Answer:
= 85.7 ° C
Explanation:
For this exercise we will use the calorimetry heat ratios, let's start with the heat lost by the evaporation of coffee, since it changes from liquid to vapor state
Q₁ = m L
Where m is the evaporated mass (m = 2.00 103-3kg) and L is 2.26 106 J / kg, where we use the latent heat of the water
Q₁ = 2.00 10⁻³ 2.26 10⁶
Q1 = 4.52 10³ J
Now the heat of coffee in the cup, which does not change state is
Q coffee = M
(
-
)
Since the only form of energy transfer is terminated, the heat transferred is equal to the evaporated heat
Qc = - Q₁
M ce (
-
) = - Q₁
The coffee dough left in the cup after evaporation is
M = 250 -2 = 248 g = 0.248 kg
-Ti = -Q1 / M
= Ti - Q1 / M 
Since coffee is essentially water, let's use the specific heat of water,
= 4186 J / kg ºC
Let's calculate
= 90.0 - 4.52 103 / (0.248 4.186 103)
= 90- 4.35
= 85.65 ° C
= 85.7 ° C
Answer:
The z-component of the force is
Explanation:
From the question we are told that
The charge on the particle is
The magnitude of the magnetic field is 
The velocity of the particle toward the x-direction is 
The velocity of the particle toward the y-direction is

The velocity of the particle toward the z-direction is

Generally the force on this particle is mathematically represented as

So we have

substituting values
So the z-component of the force is
Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).
Answer:
The correct option is (c).
Explanation:
Given that,
The energy of a photon is, 
We need to tell the color of this light. We know that, the energy of a photon is given by :

Where
c is the speed of light

The wavelength of yellow light is approx 580 nm. Hence, we can say that this photon corresponds to yellow light.
The change in temperature here corresponds to a sensible heat. The amount of energy required can be calculated by multiplying the specific heat capacity, the amount of the substance and the corresponding change in temperature.
Heat required = mCΔT
Heat required = 0.368 kg (0.0920 cal/g°C) (60 - 23)°C
Heat required = 1.25 cal