / Δx = Ffr

173382 J / 415 m = Ffr

Ffr = 417.8 N

The magnitude of the force of friction is 417.8 N

You might be interested in
A single-phase transformer has a turns ratio of 20,000/5,000. If a DC voltage of 50 V is applied to the primary winding, what wi
Kipish [7]

Sneaky question.

The secondary voltage will be zero.

Secondary voltage is the result of CHANGES in the primary voltage. That means the primary is energized with AC. The transformer in this question is energized with DC, which doesn't change. So there is no secondary voltage, this transformer doesn't work, and what you get out of it is: Smoke !

6 0
3 years ago
Question 1<br> The orange line is measuring the
11111nata11111 [884]

Answer:

Approximately 4.5 square miles

Explanation:

(surrounding van Nuys and sepulveda stations) and one measuring approximately 2.85 square miles ( surrounding north Hollywood Station)

4 0
3 years ago
A box is sliding down an incline tilted at a 11.1° angle above horizontal. The box is initially sliding down the incline at a sp
raketka [301]

Answer:s=0.68 m

Explanation:

Given

Inclination \theta =11.1^{\circ}

Speed of block(u)=1.6 m/s

Coefficient of kinetic Friction \mu _k=0.39

deceleration provided by friction=g\sin \theta -\mu _kg\cos \theta [/tex]

Using v^2-u^2=2as

Final velocity v=0

0-1.6^2=2(g\sin \theta -\mu _kg\cos \theta )s

s=\frac{-1.6^2}{2\cdot (9.8\sin 11.1-0.39\times 9.8\times \cos 11.1)}

s=0.68 m

5 0
3 years ago
Convert 8 light years to Astronomical Units
marusya05 [52]

Answer:

505929 AU

Explanation:

As you may know, one light-year is equivalent to approximately 63241.1 Astronomical Units. To get your answer, simply multiply 63241.1 * 8 to get ≈505929 AU

5 0
3 years ago
In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.
leonid [27]

Answer:

(a): F_e = 8.202\times 10^{-8}\ \rm N.

(b): F_g = 3.6125\times 10^{-47}\ \rm N.

(c): \dfrac{F_e}{F_g}=2.27\times 10^{39}.

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053\times 10^{-9} m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges q_1 and q_2 respectively is given by

F_e = \dfrac{k|q_1||q_2|}{r^2}

where,

  • k = Coulomb's constant = 9\times 10^9\ \rm Nm^2/C^2.
  • r = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, q_1 = +1.6\times 10^{-19}\ C.

The charge on the electron, q_2 = -1.6\times 10^{-19}\ C.

These two are separated by the distance, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.

Part (b):

The gravitational force of attraction between two objects of masses m_1 and m_1 respectively is given by

F_g = \dfrac{Gm_1m_2}{r^2}.

where,

  • G = Universal Gravitational constant = 6.67\times 10^{-11}\ \rm Nm^2/kg^2.
  • r = distance of separation between the masses.

For the given system,

The mass of proton, m_1 = 1.67\times 10^{-27}\ kg.

The mass of the electron, m_2 = 9.11\times 10^{-31}\ kg.

Distance between the two, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.

The ratio \dfrac{F_e}{F_g}:

\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.

6 0
3 years ago
Other questions:
  • You are asked to calculate an object's velocity, in order to do so you must know the object's ?
    12·1 answer
  • I would appreciate some help.
    11·1 answer
  • Albert observes that Emmy's watch is ticking eanwhile, Emmy observes Albert's watch and compares it with her watch. Emmy observe
    6·1 answer
  • What did the greeks call the mineral they found and why?
    6·1 answer
  • Kayla drew a diagram to compare convex and concave lenses which labels belong in the areas marked XYZ
    5·1 answer
  • An element can be identified by its spectrum. True False
    7·2 answers
  • A boy falls and hits his head with an impulse of 20N-s. On cement, the boy would hit for
    8·1 answer
  • Help mee (:
    8·1 answer
  • The number of windings on the primary coil of a transformer
    8·1 answer
  • Two objects of equal mass are a distance of 5.0 m apart and attract each other with a gravitational force of 3.0 x 10^-7 N find
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!