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3 years ago

What is a unit vector?

Physics

1 answer:

jekas [21]3 years ago

8 0

Answer:

A unit vector is (D) a vector with a magnitude of one

Explanation:

Unit Vector Information:

A unit vector is a vector with a magnitude of one. Vectors of different magnitudes can be scaled off the unit vector. Also, they can be used in two, three, or more dimensions.

Other Answers:

A. a vector with all three directions or dimensions that are given

This is the definition of a 3D vector. This is a line segment in three-dimensional space running from point A (tail) to point B (head). Each vector has a magnitude (or length) and direction.

B. a vector with only one dimension

This is the definition of an one-dimensional vector. This type of vector has no dimension, but a number of components, therefore it behaves like a real number.

C. a vector that has an x and y components that are equal in magnitude

(Unsure, I believe this would be defined as an equal vector. Correct me if I'm wrong and I will edit it.)

I hope this helps!

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Use your understanding of heat loss to ESTIMATE the cost of the lost energy through one standard window during an average summer

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Answer:

The cost of energy is $ 0.34.

Explanation:

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The commercial unit of energy is kWh.

Cost of 1 kWh energy = $ 0.17

energy loss by standard window is 2 kWh .

So, the cost of lost of energy is

Cost = $ 0.17 x 2 = $ 0.34

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3 years ago

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4 years ago

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A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ (we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

4 0

3 years ago

Which statement about stage 3 NREM is accurate?

cestrela7 [59]

Answer:

hope

that helps

Explanation:

D. The sleeper's heart rate, blood pressure, and breathing rate drop to their lowest levels.

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2 years ago

Determine the magnitude of the resultant force acting on a 1.5 −kg particle at the instant t=2 s, if the particle is moving alon

Phoenix [80]

Answer:

F = 63N

Explanation:

M= 1.5kg , t= 2s, r = (2t + 10)m and

Θ = (1.5t² - 6t).

magnitude of the resultant force acting on 1.5kg = ?

Force acting on the mass =

∑Fr =MAr

Fr = m(∇r² - rθ²) ..........equation (i)

∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)

The horizontal path is defined as

r = (2t + 10)

dr/dt = 2, d²r/dt² = 0

Angle Θ is defined by

θ = (1.5t² - 6t)

dθ/dt = 3t, d²θ/dt² = 3

at t = 2

r = (2t + 10) = (2*(2) +10) = 14

but dr/dt = 2m/s and d²r/dt² = 0m/s

θ = (1.5(2)² - 6(2) ) = -6rads

dθ/dt =3(2) - 6 = 0rads

d²θ/dt = 3rad/s²

substituting equation i into equation ii,

Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)

∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]

∑F = 1.5(14*3+0) = 63N

F = √(Fr² +FΘ²) = √(0² + 63²) = 63N

7 0

3 years ago