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g100num [7]
2 years ago
5

How are traits influenced by the environment?

Physics
1 answer:
MAXImum [283]2 years ago
7 0

Answer:

i dont k

Explanation:

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How much work is done by the force lifting a 0.1-kilogram hamburger vertically upward at a constant velocity 0.3 meter from a ta
pychu [463]

Answer:

0.2943 Nm

Explanation:

Work done is given a the product of force and diatance moved and expressed by the formula

W=Fd

Here W represent work, F is applied force and d is perpendicular distance

Also, we know that F=mg where m is the mass of an object and g is acceleration due to gravity. Substituting this back into the initial equation then

W=mgd

Taking acceleration due to gravity as 9.81 m/s2 and substituting mass with 0.1 kg and distance with 0.3 m then

W=0.1*9.81*0.3=0.2943 Nm

5 0
3 years ago
If the atomic radius of Copper is 0.128 nm, calculate the volume of its unit cell in cubic nanometers. Round off the answer to t
hodyreva [135]

Explanation:

Below is an attachment containing the solution.

4 0
3 years ago
How are wave properties and energy related?<br> give me examples
klasskru [66]

Answer:

They don’t ‘represent’ anything, they are properties of the wave.

Depending on the type of wave, we experience them as various phenomena. For example, with a sound wave we experience frequency (or wavelength, which is just another way to describe the same property) as the pitch of the sound. We experience amplitude as the loudness of the sound, although due to the characteristics of the ear, frequency also effects perceived loudness.

If the wave is a light wave, we experience the frequency (wavelength) as the colour of the light, and the amplitude as the brightness of the light.

For many waves, we don’t perceive them at all (e.g. radio waves).

For ocean waves, frequency is the time for each peak or trough to reach us, and amplitude is how tall the wave is.

6 0
2 years ago
Un neumático sin cámara, soporta una presión de 1.5 atm cuando la temperatura ambiente es de 300°K. ¿Qué presión llegará a sopor
arlik [135]

Answer:

El neumático soportará una presión de 1.7 atm.

Explanation:

Podemos encontrar la presión final del neumático usando la ecuación del gas ideal:

PV = nRT

En donde:

P: es la presión

V: es el volumen

n: es el número de moles del gas

R: es la constante de gases ideales

T: es la temperatura

Cuando el neumático soporta la presión inicial tenemos:

P₁ = 1.5 atm

T₁ = 300 K

V_{1} = \frac{nRT_{1}}{P_{1}}  (1)  

La presión cuando T = 67 °C es:

P_{2} = \frac{nRT_{2}}{V_{2}}   (2)

Dado que V₁ = V₂  (el volumen del neumático no cambia), al introducir la ecuación (1) en la ecuación (2) podemos encontrar la presión final:

P_{2} = \frac{nRT_{2}}{V_{2}} = \frac{nRT_{2}}{\frac{nRT_{1}}{P_{1}}} = \frac{P_{1}T_{2}}{T_{1}} = \frac{1.5 atm*(67 + 273)K}{300 K} = 1.7 atm  

Por lo tanto, si en el transcurso de un viaje las ruedas alcanzan una temperatura de 67 ºC, el neumático soportará una presión de 1.7 atm.

Espero que te sea de utilidad!

4 0
3 years ago
The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal cont
ivann1987 [24]

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

c is the specific heat capacity

ΔT is the change in temperature

The heat transferred by the  object A per unit mass is given by;

Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}

But heat capacity of object B is twice that of object A

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K

Therefore, the final temperature of both objects is 400 K.

4 0
2 years ago
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