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Vikentia [17]
3 years ago
12

A car travels 40 miles in 30 minutes.

Physics
1 answer:
lukranit [14]3 years ago
4 0

Answer:

(a)Average velocity ,v =128.74 Km/hr

(b)Kinetic Energy , K=958546.875 Joule

(c)Distance, s=268.8m

(d)Acceleration, a= - 2.38 m/s^2

<u>Explanation</u>:

<u>Given</u>:

Distance travelled = 40 miles

Time taken = 30 minutes.

(A) The average velocity in kilometres/hour

Converting 40 miles into km ,

we know that,

1 mile = 1.60934

40 miles =  40 x 1.60934

so 40 miles  =  64.3738 Km

similarly converting 30 minutes into hours

1 minute = \frac{1}{60}hours

30 minute = \frac{30}{60}hours

30 minute = \frac{1}{2}hours

Now

Average velocity = \frac{Speed}{time}

Substituting Values,

Average velocity = \frac{64.3738}{\frac[1}{2}}

Average velocity = 64.3738 \times 2

Average velocity =128.74 Km/hr

(B) If the car weighs 1.5 tons, what is its If the car weighs 1.5 tons, what is its kinetic energy in joules (Note: you will need to convert your velocity to m/s)? in joules (Note: you will need to convert your velocity to m/s)?

Converting 1.5 tons into kg we get

1 ton = 1000 kg

so 1.5 ton =1500 kg

converting  velocity to m/s

128.74  \times \frac{5}{18}

=>35.75 m/s----------------------------------------------------------(1)

kinetic energy  K= \frac{1}{2}mv^2

Substituting the values,

K= \frac{1}{2}1500(35.75)^2

K= \frac{1}{2}1500(1278.06)

K= \frac{1500 \times (1278.06)}{2}

K= \frac{1917093.75}{2}

K=958546.875 Joule---------------------------------------------(2)

(c)When the driver applies the brake, it takes 15 seconds to stop. How far does the car travel (in meters) while stopping

Lets use Distance formula,

S= ut+\frac{1}{2}at^2

Substituting the known values,

s= ut+\frac{1}{2}at^2

s= (37.75)(15)+\frac{1}{2}a(15)^2

s=566.25+\frac{1}{2}a(225)

s=566.25+\frac{(225a)}{2}-------------------------------------(3)

(D) What is the average acceleration of the car (in m/s2) during braking?

Using the formula

v=u +at

re arranging the formula we get,

a = \frac{v - u}{t}

Substituting the values

a = \frac{0 - 35.75}{15}

a = \frac{- 35.75}{15}

a= - 2.38 m/s^2----------------------------------------(4)

Now substituting 4 in 3 we get

s=566.25+\frac{(225( - 2.38)}{2}

s=566.25+\frac{-535.5}{2}

s=536.25-267.75

s=268.8m--------------------------------------------------------------(5)

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Answer:

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8 0
3 years ago
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
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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

6 0
3 years ago
*PLEASE HELP***
Delicious77 [7]
I am not 100% sure but i would say D, since it is a smaller object than the test thus needing a shorter wavelength
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