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Alchen [17]
3 years ago
12

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 77 km/h and 106 km/h for a Toyota Camry? (Drag area = 0.70 m2 and drag coefficient = 0.28.) at 77 km/h N at 106 km/h N
(b) What are the drag forces in newtons at 77 km/h and at 106 km/h for a Hummer H2? (Drag area = 2.44 m2 and drag coefficient = 0.57.) at 77 km/h N at 106 km/h N Supporting Materials
Physics
1 answer:
sweet [91]3 years ago
6 0

Answer:

a). 53.75 N and 101.92 N

b). 381.44 N and 723.25 N

Explanation:

V= 77 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 21.38 \frac{m}{s} \\V=106 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 29.44 \frac{m}{s}

a).

ρ= 1.2 \frac{kg}{m^{3} }, A_{t}= 0.7 m^{2}, D_{t}= 0.28

F_{t1} = \frac{1}{2} * D_{t} * A_{t}* p_{t}* v_{t}^{2}

F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 53.75 N

F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 101.92 N

b).

ρ= 1.2 \frac{kg}{m^{3} }, A_{h}= 2.44 m^{2}, D_{h}= 0.57

F_{t1} = \frac{1}{2} * D_{h} * A_{h}* p_{h}* v_{h}^{2}

F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 381.44 N

F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 723.25 N

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Two facing surfaces of two large parallel conducting plates separated by 8.5 cm have uniform surface charge densities such that
elena-s [515]

Answer:

positive plate

E = 5.764 KV / m

W = 490eV or 7.85 * 10^-17 J

E_p = 4.74 *10^(-12) eV

E_k = 490 eV

Explanation:

part a

The potential difference between two plates = 490 V

Distance between two plates = 8.5 cm

Answer: The positive plate is at higher potential because of convention.

part b

Electric Field between the plates

E = V / d

E = 490 / 0.085 = 5.764 KV / m

Answer: Electric Field between the plates E = 5.764 KV / m

part c

Work done by electric field

W = V*q

W = 490 * 1.602*10^-19

W = 7.85 * 10^-17 J

or W = 490 eV

Answer: Work done by electric field W = 490eV or 7.85 * 10^-17 J

part d

Potential Energy of an electron gained:

E_p = m_e * g * d / (1.602*10^-19)

E_p =  9.109*10^-31* 9.81 * 0.085 / (1.602*10^-19)

E_p = 4.74 *10^(-12) eV

Very very small E_p approximately 0

Answer: Potential Energy of an electron gained E_p = 4.74 *10^(-12) eV or 0.

part e

Kinetic Energy of an electron gained:

W - E_p = E_k

E_k = 490eV - 4.74*10^(-12)eV

E_k = 490 eV

Answer: Kinetic Energy of an electron gained E_k = 490 eV

7 0
3 years ago
I will mark you brainlist. How can you use a tuning fork to tune a piano?
Phoenix [80]

A tuning fork's job is to establish a single note that everybody can tune to.

Most tuning forks are made to vibrate at 440 Hz, a tone known to musicians as "concert A." To tune a piano, you would start by playing the piano's "A" key while ringing an "A" tuning fork. If the piano is out of tune, you'll hear a distinct warble between the note you're playing and the note played by the tuning fork; the further apart the warbles, the more out-of-tune the piano. By either tightening or loosening the piano's strings, you reduce the warble until it's in line with the tuning fork. Once the "A" key is in tune, you would then adjust all of the instrument's 87 other keys to match. The method is much the same for most other instruments. Whether you're tuning a clarinet or guitar, simply play a concert A and adjust your instrument accordingly

Explanation:

It can be a bit tricky to hold a tuning fork while manipulating an instrument, which is why some musicians decide to clench the base of a ringing tuning fork in their teeth. This has the unique effect of transmitting sound through your bones, allowing your brain to "hear" the tone through your jaw. According to some urban legends, touching your teeth with a vibrating tuning fork is enough to make them explode. It's a myth, obviously, but if you have a cavity or a chipped tooth, you'll quickly find this method to be unbelievably painful.

Luckily, you can also buy tuning forks that come mounted on top of a resonator, a hollow wooden box designed to amplify a tuning fork's vibrations. In 1860, a pair of German inventors even devised a battery-powered tuning fork that musicians didn't need to ring again and again

6 0
2 years ago
A 3 N force pushes on a object for 20 meters. Find the work done
astraxan [27]
W = Fx

w = 3.20 = 60 N.m
6 0
3 years ago
What makes the north star, polaris, special?
Mama L [17]
It's the only star in the sky (visible from the northern hemisphere) that never seems to move. It stays at almost exactly the same point in the sky, while the other stars all circle around it once a day.
7 0
3 years ago
A 57-gram tennis ball moving at 70 miles per hour is hit with a 110-gram tennis racquet moving at 60 miles per hour. Which state
Whitepunk [10]

Answer:

D. Because mass and energy are both conserved, the total amounts of mass and energy are the same before and after impact.

Explanation:

As we know that, the energy in motion is Kinetic Energy mathematically given as:

KE=\frac{1}{2} m.v^2

<u>Now, according to the law of conservation of energy:</u>

\frac{1}{2} m_1.u_1^2+\frac{1}{2} m_2.u_2^2=\frac{1}{2} m_1.v_1^2+\frac{1}{2} m_2.v_2^2

where:

m_1\ \&\ m_2 mass of racquet and ball respectively and u_1\ \&\ u_2 are their respective initial velocities.

v_1\ \&\ v_2 are the respective final velocities.

<u>Also the law of conservation of momentum is applicable in this case:</u>

m_1.u_1+ m_2.u_2= m_1.v_1+m_2.v_2

In this case the velocity of the lighter mass will get increases in the final condition.

5 0
2 years ago
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