Answer:
232 J/K
Explanation:
The amount of heat gained by the air = the amount of heat lost by the tea.
q_air = -q_tea
q = -mCΔT
q = -(0.250 kg) (4184 J/kg/ºC) (20.0ºC − 85.0ºC)
q = 68,000 J
The change in entropy is:
dS = dQ/T
Since the room temperature is constant (isothermal):
ΔS = ΔQ/T
Plug in values (remember to use absolute temperature):
ΔS = (68,000 J) / (293 K)
ΔS = 232 J/K
Answer:
(2) −1 e
Explanation:
A quark is the lightest elementary particles which form hadron such as proton and neutron. A quark has fractional charge.
Up, charm and top quarks have
charge where as down, strange and bottom quarks have
charge.
The antiparticle of up quark is antiup quark and has charge
charge.
The antiparticle of down quark is antidown quark and has charge
charge.
An antibaryon is composed of two anti-up quark and one anti-down quark.
Net charge of the anti-baryon is:
Thus, antibaryon has -1e charge.
Answer: <u>Option A: </u>The gas and food are examples of energy.
Explanation:
Work and energy are inter-related. Energy is required to do the work. An equal amount of energy is converted into work. Work and energy have same units. The SI unit is Joule.
The gas and food are sources of energy. They act as fuel. This energy is utilized to perform work. The gas is used to run the car. The food is metabolized inside the body which is a source of energy and utilized to perform every day work.
Hence, The gas and food are examples of energy.
In atoms, electrons surround the nucleus in specific energy levels.
Answer: Option B
<u>Explanation:
</u>
Atom, actually considered as the tiny part that is ever present in this universe. Many theories and experiments were conducted, for studying what was present inside of an atom, and many theories came into light.
And finally, Bohr-Sommerfeld Theory stated the final conclusion, that the positive charge is at the centre of the nucleus, and all the electrons revolve in their specific energy levels. If an atom wants to move to lower energy state, it should emit energy, whereas for going to higher state, it should gain energy.
Answer:
xf = 5.68 × 10³ m
yf = 8.57 × 10³ m
Explanation:
given data
vi = 290 m/s
θ = 57.0°
t = 36.0 s
solution
firsa we get here origin (0,0) to where the shell is launched
xi = 0 yi = 0
xf = ? yf = ?
vxi = vicosθ vyi = visinθ
ax = 0 ay = −9.8 m/s
now we solve x motion: that is
xf = xi + vxi × t + 0.5 × ax × t² ............1
simplfy it we get
xf = 0 + vicosθ × t + 0
put here value and we get
xf = 0 + (290 m/s) cos(57) (36.0 s)
xf = 5.68 × 10³ m
and
now we solve for y motion: that is
yf = yi + vyi × t + 0.5 × ay × t
² ............2
put here value and we get
yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s) ²
yf = 8.57 × 10³ m