The final temperature of the seawater-deck system is 990°C.
<h3>What is heat?</h3>
The increment in temperature adds up the thermal energy into the object. This energy is Heat energy.
The deck of a small ship reaches a temperature Ti= 48.17°C seawater on the deck to cool it down. During the cooling, heat Q =3,710,000 J are transferred to the seawater from the deck. Specific heat of seawater= 3,930 J/kg°C.
Suppose for 1 kg of sea water, the heat transferred from the system is given by
3,710,000 = 1 x 3,930 x (T - 48.17)
T = 990°C to the nearest tenth.
The final temperature of the seawater-deck system is 990°C.
Learn more about heat.
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Its larger and if u where wondering to positive ions are smaller
If the kinetic energy of each ball is equal to that of the other,
then
(1/2) (mass of ppb) (speed of ppb)² = (1/2) (mass of gb) (speed of gb)²
Multiply each side by 2:
(mass of ppb) (speed of ppb)² = (mass of gb) (speed of gb)²
Divide each side by (mass of gb) and by (speed of ppb)² :
(mass of ppb)/(mass of gb) = (speed of gb)²/(speed of ppb)²
Take square root of each side:
√ (ratio of their masses) = ( 1 / ratio of their speeds)²
By trying to do this perfectly rigorously and elegantly, I'm also
using up a lot of space and guaranteeing that nobody will be
able to follow what I have written. Let's just come in from the
cold, and say it the clear, easy way:
If their kinetic energies are equal, then the product of each
mass and its speed² must be the same number.
If one ball has less mass than the other one, then the speed²
of the lighter one must be greater than the speed² of the heavier
one, in order to keep the products equal.
The pingpong ball is moving faster than the golf ball.
The directions of their motions are irrelevant.
Answer:
For H2O, there is one atom of oxygen and two atoms of hydrogen. A molecule can be made of only one type of atom. In its stable molecular form, oxygen exists as two atoms and is written O2. to distinguish it from an atom of oxygen O, or ozone, a molecule of three oxygen atoms, O3.
Explanation:
Answer:
The momentum of the photon is 1.707 x 10⁻²² kg.m/s
Explanation:
Given;
kinetic of electron, K.E = 100 keV = 100,000 eV = 100,000 x 1.6 x 10⁻¹⁹ J = 1.6 x 10⁻¹⁴ J
Kinetic energy is given as;
K.E = ¹/₂mv²
where;
v is speed of the electron
![K.E = \frac{1}{2}mv^2\\\\mv^2 = 2K.E \\\\v^2 = \frac{2K.E}{m} \\\\v = \sqrt{\frac{2K.E}{m}} \\\\but \ momentum ,P = mv\\\\(v)m = (\sqrt{\frac{2K.E}{m}})m\\\\P_{photon} = (\sqrt{\frac{2K.E}{m_e}})m_e\\\\P_{photon} = (\sqrt{\frac{2\times 1.6\times 10^{-14}}{9.11\times10^{-31}}})(9.11\times 10^{-31})\\\\P_{photon} = 1.707 \times 10^{-22} \ kg.m/s](https://tex.z-dn.net/?f=K.E%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%5C%5C%5C%5Cmv%5E2%20%3D%202K.E%20%5C%5C%5C%5Cv%5E2%20%3D%20%5Cfrac%7B2K.E%7D%7Bm%7D%20%5C%5C%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B2K.E%7D%7Bm%7D%7D%20%5C%5C%5C%5Cbut%20%5C%20momentum%20%2CP%20%3D%20mv%5C%5C%5C%5C%28v%29m%20%3D%20%28%5Csqrt%7B%5Cfrac%7B2K.E%7D%7Bm%7D%7D%29m%5C%5C%5C%5CP_%7Bphoton%7D%20%3D%20%20%28%5Csqrt%7B%5Cfrac%7B2K.E%7D%7Bm_e%7D%7D%29m_e%5C%5C%5C%5CP_%7Bphoton%7D%20%3D%20%20%28%5Csqrt%7B%5Cfrac%7B2%5Ctimes%201.6%5Ctimes%2010%5E%7B-14%7D%7D%7B9.11%5Ctimes10%5E%7B-31%7D%7D%7D%29%289.11%5Ctimes%2010%5E%7B-31%7D%29%5C%5C%5C%5CP_%7Bphoton%7D%20%3D%201.707%20%5Ctimes%2010%5E%7B-22%7D%20%5C%20kg.m%2Fs)
Therefore, the momentum of the photon is 1.707 x 10⁻²² kg.m/s