Anyone good at chemistry ? Maybe you can help me out?

By what factor will the kinetic energy change if the speed of the baseball is decreased to 54.8 mi/h? Express your answer as an

Serggg [28]

Answer:

The Kinetic Energy is approximately 3 times decreased

Explanation:

A baseball weighs 5.13 oz.

a)What is the kinetic energy, in joules, of this baseball when it is thrown by a major league pitcher at 95.o mi/h?

b) By what factor will the kinetic energy change if the speed of the baseball is decreased to 54.8 mi/h? Express your answer as an integer.

Kinetic Energy (KE)=0.5×mass×velocity ^ 2

Kinetic Energy (KE)=0.5×mass × velocity ^ 2

Joules = kg×m^2/s^2

1 mile = 1609.344 meters

1 hour = 3600 sec

1 Oz = 28.34952 g = 0.02834952 kg

a) KE=0.5×m×v^2

=0.5×(5.13 oz × 0.02834952 kg/1 ounce)×(95 miles/h × 1609.344 m/1 mile × 1 hr/3600 s)^2

=130.761 kg×m^2/s^2 = 130.761 Joules

b) KE=0.5×m×v^2

=0.5×(5.13 oz × 0.02834952 kg/1 ounce)×(54.8 miles/h × 1609.344 m/1 mile × 1 hr/3600 s)^2

=43.51028 kg×m^2/s^2 = 43.51028 Joules

= 130.761 / 43.51028 = 3.00528,

As such the Kinetic Energy is approximately 3 times decreased

4 0

3 years ago

A lithium flame has a characteristic red color due to emissions of wavelength 671 nm. What is the mass equivalence of 1 mol of p

e-lub [12.9K]

Answer:

1.98x10⁻¹² kg

Explanation:

The energy of a photon is given by:

E= hc/λ

h is Planck's constant, 6.626x10⁻³⁴ J·s

c is the speed of light, 3x10⁸ m/s

and λ is the wavelenght, 671 nm (or 6.71x10⁻⁷m)

E = 6.626x10⁻³⁴ J·s * 3x10⁸ m/s ÷ 6.71x10⁻⁷m = 2.96x10⁻¹⁹ J

Now we multiply that value by Avogadro's number, to calculate the energy of 1 mol of such protons:

1 mol = 6.023x10²³ photons

2.96x10⁻¹⁹ J * 6.023x10²³ = 1.78x10⁵ J

Finally we calculate the mass equivalence using the equation:

E=m*c²

m=E/c²

m = 1.78x10⁵ J / (3x10⁸ m/s)² = 1.98x10⁻¹² kg

6 0

3 years ago

Igneous rock___to form magma a)melts b)weathers c)compacts d)crystallizes NEED HELP ASAP

dangina [55]

Answer:

should be A)Melts.Because magma is liquid i think

5 0

3 years ago

Read 2 more answers

A magnesium (Mg) atom gives one electron to two atoms of another element, then it takes on a 2+ charge. The two atoms of the oth

Setler [38]

Answer:

See below

Explanation:

You missed the following elements, but any element with 7 valence electrons could behave like that: Cl, Br, I, F, etc

3 0

3 years ago

Read 2 more answers

Alice and Bob are experimenting with two moles of neon, a monatomic gas, that starts out at conditions of standard temperature a

Archy [21]

Answer:

29273.178 joules have been added to the gas for the entire process.

Explanation:

The specific heats of monoatomic gases, measured in joules per mol-Kelvin, are represented by the following expressions:

Isochoric (Constant volume)

$c_{v} = \frac{3}{2}\cdot R_{u}$ (1)

Isobaric (Constant pressure)

$c_{p} = \frac{5}{2}\cdot R_{u}$ (2)

Where $R_{u}$ is the ideal gas constant, measured in pascal-cubic meters per mol-Kelvin.

Under the assumption of ideal gas, we notice the following relationships:

1) Temperature is directly proportional to pressure.

2) Temperature is directly proportional to volume.

Now we proceed to find all required temperatures below:

(i) Alice heats the gas at constant volume until its pressure is doubled:

$\frac{T_{2}}{T_{1}} = \frac{P_{2}}{P_{1}}$ (3)

( $\frac{P_{2}}{P_{1}} = 2$ , $T_{1} = 273.15\,K$ )

$T_{2} = \frac{P_{2}}{P_{1}} \times T_{1}$

$T_{2} = 2\times 273.15\,K$

$T_{2} = 546.3\,K$

(ii) Bob further heats the gas at constant pressure until its volume is doubled:

$\frac{T_{3}}{T_{2}} =\frac{V_{3}}{V_{2}}$ (4)

( $\frac{V_{3}}{V_{2}} = 2$ , $T_{2} = 546.3\,K$ )

$T_{3} = \frac{V_{3}}{V_{2}}\times T_{2}$

$T_{3} = 2\times 546.3\,K$

$T_{3} = 1092.6\,K$

Finally, the heat added to the gas ( $Q$ ), measured in joules, for the entire process is:

$Q = n\cdot [c_{v}\cdot (T_{2}-T_{1})+c_{p}\cdot (T_{3}-T_{2})]$ (5)

If we know that $R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}$ , $n = 2\,mol$ , $T_{1} = 273.15\,K$ , $T_{2} = 546.3\,K$ and $T_{3} = 1092.6\,K$ , the heat added to the gas for the entire process is:

$c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{v} = 12.471\,\frac{J}{mol\cdot K}$

$c_{p} = \frac{5}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{p} = 20.785\,\frac{J}{mol\cdot K}$

$Q = (2\,mol)\cdot \left[\left(12.471\,\frac{J}{mol\cdot K} \right)\cdot (536.3\,K-273.15\,K)+\left(20.785\,\frac{J}{mol\cdot K} \right)\cdot (1092.6\,K-546.3\,K )\right]$

$Q = 29273.178\,J$

29273.178 joules have been added to the gas for the entire process.

6 0

2 years ago