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photoshop1234 [79]
3 years ago
9

Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resi

stance. But driving slowly in dry sand is another story. If a 1500 kg car is driven in sand at 4.9 m/s , the coefficient of rolling friction is 0.060. In this case, nearly all of the energy that the car uses to move goes to overcoming rolling friction, so you can ignore air drag in this problem.
Required:
a. What propulsion force is needed to keep the car moving forward at a constant speed?
b. What power is required for propulsion at 5.0 m/s?
c. If the car gets 15 mpg when driving on sand, what is the car's efficiency? One gasoline contains 1.4×10 ^8 J of chemical energy.
Physics
1 answer:
trapecia [35]3 years ago
7 0

Answer:

a)  F_p=882N

b)  P=4410W

c)  V_p'=24135 ,n=15.2\%

Explanation:

From the question we are told that:

Mass M=1500kg

Velocity v=4.9m/s

Coefficient of Rolling Friction \mu=0.06

a)

Generally the equation for The Propulsion Force is mathematically given by

 F_p=\mu*mg

 F_p=0.06*1500*9.81

 F_p=882N

b)

Therefore Power Required at

 V_p=5.0m/s

 P=F_p*V_p

 P=882*5

 P=4410W

c)

 V_p' =15mpg

 V_p'=15*\frac{1609}

 V_p'=24135

Generally the equation for Work-done is mathematically given by

 W=F_p*V_p'

 W=882*15*1609

 W=2.13*10^7

Therefore

Efficiency

 n=\frac{W}{E}*100\%

Since

Energy in one gallon of gas is

 E=1.4*10^8J

Therefore

 n=\frac{2.1*10^7}{1.4*10^8}*100\%

 n=15.2\%

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STatiana [176]

Answer:

See explanation for step by step explanations.

Explanation:

Let X be the mass of a randomly selected resistor

The probability density function is given as

, 80<x<120

=0, otherwise

a) By using the given pdf we have to find the probability of the resistors having resistance less than 90

=

=

=0.0625

the probability of the resistors having resistance less than 90 is 0.0625

Step 2 of 4<

/p>

b) we have to find the mean resistance

Mean=

=

=

=106.67

The mean resistance is 106.67

5 0
3 years ago
Give an example of when your average speed would be higher than instantaneous speed
brilliants [131]

When I drive to the office, I drive through two school zones,
and four intersections that are controlled by traffic lights.

My average speed for the trip is higher than my instantaneous
speed is at any point in the school zones, or at any time when
I'm waiting for a red light to change.

3 0
3 years ago
A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is . The battery is rem
Viktor [21]

Answer:

A

The energy dissipated in the resistor {U_k} = \frac{U}{k}

B

The energy dissipated in the resistor{U_k} = kU

Explanation:

In order to gain a good understanding of the solution above it is necessary to understand that the concept required to solve the question is energy stored in the parallel plate capacitor.

Initially, take the first case. In that, according to the formula for energy stored in parallel plate capacitor with the dielectric inserted between the two plates, find the energy stored. Then, find the energy stored in the parallel plate capacitor when no dielectric is present. Then, write the equation of energy stored in the capacitor with the dielectric present in the form of the energy stored in the capacitor without the dielectric present. The equation must not be in the form of voltage as battery is removed in this case.

For part B, use the equation of the energy dissipated in the resistor. Write it in the form of the equation for energy stored in the parallel plate capacitor without dielectric in it. The equation must be in the form of voltage as battery is kept connected. Looking at the fundamentals

The energy stored in the parallel plate capacitor with the dielectric is given by,

                 U _k = \frac{1}{2} \frac{q ^2}{kC}

Here, the energy stored in the capacitor will be equal to the energy dissipated in the resistor. In this equation, Uk is the energy dissipated in the resistor, q is charge, k is the dielectric constant, and C is the capacitance.

Now, the equation of the energy stored in the parallel plate capacitor without dielectric is,

​ U= \frac{1}{2} \frac{q ^2}{C}

In this equation, U is the energy stored in the parallel plate capacitor without dielectric, q is charge, and C is the capacitance.

For part B, the battery is still connected. Thus, the equation q = CV is used to modify the above equation.

Thus, the energy stored in the parallel plate capacitor with the dielectric is given by,

U_ k = \frac{1}{2} \frac{k ^{2} C^ 2 V ^2}{kC} \\\\= \frac{1}{2}  kCV ^2

In this equation, Uk is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

The equation of the energy stored in the parallel plate capacitor without dielectric is,

U= \frac{1}{2} \frac{C^ 2 V ^2}{C} \\\\= \frac{1}{2} CV ^2

In this equation, U is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

(A)

The equation for energy dissipated in the resistor is,

 U _k = \frac{1}{2} \frac{q ^2}{kC}

Substitute U = \frac{1}{2}\frac{{{q^2}}}{C}  in the equation of {U_k}

U _k = \frac{1}{2} (\frac{1}{k} )\frac{q ^2}{C} \\\\= (\frac{1}{k} ) \frac{q^2}{C}\\\\ U_{k} = \frac{U}{k}

Note :

If the resistance relates to the capacitor, the energy stored in the capacitor is dissipated through the resistance. Thus, by substituting the equation of U, the expression is found out.

(B)

The equation for energy dissipated in the resistor is

U_{k} = \frac{1}{2}kCV^2

Here, V is voltage in the circuit.

Substitute U =\frac{1}{2} CV^2 in the equation of {U_k}

So,

        U_{k} = \frac{1}{2} kCV^2\\

       = k(\frac{1}{2} CV^2)

       U_{k} = kU

4 0
3 years ago
A current of 0.92 a flows in a wire. how many electrons are flowing past any point in the wire per second? the charge on one ele
Fantom [35]
The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, \Delta t:
I= \frac{Q}{\Delta t}
First we need to find the net charge flowing at a certain point of the wire in one second, \Delta t=1.0 s. Using I=0.92 A and re-arranging the previous equation, we find
Q=I \Delta t= (0.92 A)(1.0 s)=0.92 C

Now we know that each electron carries a charge of e=1.6 \cdot 10^{-19} C, so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
N= \frac{Q}{q} = \frac{0.92 C}{1.6 \cdot 10^{-19} C}=5.75 \cdot 10^{18}
3 0
3 years ago
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4 0
3 years ago
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