Answer: 3.21 N


For weight, we will multiply by 

Hence, the rock would weigh 3.21 N.
Answer:
30 miliAmps
Explanation:
Step 1:
Obtaining an expression to solve the question. This is illustrated below:
From ohm's law,
V = IR
Were:
V is the voltage.
I is the current.
R is resistance.
From the question given, we were told that the resistance is constant. Therefore the above equation can be written as shown below:
V = IR
V/I = constant
V1/I1 = V2/I2
V1 is initial voltage.
V2 final voltage.
I1 is initial current.
I2 final current.
Step 2:
Data obtained from the question. This include the following:
Initial voltage (V1) = V
Initial current (I1) = 60 miliAmps
Final voltage (V2) = one-half of the original voltage = 1/2V = V/2
Final current (I2) =..?
Step 3:
Determination of the new current. This can be obtained as follow:
V1/I1 = V2/I2
V/60 = (V/2) / I2
Cross multiply to express in linear form
V x I2 = V/2 x 60
V x I2 = V x 30
Divide both side by V
I2 = (V x 30)/V
I2 = 30mA.
Therefore, the new current is 30miliAmps
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Answer:
anything that takes up space.
Answer:
The initial temperature of the hot water is
(assuming that no heat was lost to the surroundings.)
Explanation:
Let
denote the mass of the hot water.
The question states that the mass of the water at
is three times the mass of the hot water. If the mass of the hot water is
, the mass of the cold water would be
.
Let
denote the specific heat capacity of water. Let
denote the mass of some water. The energy required to change the temperature of that much water by
(without state change) would be:
.
The temperature change for the cold water was:
.
Energy required to raise the temperature of water with mass
from
to
:
.
On the other hand, if the initial temperature of the hot water is
(where
,) the temperature change would be:
.
Calculate the energy change involved:
.
If no energy was lost to the surroundings,
should be equal to
. That is:
.
Simplify and solve for
:
.
.
Therefore, the initial temperature of the hot water would be
.