What is the net worth of a car moving 30 miles per hour

Please help me on 3,4 and 6. Thanks

rusak2 [61]

3. The second piston will experience the same force as compared with the first. This is because since the pressure is the same everywhere inside the fluid system, the force is proportional to the surface area. We are told that both the first and the second piston have the same surface area, therefore, they will both experience the same force/pressure.

4. The situation is much the same as number 3 above, with the exception that the second piston is twenty times larger than the first. Again, since the pressure is the same everywhere inside the fluid system, the force is proportional to the surface area. We are told that the second piston is 20 times larger than the first, therefore, the larger piston will experience 20 times larger the force of the small one.

6. The answer is TRUE. The hydraulic braking system of most cars makes use of a vacuum servo (or booster), which is located between the brake pedal and the master cylinder piston. This vacuum servo amplifies the force applied from the brake pedal.

3 0

4 years ago

A resistor with an unknown resistance is connected in parallel to a 13 Ω resistor. When both resistors are connected in parallel

larisa86 [58]

Answer:

R2 = 10.31Ω

Explanation:

For two resistors in parallel you have that the equivalent resistance is:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\\$ (1)

R1 = 13 Ω

R2 = ?

The equivalent resistance of the circuit can also be calculated by using the Ohm's law:

$I=\frac{V}{R_{eq}}\\\\R_{eq}=\frac{V}{I}$ (2)

V: emf source voltage = 23 V

I: current = 4 A

You calculate the Req by using the equation (2):

$R_{eq}=\frac{23V}{4A}=5.75\Omega$

Now, you can calculate the unknown resistor R2 by using the equation (1):

$\frac{1}{R_2}=\frac{1}{R_{eq}}-\frac{1}{R_1}\\\\R_2=\frac{R_{eq}R_1}{R_1-R_{eq}}\\\\R_2=\frac{(5.75\Omega)(13\Omega)}{13\Omega-5.75\Omega}=10.31\Omega$

hence, the resistance of the unknown resistor is 10.31Ω

8 0

3 years ago

Read 2 more answers

A car is traveling at 39.7 mi/h on a horizontal highway. The acceleration of gravity is 9.8 m/s 2 . If the coefficient of fricti

777dan777 [17]

Answer:

The minimum distance in which the car will stop is

x=167.38m

Explanation:

$39.7\frac{mi}{h}*\frac{1km}{0.621371mi}*\frac{1000m}{1km}*\frac{1h}{3600s}=17.747\frac{m}{s}$

∑F=m*a

∑F=u*m*g

The force of friction is the same value but in different direction of the force moving the car so it can stop so

$F=m*a\\a=\frac{F}{m}\\a=\frac{u*m*g}{m}\\a=u*g\\a=0.096*-9.8\frac{m}{s^{2} }$

$a=-0.9408 \frac{m}{s^{2}}$

$v_{f}^{2}=v_{o}^{2}+2*a*(x_{f}-x_{o})\\v_{f}=0 \\x_{o}=0\\0=v_{o}^{2}+2*a*x_{f}\\x_{f}=\frac{v_{o}^{2}}{2*a} \\x_{f}=\frac{(-17.747\frac{m}{s})^{2}}{2*(-0.9408)} \\x_{f}=167.38m$

4 0

3 years ago

Water is boiled at 1 atm pressure in a 25-cm-internal- diameter stainless steel pan on an electric range. if it is observed that

patriot [66]

3933 watts At 100 C (boiling point of water), it's density is 0.9584 g/cm^3. The volume of water lost is pi * 12.5^2 * 10 = 4908.738521 cm^3 The mass of water boiled off is 4908.738521 * 0.9584 = 4704.534999 grams. Rounding to 4 significant figures gives me 4705 grams of water. The heat of vaporization for water is 2257 J/g. So the total energy applied is 2257 J/g * 4705 g = 10619185 J Now we need to divide that by how many seconds we've spent boiling water. That would be 45 * 60 = 2700 seconds. Finally, the rate of heat transfer in Joules per second will be the total number of joules divided by the total number of seconds. So 10619185 J / 2700 s = 3933 J/s = 3933 (kg m^2/s^2)/s = 3933 (kg m^2/s^3) = 3933 watts

3 0

3 years ago

Which is more reliable—using a manual stop watch or using light gates? Explain.

Kitty [74]

Light gates are more reliable. When using a manual stop watch, it is difficult to stop it at an exact time. A light gate is able to detect when an object passes through a 'gate' with the infrared transmitter and receiver.

7 0

4 years ago