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aksik [14]
3 years ago
7

Suppose a proton ( = 1. 67×10^−27 kg) is confined to a box of width = 1. 00×10^−14 m (a typical nuclear radius).

Physics
1 answer:
dimulka [17.4K]3 years ago
3 0

Answer:

22e837281949222324

Explanation:

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An insulated beaker with negligible mass contains 0.250 kg of water at 75.0C. How many kilograms of ice at -20.0C must be droppe
kkurt [141]

Answer:

The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg

Explanation:

Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water

Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C

To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.

Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C

Latent heat of ice = L = 334000 J/kg

Specific heat capacity of water = C = 4186 J/kg.°C

Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m

Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J

543600 m = 36627.5

m = 0.0674 kg = 67.4 g of ice.

3 0
3 years ago
Round your answers to one decimal place.this parallel circuit has two resistors at 15 and 40 ohms. what is the total resistance?
lubasha [3.4K]
1) The equivalent resistance of two resistors in parallel is given by:
\frac{1}{R_{eq}}= \frac{1}{R_1}+ \frac{1}{R_2}
so in our problem we have
\frac{1}{R_{eq}} =  \frac{1}{15 \Omega}+ \frac{1}{40 \Omega}=0.092 \Omega^{-1}
and the equivalent resistance is
R_{eq} =  \frac{1}{0.092 \Omega^{-1}}=10.9 \Omega

2) If we have a battery of 12 V connected to the circuit, the current in the circuit will be given by Ohm's law, therefore:
I= \frac{V}{R_{eq}}= \frac{12 V}{10.9 \Omega}=1.1 A
4 0
3 years ago
Read 2 more answers
Identify the three types of isometric transformations. A. reflection, rotation, translation B. reflection, rotation, dilation C.
inessss [21]
<span>reflection, rotation, translation</span>
4 0
3 years ago
IN WHAT SEQUENCE DOES MAGMA MOVE THROUGH A VOLCANO?
Natalka [10]
Well the volcanos come from the lower crust and moves up with changes the climate
6 0
4 years ago
A positively charged metal sphere, A, is held close to but not touching and identical uncharged sphere, sphere B. Sphere A is no
Yuri [45]

Answer:

The sphere C carries no net charge.

Explanation:

  • When brougth close to the charged sphere A, as charges can move freely in  a conductor, a charge equal and opposite to the one on the sphere A, appears on the sphere B surface facing to the sphere A.
  • As sphere B must remain neutral (due to the principle of conservation of charge) an equal charge, but of opposite sign, goes to the surface also, on the opposite part of the sphere.
  • If sphere A is removed, a charge movement happens in the sphere B, in such a way, that no net charge remains on the surface.
  • If in such state, if  the sphere B (assumed again uncharged completely, without any local charges on the surface), is touched by an initially uncharged sphere C, due to the conservation of  charge principle, no net  charge can be built on sphere C.
3 0
3 years ago
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