Answer:
The mass of the object involved and the value of the gravitational acceleration
Explanation:
- Gravitational potential energy is defined as the energy possessed by an object in a gravitational field due to its position with respect to the ground:
where m is the mass of the object, g is the gravitational acceleration and h is the heigth of the object with respect to the ground.
- Elastic potential energy is defined as the energy possessed by an elastic object and it is given as:
where k is the spring constant of the elastic object, while x is the compression/stretching of the spring with respect to the equilibrium position.
As we can see from the equations, both types of energy depends on the relative position of the object/end of the spring with respect to a certain reference position (h in the first formula, x in the second formula), but gravitational potential energy also depends on m (the mass) and g (the gravitational acceleration) while the elastic energy does not.
Some guidance notes which may help.To calculate the current flow, Ohm's law can be used. This can be written as current=voltage/resistance, or I=V/R. V is 1.5V.R for the copper wire quoted would be calculated as R = resistivity x length/cross sectional area. The area would be calculated from the formula area = pi x diameter squared/4So, R=resistivity x length divided by (pi x diameter squared/4)Until is the resistivity of copper is known, that's about as far as can be gone.Any further questions, please ask.
R = U : I. U is in Voltage and I is in Ampère. That gives you R = 36 : 8 = 4,5 Ohm
S = ut + 0.5at^2
<span>10 = 0 + 0.5(9.81)t^2 {and if g = 10 then t^2 = 2 so t ~1.414} </span>
<span>t^2 ~ 2.04 </span>
<span>t ~ 1.43 seconds</span>
Answer:
The voltage across the capacitor is 1.57 V.
Explanation:
Given that,
Number of turns = 10
Diameter = 1.0 cm
Resistance = 0.50 Ω
Capacitor = 1.0μ F
Magnetic field = 1.0 mT
We need to calculate the flux
Using formula of flux
Put the value into the formula
We need to calculate the induced emf
Using formula of induced emf
Put the value into the formula
Put the value of emf from ohm's law
We know that,
We need to calculate the voltage across the capacitor
Using formula of charge
Put the value into the formula
Hence, The voltage across the capacitor is 1.57 V.
