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LiRa [457]
3 years ago
10

To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the r

oad (the rolling resistance). The density of air is 1.2 kg/m3. (a) What are the drag forces in newtons at 80 km/h and 105 km/h for a Toyota Camry? (Drag area = 0.70 m2 and drag coefficient = 0.28.) at 80 km/h N at 105 km/h N (b) What are the drag forces in newtons at 80 km/h and at 105 km/h for a Hummer H2? (Drag area = 2.44 m2 and drag coefficient = 0.57.) at 80 km/h N at 105 km/h N
Physics
1 answer:
bogdanovich [222]3 years ago
4 0

Answer:

Toyota Camry

     F_d = 51.852 N

     F_d = 100.042 N

Hummer H2

     F_d = 412.0888 N

     F_d = 8351.755 N

Explanation:

Given:

- The density of air p_air = 1.2 kg/m^3

- The drag force equals car's engine force.

Find:

- What are the drag forces in newtons at 80 km/h and 105 km/h for a Toyota Camry? (Drag area = 0.70 m2 and drag coefficient = 0.28.)

- What are the drag forces in newtons at 80 km/h and at 105 km/h for a Hummer H2? (Drag area = 2.44 m2 and drag coefficient = 0.57.)

Solution:

- The formula for drag force is given as follows:

                                 F_d = 0.5*C_d*p_air*A*V^2

Where,

A : The drag Area  m^2

C_d: The drag coefficient

V: Velocity  m/s

a)  Toyota Camry

                  C_d = 0.28 , A = 0.70 m^2 , V = 80 km/h = 22.222 m/s

Then compute the F_d drag force:

                  F_d = 0.5*0.28*1.2*0.70*(22.22)^2

                  F_d = 51.852 N

                 C_d = 0.28 , A = 0.70 m^2 , V = 105 km/h = 29.1667 m/s

Then compute the F_d drag force:

                  F_d = 0.5*0.28*1.2*0.70*(29.1667)^2

                  F_d = 100.042 N

b)   Hummer H2

                 C_d = 0.57 , A = 2.44 m^2 , V = 80 km/h = 22.222 m/s

Then compute the F_d drag force:

                  F_d = 0.5*0.57*1.2*2.44*(22.22)^2

                  F_d = 412.0888 N

                 C_d = 0.28 , A = 0.70 m^2 , V = 105 km/h = 29.1667 m/s

Then compute the F_d drag force:

                  F_d = 0.5*0.57*1.2*2.44*(29.1667)^2

                  F_d = 8351.755 N

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Answer:

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Explanation:

in this question we have given

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height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as,W_{1}=Force \times displacement

=\int\limits^{15}_0 {.2x} \, dx ..............(1)

=.1\times 15^2

=22.5 Nm

work done in lifting the sand is given as,W_{2}=Force \times displacement

W_{2}=\int\limits^{15}_0 F \, dx.................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m=\frac{20-18}{15-0}

m=.133

Therefore,

F=.133x+2

Put value of F in equation 2

W_{2}=\int\limits^{15}_0 (.133x+2) \, dx

W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm

Therefore,

work done lifting the bucket (sand and rope) to the top of the building,W=W_{1}+W_{2}

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4 0
3 years ago
A kid on a playground swing makes a complete to-and-fro swing each 2 seconds.
sashaice [31]

Answer:

The frequency of the swing: 1/2 Hertz

The period is: 2 Seconds

Explanation:

The time the kid takes to make a complete to-and-fro swing = 2 seconds

The period, T, is the time it takes to make one complete cycle of an oscillatory motion, therefore, we have;

The frequency, f, is the number of cycles completed each second, therefore, we have;

The time for 1 cycle = 2 seconds

2 seconds = 1 cycle

Dividing both sides by 2 gives;

2/2 seconds = 1/2 cycles

2/2 = 1

In 2/2 = 1 seconds The number of cycles completed = 1/2 cycles

Therefore, the number of cycles completed per (one) second = 1/2 cycles

Therefore the frequency of the swing, f = 1/2 cycle/seconds = 1/2 Hertz

The period, T, is the time it takes to complete one to-and-fro swing which is one cycle which is 2 seconds

Therefore, the period is 2 Seconds.

4 0
3 years ago
Find the hiker’s gravitational potential energy if the cliff is 60m high
Furkat [3]

Answer:

Potential energy is U=mgh

Explanation:

The potential energy depends on the mass, the acceleration of gravity g and the height at which the object or person is.

Potential energy  U=mgh

In this case we would need to know the exact mass of the hiker in order to calculate the potential energy.

But we know the values of g and h

g=9.81m/s^2

h=60m

So, the potential energy

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m is the mass of the hiker, wich is not in the description of the problem.

4 0
3 years ago
An energy source forces a constant current of 2A to flow through a light bulbfilament for twenty seconds. If 4.6 kJ is given off
QveST [7]

Answer:

The voltage drop across the bulb is 115 V

Explanation:

The voltage drop equation is given by:

V=\frac{\Delta W}{\Delta q}

Where:

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We need to use the definition of electric current to find Δq

I=\frac{\Delta q}{\Delta t}

Where:

I is the current (2 A)

Δt is the time (20 s)

2=\frac{\Delta q}{20}

q=40 C

Then, we can put this value of charge in the voltage equation.

V=\frac{4600}{40}=115 V

Therefore, the voltage drop across the bulb is 115 V.

I hope it helps you!

6 0
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A specimen of steel has a rectangular cross section 20 mm wide and 40 mm thick, an elastic modulus of 207 GPa, and a Poisson’s r
katrin2010 [14]

Answer:

There's a decrease in width of 2.18 × 10^(-6) m

Explanation:

We are given;

Shear Modulus;E = 207 GPa = 207 × 10^(9) N/m²

Force;F = 60000 N.

Poisson’s ratio; υ =0.30

We are told width is 20 mm and thickness 40 mm.

Thus;

Area = 20 × 10^(-3) × 40 × 10^(-3)

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Now formula for shear modulus is;

E = σ/ε_z

Where σ is stress given by the formula Force(F)/Area(A)

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E = (F/A)/ε_z

ε_z = (F/A)/E

ε_z = (60,000/(8 × 10^(-4)))/(207 × 10^(9))

ε_z = 3.62 × 10^(-4)

Now, formula for lateral strain is;

ε_x = - υ × ε_z

ε_x = -0.3 × 3.62 × 10^(-4)

ε_x = -1.09 × 10^(-4)

Now, change in width is given by;

Δw = w_o × ε_x

Where w_o is initial width = 20 × 10^(-3) m

So; Δw = 20 × 10^(-3) × -1.09 × 10^(-4)

Δw = -2.18 × 10^(-6) m

Negative means the width decreased.

So there's a decrease in width of 2.18 × 10^(-6) m

6 0
3 years ago
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