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2 years ago

What is cheetah speed and strength?

Physics

1 answer:

pav-90 [236]2 years ago

3 0

Answer:

Speed: 109.4–120.7 km/h (68.0–75.0 mph)

Strength: Couldn't find out.

Explanation:

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What would the final volume of 40 L of gas at 80 pascals be if the pressure increases to 130 pascals?

melomori [17]

Answer:

Final volume, V2 = 24.62 L

Explanation:

Given the following data;

Initial volume = 40 L

Initial pressure = 80 Pa

Final pressure = 130 Pa

To find the final volume V2, we would use Boyles' law.

Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.

Mathematically, Boyles law is given by;

$PV = K$

$P_{1}V_{1} = P_{2}V_{2}$

Substituting into the equation, we have;

$80 * 40 = 130V_{2}$

$3200 = 130V_{2}$

$V_{2} = \frac {3200}{130}$

$V_{2} = 24.62$

Final volume, V2 = 24.62 Liters

5 0

2 years ago

All waves have wavelength, frequency, rest point, and speed. True or False?

nexus9112 [7]

Answer:

it is true

Explanation:

4 0

2 years ago

Read 2 more answers

An proton-antiproton pair is produced by a 2.20 × 10 3 MeV photon. What is the kinetic energy of the antiproton if the kinetic e

timama [110]

Answer:

K = 80.75 MeV

Explanation:

To calculate the kinetic energy of the antiproton we need to use conservation of energy:

$E_{ph} = E_{p} + E_{ap} = E_{0p} + K_{p} + E_{0ap} + K_{ap} = m_{0p}c^{2} + K_{p} + m_{0ap}c^{2} + K_{ap}$

where $E_{ph}$ : is the photon energy, $E_{0p} and E_{0ap}$ : are the rest energies of the proton and the antiproton, respectively, equals to m₀c², $K_{p} and K_{ap}$ : are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.

Therefore the kinetic energy of the antiproton is:

$K_{ap} = E_{ph} - m_{0p}c^{2} - K_{p} - m_{0ap}c^{2}$

The proton mass is equal to the antiproton mass, so:

$K_{ap} = E_{ph} - 2m_{0p}c^{2} - K_{p}$

$K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2} - 242.85MeV$

$K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2}(\frac{1eV}{1.602 \cdot 10^{-19}J})(\frac{1 MeV}{10^{6}eV}) - 242.85MeV$