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3 years ago

A baseball has a mass of 0.45 kg and is thrown with a speed of 25 m/s. what is the momentum of the baseball?

2 answers:

amm18123 years ago

7 0

The momentum of a particle is defined as the product of its mass times its velocity. It is a vector quantity. Therefore, the momentum of the baseball would be as follows:

Momentum = mass x velocity
Momentum = 0.45 kg (25 m/s)
Momentum = 11.25 kg m/s

Ray Of Light [21]3 years ago

4 0

The answer given is correct, but it would be rounded to 11 on the site.

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lara31 [8.8K]

Answer:

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Explanation:

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3 years ago

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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:
$E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}=+1.053 V$

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3 years ago

A linear accelerator uses alternating electric fields to accelerate electrons to close to the speed of light. A small number of

kobusy [5.1K]

Answer:

8.1 x 10^13 electrons passed through the accelerator over 1.8 hours.

Explanation:

The total charge accumulated in 1.8 hours will be:

Total Charge = I x t = (-2.0 nC/s)(1.8 hrs)(3600 s/ 1 hr)

Total Charge = - 12960 nC = - 12.96 x 10^(-6) C

Since, the charge on one electron is e = - 1.6 x 10^(-19) C

Therefore, no. of electrons will be:

No. of electrons = Total Charge/Charge on one electron

No. of electrons = [- 12.96 x 10^(-6) C]/[- 1.6 x 10^(-19) C]

No. of electrons = 8.1 x 10^13 electrons

6 0

3 years ago

In the 25 ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of int

Aleonysh [2.5K]

Answer:

a) 8.33 x 10^-6 Pa

b) 8.23 x 10^-11 atm

c) 1.67 x 10^-5 Pa

d) 1.65 x 10^-10 atm

Explanation:

Intensity of the light $I$ = 2500 W/m^2

speed of light $c$ = 3 x 10^8 m/s

a) we know that the pressure for for a totally absorbing surface is given as

$P_{abs}$ = $I/c$ = 2500/(3 x 10^8) = 8.33 x 10^-6 Pa

b) 1 atm = 101325 Pa

$P_{abs}$ = (8.33 x 10^-6)/101325 = 8.23 x 10^-11 atm

c) for a totally reflecting surface

$P_{ref}$ = $2I/c$ = twice the value for totally absorbing

$P_{ref}$ = 2 x 8.33 x 10^-6 = 1.67 x 10^-5 Pa

d) 1 atm = 101325 Pa

$P_{ref}$ = 2 x 8.23 x 10^-11 = 1.65 x 10^-10 atm

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zubka84 [21]

Answer:

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