I need help with question #8 please!!

Three cars (car F, car G, and car H) are moving with the same speed and slam on their brakes. The most massive car is car F, and

Crazy boy [7]

To solve this problem it is necessary to apply the concepts related to Normal Force, frictional force, kinematic equations of motion and Newton's second law.

From the kinematic equations of motion we know that the relationship of acceleration, velocity and distance is given by

$v_f^2=v_i^2+2ax$

Where,

$v_f =$ Final velocity

$v_i =$ Initial Velocity

a = Acceleration

x = Displacement

Acceleration can be expressed in terms of the drag coefficient by means of

$F_f = \mu_k (mg) \rightarrow$ Frictional Force

$F = ma \rightarrow$ Force by Newton's second Law

Where,

m = mass

a= acceleration

$\mu_k =$ Kinetic frictional coefficient

g = Gravity

Equating both equation we have that

$F_f = F$

$\mu_k mg=ma$

$a = \mu_k g$

Therefore,

$v_f^2=v_i^2+2ax$

$0=v_i^2+2(\mu_k g)x$

Re-arrange to find x,

$x = \frac{v_i^2}{2(-\mu_k g)}$

The distance traveled by the car depends on the coefficient of kinetic friction, acceleration due to gravity and initial velocity, therefore the three cars will stop at the same distance.

3 0

2 years ago

Four astronauts are in a spherical space station. (a) If, as is typical, each of them breathes about 500 cm

mario62 [17]

Answer:

A) if each astronaut breathes about 500 cm³, the total volume of air breathed in a year is 14716.8m³.

B) The Diameter of this spherical space station should be 30.4m

Explanation:

The breathing frequency (according to Rochester encyclopedia) is about 12-16 breath per minute. if we take the mean value (14 breath per minute), we can estimate the total breaths of a person along a year:

$f_b=14\frac{br}{min} \cdot \frac{60min}{1hr} \frac{24hr}{1day}\frac{365day}{1year}=29433600\frac{br}{year}$

If we multiply this for the number of people in the station and the volume each breath needs, we obtain the volume breathed in a year.

The volume of a sphere is:

$V_{sph}=\frac{4\pi}{3}r^3$

So the diameter is:

$D=2r=2\sqrt[3]{\frac{3V_{sph}}{4\pi}} =30.4m$

6 0

2 years ago

A volumen constante un gas ejerce una presión de 880 mmHg a 20o Celsius dentro de una olla a presión ¿Qué temperatura habrá si e

jasenka [17]

Answer: Hence, the final temperature is 350 K

Explanation :

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=880mmHg\\T_1=20^0C=(20+273)K=293K\\P_2=1050mmHg\\T_2=?$

Putting values in above equation, we get:

$\frac{880mmHg}{293K}=\frac{1050mmHg}{T_2}\\\\T_2=350K$

Hence, the final temperature is 350 K

8 0

2 years ago

Choose all the answers that apply.

Serhud [2]

<span>According to Newton's first law of motion:

-- objects at rest will remain at rest unless acted upon by an outside force

-- objects in motion will remain in motion unless acted upon by an outside force

</span>

8 0

2 years ago

What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?

alukav5142 [94]

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

R1 = 633

RT = 205

Required

Determine R2

Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

1/R2 = 1/205 - 1/633

Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

R2 = 303 --- approximated

5 0

2 years ago