How to find power in physics?

If A > B, under what condition is |A-BI=|A|- IB|? a. Vectors A and B are in opposite directions b. Vectors A and B are in the

kupik [55]

Answer:

b) Vectors A and B are in the same direction.

Explanation:

To understand this problem we will say that vector A has a magnitude of 5 units and vector B a magnitude of 3 units. In the subtraction of vectors the initial parts of vectors always bind together. And the vector resulting from the subtraction is traced from the end of the second vector (B) to the end of the first vector (A).

The length of the resultant vector will be 5 - 3 = 2

In the attached image, we analyze case a), b), and d)

For a)

As we can see in the attached image the resultant vector has a length of 8 units.

For d)

As we can see in the attached image the resultant vector has a length of 5.83 units.

For b)

The resultant vector has a length of 2 units.

Therefore the case given in b) is true

5 0

4 years ago

Classify each of the following chemical reactions. Upper S plus upper O Subscript 2 right arrow upper S upper O subscript 2. Upp

vodka [1.7K]

Check attached image.

4 0

3 years ago

A biconvex lens is formed by using a piece of plastic(n=1.70).

creativ13 [48]

Answer:

$f =17.15\ cm$

Explanation:

given,

refractive index of lens, n = 1.70

Radius of curvature of front surface. R₁ = 20 cm

Radius of curvature of the back surface, R₂ = 30 cm

focal length= ?

$\dfrac{1}{f}=(n-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})$

R₁ = +20 cm

R₂ = -30 cm

n = 1.70

$\dfrac{1}{f}=(1.70-1)(\dfrac{1}{20}-\dfrac{1}{-30})$

$\dfrac{1}{f}=0.70 \times 0.0833$

$f = \dfrac{1}{0.7 \times 0.0833}$

$f =17.15\ cm$

the focal length of the lens is equal to 17.15 cm

3 0

3 years ago

A mass free to vibrate on a level, frictionless surface at the end of a horizontal spring is pulled 35 cm from its equilibrium p

saul85 [17]

Answer:

0.67 s

Explanation:

This is a simple harmonic motion (SHM).

The displacement, $x$ , of an SHM is given by

$x = A\cos(\omega t)$

A is the amplitude and $\omega$ is the angular frequency.

We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or $\frac{\pi}{4}$ radian.