About how much time elapses between high tides on the same day?

The specific heat of granite is 800 J/(kg·°C). How much heat does it take to raise 1 kg of granite 4°C?

Triss [41]

Q=mcΔt
Q= 1kg * 800J/kg°C*4°C
Q= 3200J

7 0

3 years ago

A photon of wavelength 192 nm strikes an aluminum surface along a line perpendicular to the surface and releases a photoelectron

alex41 [277]

Answer:

$KE=3.529\times10^{−27}\ J$

Explanation:

Given that

Wavelength λ=192 nm

So energy of photon,E

$E=\dfrac{hC}{\lambda }$

Now by putting the values

$h=6.6\times 10^{-34}\ m^2.kg/s$

$C=3\times 10^{8}\ m/s$

$E=\dfrac{6.6\times 10^{-34}\times 3\times 10^{8}}{192\times 10^{-9} }$

$E=1.03\times 10^{-18} J$

We know that

Kinetic energy given as

$KE=\dfrac{P^2}{2m}$

$KE=\dfrac{E^2}{2mC^2}$

$KE=\dfrac{(1.03\times 10^{-18})^2}{2\times 1.67\times 10^{-27}(3\times 10^8)^2}$

$KE=3.529\times10^{−27}\ J$

5 0

3 years ago

A room is heated as a result of solar radiation coming in through the windows. Is this a heat or work interaction for the room?

taurus [48]

Answer:

who r u? I am u. i am me. no sir, you are you. wait idk.

3 0

3 years ago

A climber pulls down on a rope causing his body to rise up with the rope? Which law of motion is it?

12345 [234]

This would be called the law of action-reaction. This states that every action will have an equal and opposite reaction. The action in the example is pulling down on the rope. The opposite and equal reaction is the climers body moving upward. The same law can be applied to a rocket. The action is the engines pushing down and the reaction is the rocket going up. :D

4 0

3 years ago

What is the final temperature when a 3.0 kg gold bar at 99 0C is dropped into 0.22 kg of water at 25oC?

liq [111]

Answer:

46.9 C

Explanation:

The heat released by the gold bar is equal to the heat absorbed by the water:

$m_g C_g (T_g-T_f)=m_w C_w (T_f-T_w)$

where:

$m_g = 3.0 kg$ is the mass of the gold bar

$C_g=129 J/kg C$ is the specific heat of gold

$T_g=99 C$ is the initial temperature of the gold bar

$m_w = 0.22 kg$ is the mass of the water

$C_w=4186 J/kg C$ is the specific heat of water

$T_w=25 C$ is the initial temperature of the water

$T_f$ is the final temperature of both gold and water at equilibrium

We can re-arrange the formula and solve for T_f, so we find:

$m_g C_g T_g -m_g C_g T_f = m_w C_w T_f - m_w C_w T_w\\m_g C_g T_g +m_w C_w T_w= m_w C_w T_f +m_g C_g T_f \\T_f=\frac{m_g C_g T_g +m_w C_w T_w}{m_w C_w + m_g C_g}=\\=\frac{(3.0)(129)(99)+(0.22)(4186)(25)}{(0.22)(4186)+(3.0)(129)}=\frac{38313+23023}{921+387}=\frac{61336}{1308}=46.9 C$

5 0

3 years ago