Answer:
P = 4.5 watts
Explanation:
Given that,
EMF of the circuit, E = 3 volt
The resistance of the resistors, R = 2 ohms
We need to find the power of this circuit. The relation between power, emf and resistance is given by the formula as follows :
Substitute all the values,
So, the power of this circuit is equal to 4.5 watts.
It is called condensation. Hope this helped!
Answer:
a) 378Ns
b) 477.27N
Explanation:
Impulse is the defined as the product of the applied force and time taken. This is expressed according to the formula
I = Ft = m(v-u)
m is the mass = 70kg
v is the final velocity = 5.4m/s
u is the initial velocity = 0m/s
Get the impulse
I = m(v-u)
I = 70(5.4-0)
I = 70(5.4)
I = 378Ns
b) Average total force is expressed as
F = ma (Newton's second law)
F = m(v-u)/t
F = 378/0.792
F = 477.27N
Hence the average total force experienced by a 70.0-kg passenger in the car during the time the car accelerates is 477.27N
The force between the two objects is 19.73 nN.
<u>Explanation:
</u>
Any force acting between two objects tends to be directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects. And this kind of attraction force between two objects is termed as gravitational force.
So if we consider 
and 
as the masses of both objects and let d be the distance of separation of two objects. Then the force between the two objects can be determined as below:
As gravitational constant 
, = 20 kg and = 100 kg, while d = 2.6 m, then
Thus, we get finally,
As we know, nano denoted by letter 'n' equals to 
So the force acting between two objects is 19.73 nN.
Answer:
a)1815Joules b) 185Joules
Explanation:
Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;
F = ke where;
F is the applied force
k is the elastic constant
e is the extension of the material
From the formula, k = F/e
F1/e1 = F2/e2
If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;
k = 60/0.5
k = 120N/m
a) To get the work done in stretching the spring 5.5m from its position,
Work done by the spring = 1/2ke²
Given k = 120N/m, e = 5.5m
Work done = 1/2×120×5.5²
Work done = 60× 5.5²
Work done = 1815Joules
b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;
Work done = 1/2ke²
Work done =1/2× 120×1.5²
Works done = 60×1.5²
Work done = 135Joules
