Ask question

Ask question

All categories

sertanlavr [38]

2 years ago

6

Consider the vector b⃗ with magnitude 4.00 m at an angle 23.5∘ north of east. what is the x component bx of this vector? express

your answer in meters to three significant figures.

1 answer:

BlackZzzverrR [31]2 years ago

6 0

Decomposing the vector b on the x-axis and the y-axis, we get a rectangle triangle where the two sides are bx (x-axis) and by (y-axis), and b is the hypothenuse.
The component in x, bx, is equal to the product between the hypothenuse and the cosine of the angle between b and the x-axis, which is $23.5 ^{\circ}$ :
$b_x = b \cos (23^{\circ})=(4.00 m)(\cos (23^{\circ}))=3.68 m$

You might be interested in

Lorenzo is making a prediction. “I learned that nonmetals increase in reactivity when moving from left to right. So I predict th

nadezda [96]

That prediction is not correct because Xenon is extremely stable; column 18 of the periodic table contains the noble gasses, which are stable because their outer-most energy levels are completely filled. Having the octet (8) of valence electrons means that the element no longer needs to lose or gain electrons to gain stability.

The column 17 elements are unstable because they only have one valence electron short of the stable octet configuration of the noble gasses.

6 0

2 years ago

Read 2 more answers

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r

mina [271]

Answer:

t = 103.45 n m

Explanation:

given,

refractive index of cornea = 1.38

refractive index of eye drop = 1.45

wavelength of refractive index = 600 nm

refractive index of eye drop is greater than refractive index of cornea and the air.

Formula used in this case

for constructive interference

$2 n t = (m + \dfrac{1}{2})\lambda$

At m = 0 for the minimum thickness, so

$2\times 1.45 \times t = (0 + 0.5)\times 600$

$2.9 \times t =300$

$t =\dfrac{300}{2.9}$

t = 103.45 n m

the minimum thickness of the film of eyedrops t = 103.45 n m

6 0

2 years ago

If two waves with identical crests and troughs meet, what is happening? The wave is reflecting. Constructive interference is occ

Kazeer [188]

The correct answer would be that destructive interference is happening. In this interference, the crest of a wave meets a trough of another wave resulting to an amplitude that is lower. The opposite is called the constructive interference. Hope this answers the question.

7 0

1 year ago

Read 2 more answers

What force would be needed to accelerate a 0.040-kg golf ball at 20.0 m/s?

Naily [24]

Answer:

any amount of force will do it as time is not mentioned here

5 0

2 years ago

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0

1 year ago