Answer:
2.73×10¯³⁴ m.
Explanation:
The following data were obtained from the question:
Mass (m) = 0.113 Kg
Velocity (v) = 43 m/s
Wavelength (λ) =?
Next, we shall determine the energy of the ball. This can be obtained as follow:
Mass (m) = 0.113 Kg
Velocity (v) = 43 m/s
Energy (E) =?
E = ½m²
E = ½ × 0.113 × 43²
E = 0.0565 × 1849
E = 104.4685 J
Next, we shall determine the frequency. This can be obtained as follow:
Energy (E) = 104.4685 J
Planck's constant (h) = 6.63×10¯³⁴ Js
Frequency (f) =?
E = hf
104.4685 = 6.63×10¯³⁴ × f
Divide both side by 6.63×10¯³⁴
f = 104.4685 / 6.63×10¯³⁴
f = 15.76×10³⁴ Hz
Finally, we shall determine the wavelength of the ball. This can be obtained as follow:
Velocity (v) = 43 m/s
Frequency (f) = 15.76×10³⁴ Hz
Wavelength (λ) =?
v = λf
43 = λ × 15.76×10³⁴
Divide both side by 15.76×10³⁴
λ = 43 / 15.76×10³⁴
λ = 2.73×10¯³⁴ m
Therefore, the wavelength of the ball is 2.73×10¯³⁴ m.
The cheetah can run 28,25 km
Answer:
f something happens to go wrong at a nuclear reactor, anyone living in a 10-mile radius of the plant may have to evacuate. This map also shows a 50-mile evacuation zone, the safe distance that the U.S. government recommended to Americans who were near
because
When the temperature of 0.50 kg of water decreases by 22 °C, the energy transferred to the surroundings from the water is -46.2 kJ.
A sample of 0.50 kg of water boils (reaches 100 °C). After a while, its temperature decreases by 22 °C.
We can calculate the energy transferred to the surroundings from the water in the form of heat (Q) using the following expression.
![Q = c \times m \times \Delta T = \frac{4200J}{kg.\° C} \times 0.50kg \times (-22\° C) \times \frac{1kJ}{1000J} = -46.2 kJ](https://tex.z-dn.net/?f=Q%20%3D%20c%20%5Ctimes%20m%20%5Ctimes%20%5CDelta%20T%20%3D%20%5Cfrac%7B4200J%7D%7Bkg.%5C%C2%B0%20C%7D%20%20%5Ctimes%200.50kg%20%5Ctimes%20%28-22%5C%C2%B0%20C%29%20%5Ctimes%20%5Cfrac%7B1kJ%7D%7B1000J%7D%20%3D%20-46.2%20kJ)
where,
- c: specific heat capacity of water
- m: mass of water
- ΔT: change in the temperature
When the temperature of 0.50 kg of water decreases by 22 °C, the energy transferred to the surroundings from the water is -46.2 kJ.
Learn more: brainly.com/question/16104165