Answer:
linear density of the string = 4.46 × 10⁻⁴ kg/m
Explanation:
given,
mass of the string = 31.2 g
length of string = 0.7 m
linear density of the string = 
linear density of the string = 
linear density of the string = 44.57 × 10⁻³ kg/m
linear density of the string = 4.46 × 10⁻⁴ kg/m
Our data are,
State 1:
State 2:
We know as well that 
To find the mass we apply the ideal gas formula, which is given by
Re-arrange for m,
Because of the pressure, temperature and volume ratio of state 1 and 2, we have to
Replacing,
For conservative energy we have, (Cv = 0.718)
Answer:
False
Explanation:
Because when you go through east
( +x axis ) then you go to west ( -x axis )
You will subtract -9 from +15
it's become +6
( I talk about the displacement not distance) ( West = - East )
I hope that it's a clear ") .
Answer:
Option B. 5 nC
Explanation:
From the question given above, the following data were obtained:
Capicitance (C) = 100 pF
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:
1 pF = 1×10¯¹² F
Therefore,
100 pF = 100 pF × 1×10¯¹² F / 1 pF
100 pF = 1×10¯¹⁰ F
Next, we shall determine the quantity of charge. This can be obtained as follow:
Capicitance (C) = 1×10¯¹⁰ F
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Q = CV
Q = 1×10¯¹⁰ × 50
Q = 5×10¯⁹ C
Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:
1 C = 1×10⁹ nC
Therefore,
5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C
5×10¯⁹ C = 5 nC
Thus, the quantity of charge is 5 nC
