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3 years ago

13

The speed of a car is increased uniformly from 11 meters per second to 19 meters per second. The average speed of the car during

this interval is

1 answer:

dsp733 years ago

6 0

15 m/s average is (11+19)/2

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You want to raise the temperature of 22kg of water from 60°C to 95°C. If electricity costs $0.08/kWh, how much would it cost you

Sladkaya [172]

Answer:

dgtgh

Explanation:

dgdfhdhfffff thdhfghjfj e

3 0

3 years ago

Calculate the standard electrode potential difference (e°) of the daniell cell (at 1 bar) if temperature is 473.15 k.

anzhelika [568]

Missing data in the text of the exercise: The molar concentration of Zinc is 10 times the molar concentration of copper.

Solution:

1) First of all, let's calculate the standard electrode potential difference at standard temperature. This is given by:
$E^0=E_{cat}^0-E_{an}^0$
where $E_{cat}^0$ is the standard potential at the cathode, while $E_{an}^0$ is the standard potential at the anode. For a Daniel Cell, at the cathode we have copper: $E_{Cu}^0=+0.34 V$ , while at the anode we have zinc: $E_{Zn}^0=-0.76 V$ . Therefore, at standard temperature the electrode potential difference of the Daniel Cell is
$E^0=+0.34 V-(-0.76 V)=+1.1 V$

2) To calculate $E^0$ at any temperature T, we should use Nerst equation:
$E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}$
where
$R=8.31 J/(K mol)$
$T=473.15 K$ is the temperature in our problem
$z=2$ is the number of electrons transferred in the cell's reaction
$F=9.65\cdot 10^4 C/mol$ is the Faraday's constant
$[Zn]$ and $[Cu]$ are the molar concentrations of zinc and in copper, and in our problem we have $[Zn]=10[Cu]$ .
Using all these data inside the equation, and using $E^0=+1.1 V$ , in the end we find:
$E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}=+1.053 V$

8 0

3 years ago

How can you describe the motion of an object in a race?

Hitman42 [59]

You can describe the motion of an object by its position, speed, direction, and acceleration

4 0

3 years ago

On a distant planet, a rock falls in 48.4 s from the top of a 1.10e+02 m cliff to the planet surface below. What is the accelera

pav-90 [236]

this can be solve using the formala of free fall

t = sqrt( 2y/ g)

where t is the time of fall

y is the height

g is the acceleration due to gravity

48.4 s = sqrt (2 (1.10e+02 m)/ g)

G = 0.0930 m/s2

The velocity at impact

V = sqrt(2gy)

= sqrt( 2 ( 0.0930 m/s2)( 1.10e+02 m)

V = 4.523 m/s

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8 0

4 years ago

A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co

algol [13]

Answer:

$\displaystyle X_T=66.6\ km$

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

$\displaystyle V_f=V_o+a\ t$

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of $0.987 m/s^2$ for 182 seconds. Then it brakes with a constant acceleration of $-0.321 m/s^2$ until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

$\displaystyle X=V_o\ t+\frac{a\ t^2}{2}$

Our data is

$\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec$

Let's compute the first distance X1

$\displaystyle X_1=0+\frac{0.987\times 182^2}{2}$

$\displaystyle X_1=16,346.7\ m$

Now, we find the speed at the end of the first period of time

$\displaystyle V_{f1}=0+0.987\times 182$

$\displaystyle V_{f1}=179.6\ m/s$

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

$\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0$

$\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}$

$\displaystyle t=559.5\ sec$

Computing the second distance

$\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}$

$\displaystyle X_2=50,243.2\ m$

The total distance is

$\displaystyle X_t=x_1+x_2=16,346.7+50,243.2$

$\displaystyle X_t=66,589.9\ m$

$\displaystyle \boxed{X_T=66.6\ km}$

8 0

4 years ago