Answer:
yes
Explanation:
The solubility of glucose at 30°C is
125 g/100 g water. Classify a solution made by adding 550 g of glucose to 400 mL of water at 30°C. Explain your classification, and describe how you could increase the amount of glucose in the solution without adding more glucose.
<span>Balloons are blown up, and then rubbed against your shirt many times. The balloon then touches the ceiling. When released, the balloon remains stuck to the ceiling. The balloon is charged by contact. The ceiling has a neutral charge. The charged balloon induces a slight surface charge on the ceiling opposite to the charge on the balloon. Balloon and ceiling electric charges are opposite in sign, so they will attract each other. Since both the balloon and the ceiling are insulators, charge can not flow from one to the other. The charge on the balloon is fixed on the balloon and the charge on the ceiling remains fixed to the ceiling. It just so happens that the<span> electrostatic force the ceiling exerts on the balloon is sufficient to hold the balloon in place (i.e. overcomes gravity, etc.).</span></span>
Here Power = Voltage * Current
So, Voltage = Power/Current
Put the values,
V = 240/2
V = 120 V
In short, Your Final Answer would be: 120 Volts
Hope this helps!
Air resistance is ignored.
g = 9.8 m/s².
At maximum height, the vertical velocity is zero.
Let h = the maximum height reached.
Let u = the vertical launch velocity.
Because ot takes 5.0 seconds to reach maximum height, therefore
(u m/s) - (9.8 m/s²)*(5 s) = 0
u = 49 m/s
The maximum height reached is
h = (49 m/s)*(5 s) - (1/2)*(9.8 m/s²)*(5 s)²
= 122.5 m
Answer: 122.5 m
1) 
The capacitance of a parallel-plate capacitor is given by:
where
is the vacuum permittivity
A is the area of each plate
d is the distance between the plates
Here, the radius of each plate is
so the area is
While the separation between the plates is
So the capacitance is
And now we can find the energy stored,which is given by:
2) 0.71 J/m^3
The magnitude of the electric field is given by
and the energy density of the electric field is given by
and using
, we find
