Answer:
Area=1.5(1.5)=2.25m^2
Force of gravity=10N
\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{Force}{Area}\end{gathered}
⟼Pressure=
Area
Force
\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{10}{2.25}\end{gathered}
⟼Pressure=
2.25
10
\begin{gathered}\\ \sf\longmapsto Pressure=4.4Pa\end{gathered}
⟼Pressure=4.4Pa
Answer:
dddddddddddddddddddddddddd
Explanation:
The solution to the questions are given as


- the direction of induced current will be Counterclock vise.
<h3>What is the direction of the
current induced in the loop, as viewed from above the loop.?</h3>
Given, $B(t)=(1.4 T) e^{-0.057 t}$




(b) 

c)
In conclusion, the direction of the induced current will be Counterclockwise.
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Answer:
It is not correct because the amplitude of the waves can be bigger than others and the graph can be going up and down
Explanation: I got the question right
Answer:
2.7ohms
Explanation:
Given parameters:
Voltage of the battery = 12V
Current = 4.5A
Unknown:
Resistance of the resistor = ?
Solution:
From Ohm's law, we know that;
V = IR
V is the voltage
I is the current
R is the resistance
So;
R =
=
= 2.7ohms