Answer:
Q = 12540 J
Explanation:
It is given that,
Mass of water, m = 50 mL = 50 g
It is heated from 0 degrees Celsius to 60 degrees Celsius.
We need to find the energy required to heat the water. The formula use to find it as follows :
Where c is the specific heat of water, c = 4.18 J/g°C
Put all the values,
So, 12540 J of energy is used to heat the water.
The sprinter’s average acceleration is 1.98 m/s²
The given parameters;
Convert the velocity of the sprinter to m/s;
The time of motion is seconds;
The sprinter’s average acceleration is calculated as follows;
Thus, the sprinter’s average acceleration is 1.98 m/s²
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Answer:Water Only
Explanation:
Given
vessel is insulated therefore no heat can be added or removed i.e. heat exchange is zero
If hot water at is mixed with cold water at
then at equilibrium vessel contains only water and final temperature of water will be between
and
Heat released by hot water is equal to heat gain by cold water .
0.125 mm . is the thickness of the sample.
<h3>What do you mean by hall voltage ?</h3>The Hall effect is the creation of a voltage difference (the Hall voltage) across an electrical conductor, which is transverse to an applied magnetic field perpendicular to the current and an electric current in the conductor. Edwin Hall made the discovery in 1879.
We need to know the material's current, magnetic field, length, number of charge carriers, and area in order to calculate the Hall voltage. The Hall voltage is computed using the formula: v=IBlneA=(100A)(1.5T)(1.0102m)(5.91028/m3)(1.61019C)(2.0105m2)=7.9106V.
lof4
First we have to plot those point Then we can use some computer program to fit those point linearly to get slope
of that graph a and interception b. We already know, from theory, that Hall's voltage AVH and magnitude of
magnetic field B are connected as
Δ
where I is current trough probe, n is concentration of charge carriers, q = 1.6 • 10¯19 C is charge of charge
carries and t is thickness of the material. We have put the data from the problem on a graph and fitted linearly and
got
a = 100 μ
b = —0.02 μV.
As we can see, our result are in agreement with theoretical assumptions because interception b is almost O, and a
is asked relation between Hall's voltage A VH and magnitude of magnetic field B. Then we can write
ΔVH = V/TB
(4) Then we can use result (4) and numbers from the textbook to calculate the thickness of the sample as
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