Answer:
No, the 50 ohm and 100 ohm resistor will not continue to operate.
Explanation:
A closed circuit is the circuit in which there is no break between the negative and the positive end of the battery.
When in this, combinational circuit the 80 ohm resistor fail then there will not any continue supply of current in the circuit due to the breakage because the electron will flow from negative end of the battery to positive end if their is no breaking in the circuit.
Therefore the 50 ohm and 100 ohm circuit will not continue to operate because of the breaking of the circuit and current will not flow.
A. Angular momentum is always conserved would be the correct answer.
This is because like linear momentum (mvmv), angular momentum (r×mvr×mv) is a conserved quantity, where rr is the vector from the center of rotation. For a skater holding a static pose, for each particle making up her body, the contribution in magnitude to the total angular momentum is given by mirivimirivi. Thus bringing in her arms reduces riri for those particles. In order to conserve angular momentum, there is then an increase in the angular velocity.
hope this helps!
Thank you for posting your question here. The total distance traveled by the truck during the 3.2 seconds interval is 83 m. Below is the solution:
d = vit + 1/2 at^2
d = (22m/ s) (3.2s) + 1/2 (2.4m/ s^2) (3.2s)^2
d = 83 m
Hope the answer helps.
Answer:
Explanation:
<u>Coulomb's Law</u>
The force between two charged particles of charges q1 and q2 separated by a distance d is given by the Coulomb's Law formula:
Where:
q1, q2 = the objects' charge
d= The distance between the objects
We know both charges are identical, i.e. q1=q2=q. This reduces the formula to:
Since we know the force F=1 N and the distance d=1 m, let's find the common charge of the spheres solving for q:
Substituting values:
This charge corresponds to a number of electrons given by the elementary charge of the electron:
Thus, the charge of any of the spheres is:
Answer:
Explanation:
Given that,
Mass of the thin hoop
M = 2kg
Radius of the hoop
R = 0.6m
Moment of inertial of a hoop is
I = MR²
I = 2 × 0.6²
I = 0.72 kgm²
Period of a physical pendulum of small amplitude is given by
T = 2π √(I / Mgd)
Where,
T is the period in seconds
I is the moment of inertia in kgm²
I = 0.72 kgm²
M is the mass of the hoop
M = 2kg
g is the acceleration due to gravity
g = 9.8m/s²
d is the distance from rotational axis to center of of gravity
Therefore, d = r = 0.6m
Then, applying the formula
T = 2π √ (I / MgR)
T = 2π √ (0.72 / (2 × 9.8× 0.6)
T = 2π √ ( 0.72 / 11.76)
T = 2π √0.06122
T = 2π × 0.2474
T = 1.5547 seconds
T ≈ 1.55 seconds to 2d•p
Then, the period of oscillation is 1.55seconds
