What type of circuit has little or no resistance?

The initial and final velocities of two blocks experiencing constant acceleration are respectively −7.45 m/s and 14.9 m/s. (a) T

ikadub [295]

Answer:

a) $a_{1}=3.7 m/s^{2}$

b) $a_{2}=3.68 m/s^{2}$

Explanation:

a) The displacement of the first object is 22.5 m, so we can use the next equation:

$v_{f}^{2}=v_{i}^{2}+2a\Delta x$

$a=\frac{v_{f}^{2}-v_{i}^{2}}{2x}$

$a=\frac{14.9^{2}-(-7.45)^{2}}{2*22.5}$

$a_{1}=3.7 m/s^{2}$

positive acceleration.

b) Using the same equation we can find the second value of the acceleration:

$a=\frac{v_{f}^{2}-v_{i}^{2}}{2x}$

$a=\frac{14.9^{2}-(-7.45)^{2}}{2*22.6}$

$a_{2}=3.68 m/s^{2}$

positive acceleration.

I hope it helps you!

8 0

3 years ago

Consider a system two point charges. One has charge +q at (x, y,z) -(a,0,0) and another of charge-q at (x, y, z) = (-a, 0,0). 5.

olga2289 [7]

Answer:

electricfield at (0,0,0) is Et = 2 k q / a²

Explanation:

For the first part see the diagram , the field lines start from the positive charge and reach the negative charge, notice that no line should cross, some lines go to infinity

For the second part we use that the electric field is a vector quantity and therefore we add the field of each charge, using the equation

E = k q / r²

Point (0,0,0)

We calculate for the charge -q which is at a distance R = a

E1 = k (-q) / a²

E1 = - kq / a²

As the test charge is positive in the field it goes to the left, attractive force

We calculate for the charge that is also at R = a

E2 = k q / a²

This field goes to the left, repulsive force

We find the total electric field

Et = E1 + E2

Et = kq / a² + kq / a²

Et = 2 k q / a²

Point (0,0, R)

We use the same equations, but with another distance, for the charge -q the distance is R = R+a and for the charge + q the distance is R = R-a

E1 = k q / (R + a)²

E2 = kq / (R-a)²

Et = kq [1 / (R + a)² + 1 / (R-a)²]

Et= kq {[(R-a)² + (R + a)²] / [(R + a)² (R-a)²]}

Et= kq {2 (R² + a²) / [(R + a)² (R-a)²]}

If we use the condition that R> a we can despise in the patents "a"

(R² + a²) = R² (1+ a² / R²) ≈ R²

(R + a)² = R² (1 + a / R)² ≈ R²

(R- a)² = R² (1-a / R)² ≈ R²

Substituting in the total electric field

Et = kq {2 R²) / [R²R²]}

Et =kq 2 / R²

7 0

3 years ago

The peak intensity of radiation from Star Beta is 350 nm. In what spectral band is this? UV, radio waves, visible light, or infa

Stels [109]

I say uv all the way

4 0

3 years ago

Read 2 more answers

Applying the Law of Conservation of Energy. If a car was released down the track from a height what happens to the potential ene

erastova [34]

Answer:

According to the law of conservation of energy, energy cannot be created or destroyed, although it can be changed from one form to another. KE + PE = constant. A simple example involves a stationary car at the top of a hill. As the car coasts down the hill, it moves faster and so it’s kinetic energy increases and it’s potential energy decreases. On the way back up the hill, the car converts kinetic energy to potential energy. In the absence of friction, the car should end up at the same height as it started.

This law had to be combined with the law of conservation of mass when it was determined that mass can be inter-converted with energy.

One can also imagine the energy transformation in a pendulum. When the ball is at the top of its swing, all of the pendulum’s energy is potential energy. When the ball is at the bottom of its swing, all of the pendulum’s energy is kinetic energy. The total energy of the ball stays the same but is continuously exchanged between kinetic and potential forms

4 0

2 years ago

Which object has the larger magnitude of its momentum?

Charra [1.4K]

The object that had the most 1000 ton weight has the most momentum

5 0

3 years ago