The unknown substance can be lithium, which has a specific heat capacity of approximately
.
Explanation:
When heat energy is supplied to a certain substance, the temperature of the substance increases according to the equation:

where
Q is the amount of energy supplied
m is the mass of the sample
is the specific heat capacity of the substance
is the change in temperature
In this problem, we have
m = 4.9 g is the mass
Q = 668.85 J is the specific heat capacity
is the change in temperature
Solving for
, we find the specific heat capacity of the substance:

Looking at tables of specific heat capacity, we can see that the unknown substance can be lithium, which has a specific heat capacity of approximately
.
Learn more about specific heat capacity:
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Good. You can do some very interesting experiments with that equipment.
A) 0
because all of the forces cancel out, so it is not moving with balanced forces.
I'm actually going ahead in the book (DC Circuits) so this isn't really homework but I figured the tag was appropriate....the name of the chapter is Ohm's Law and Watt's Law.
<span>Problem: Calculate the power dissipated in the load resistor, R, for each of the circuits.Circuit (a): V = 10V; I = 100mA; R = ?; Since I know
V and
I use formula
P = IV: P = IV = (100mA)(10V) = 1 W.</span>
The next question is what I'm not sure about:
Question: What is the power in the circuit (a) above if the voltage is doubled? (Hint: Consider the effect on current).
What I did initially was: P = IV = (100mA)(2V) = 2 W
But then I looked at the answer and it said 4 W, then I looked at the Hint again. Then I remembered in the book early on it said "If the voltage increases across a resistor, current will increase."
So question is: When solving problems I have to increase (or decrease) current (I) every time voltage (V) is increased (decreased) in a problem, right? How about the other way around, when increasing current (I), you need to increase voltage (V). I'm pretty sure that's how they got 4 W, but want to make sure before I head to the next section of the book.
P = IV = (200mA)(2V) = 4 W