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3 years ago

Which vector has a y-component with a length of 4?

Physics

1 answer:

mihalych1998 [28]3 years ago

6 0

ANSWER

The correct answer is D.

EXPLANATION

The y-component of the vector, is the same as the vertical component of the vector.

We can see from the graph that each box in the grid is one unit.

The component of vector a is

$\binom {2}{4}$

We count 4 boxes along the positive y- axis to get the y-component of vector a.

See graph in attachment.

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A light wave moves from diamond (n= 2.4) into water (n= 1.33) at an angleof 24°. what angle does it have in water?

worty [1.4K]

Answer:

n1 sin θ1 = n2 sin θ2 Snell's Law (θ1 is the angle of incidence)

sin θ2 = n1 / n2 * sin θ1

sin θ2 = 2.4 / 1.33 * sin θ1

sin θ2 = 1.80 * .407 = .734

θ2 = 47.2 deg

3 0

2 years ago

Water initially at 200 kPa and 300°C is contained in a piston–cylinder device fitted with stops. The water is allowed to cool at

netineya [11]

Answer:

Δu=1300kJ/kg

Explanation:

Energy at the initial state

$p_{1}=200kPa\\t_{1}=300^{o}\\u_{1}=2808.8kJ/kg(tableA-5)$

Is saturated vapor at initial pressure we have

$p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)$

Process 2-3 is a constant volume process

$p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)$

The overall in internal energy

Δu=u₁-u₃

We replace the values in equation

Δu=u₁-u₃

$=2808.8kJ/kg-1508.6kJ/kg\\=1300kJ/kg$

Δu=1300kJ/kg

3 0

3 years ago

Find electric field at point p which is a distance l away from the both +q and -q

denis-greek [22]

Answer:

$\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

$=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }$

Where

q is the charge
r is the distance
$E$ is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }$

F2 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

⇒

F1+F2= $\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

8 0

3 years ago

Ron is on a Ferris wheel of radius 30 ft that turns counterclockwise at a rate of one revolution every 12 seconds. The lowest po

alexgriva [62]

Answer:

$x = 30cos\frac{\pi}{6}t$

$y = 30sin\frac{\pi}{6}t + 45$

Explanation:

1 full revolution is $2\pi.$ let \theta be the angle of Ron's position.

At t = 0. $\theta = 0$

one full revolution occurs in 12 sec, so his angle at t time is

$\theta =2\pi \frac{t}{12} = \frac{\pi}{6}t$

r is radius of circle and it is given as

$x = rcos\theta$

$y = rsin\theta$

for r = 30 sec

$x = 30cos\frac{\pi}{6}t$

$y = 30sin\frac{\pi}{6}t$

however, that is centered at (0,0) and the positioned at time t = 0 is (30,0). it is need to shift so that the start position is (30,45). it can be done by adding to y

$x = 30cos\frac{\pi}{6}t$

$y = 30sin\frac{\pi}{6}t + 45$

7 0

3 years ago

I was walking with Ruby in a garden from rest in a straight line with uniform acceleration so that we covered 0.25 m in the fift

Serhud [2]

Answer:

0.02m/s^2

Explanation:

7 0

2 years ago