Answer:
The specific heat capacity can be defined as the amount of heat required to raise the temperature of 1 unit of mass by 1 unit temperature. The specific heat capacity of water is 4.186 joule/gram °C which is higher than common substances. The land has lower specific heat capacity. Thus, the land gets hot quickly than water.
This results in warming up air near the land which creates a difference in pressure across the coastal region. Sea breeze blows from sea towards landmass. Opposite happens at night, when water is still warm and land gets cooled down quickly. Then land breeze blows from landmass towards the sea. This breeze maintains a moderate temperature and windy and humid weather in the coastal regions.
We don't know how many of ANY color are in the bag right now, so there's no way to calculate an answer.
What Tom has to do is make sure that the number of marbles that are NOT blue is NINE TIMES the number of blue ones in the bag.
Answer:
In order to measure the amount of solution added in or drained out, the burette must be observed at eye level straight to the bottom of the meniscus. The liquid in the burette should be completely free of bubbles to ensure accurate measurements.
Answer:
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Explanation:
For this problem let's use Newton's second law applied to each body
Body A
X axis
T = m_A a
Axis y
N- W_A = 0
Body B
Vertical axis
W_B - T = m_B a
In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension
We write the equations
T = m_A a
W_B –T = M_B a
We solve this system of equations
m_B g = (m_A + m_B) a
a = m_B / (m_A + m_B) g
In this initial case
m_A = M
m_B = M
a = M / (1 + 1) M g
a = ½ g
Let's find the tension
T = m_A a
T = M ½ g
T = ½ M g
Now we change the mass of the second block
m_B = 2M
a = 2M / (1 + 2) M g
a = 2/3 g
We seek tension for this case
T’= m_A a
T’= M 2/3 g
Let's look for the relationship between the tensions of the two cases
T’/ T = 2/3 M g / (½ M g)
T’/ T = 4/3
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
