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2 years ago

To determine whether a celestial body is approaching or receding from earth, astronomers employ?

1 answer:

7 0

To determine whether a celestial body is approaching or receding from Earth, astronomers employ The Doppler principle.

There is a Doppler effect for both light and sound.

For instance, astronomers regularly measure the amount to which a star or galaxy's light is "stretched" towards the lower frequency, red region of the spectrum to calculate how quickly the object is travelling away from us.

A wave's frequency changes in response to an observer moving in respect to the wave source, which is known as the Doppler effect or Doppler shift.

The Doppler effect results from the fact that as the wave source moves in the direction of the observer, the crest of each succeeding wave emerges from a location that is closer to the observer than the crest of the previous wave.

As a result, each wave travels to the observer slightly faster than the one before it. As a result, the interval between successive wave crests arriving at the observer is shorter, increasing the frequency.

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Density is a physical property that relates the mass of a substance to its volume. A. Calculate the density, in g/mL , of a liqu

Angelina_Jolie [31]

Answer:

A) 0.660 g/ml

B) 1.297 ml

C) 0.272 g

Explanation:

Every substance, body or material has mass and volume, however the mass of different substances occupy different volumes. This is where density $D$ appears as a physical characteristic property of matter that establishes a relationship between the mass $m$ of a body or substance and the volume $V$ it occupies:

$D=\frac{m}{V}$ (1)

Knowing this, let's begin with the answers:

<h2 /><h2>Answer A:</h2>

Here the mass is $m=0.155g$ and th volume $V=0.000235L=0.235mL$

Solving (1) with these values:

$D=\frac{0.155g}{0.235mL}$ (2)

$D=0.660g/mL$ (3)

<h2>Answer B:</h2>

In this case the mass of a sample is $m=4.71g$ and its density is $D=3.63g/mL$ .

Isolating $V$ from (1):

$V=\frac{m}{D}$ (4)

$V=\frac{4.71g}{3.63g/mL}$ (5)

$V=1.297mL$ (5)

<h2>Answer C:</h2>

In this case the volume of a sample is $V=0.293mL$ and its density is $D=0.930g/mL$ .

Isolating $m$ from (1):

$m=D.V$ (6)

$m=(0.930g/mL)(0.293mL)$ (7)

$m=0.272g$ (8)

4 0

3 years ago

Read 2 more answers

A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (

SIZIF [17.4K]

The solution to the questions are given as

$t=40.39 \mathrm{sec}$
$\varepsilon &=(0.12v)e^{0.057t}$
the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}$

$\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$$

$\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}$

$\varepsilon &=(0.12v)e^{0.057t}$

(b) $Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$$

$\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}$

In conclusion, the direction of the induced current will be Counterclockwise.

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4 0

2 years ago

A circular loop of wire of cross-sectional area 0.12 m2 consists of 200 turns, each carrying 0.50 A. It is placed in a magnetic

defon

Answer:

0.52 Nm

Explanation:

A = 0.12 m^2, N = 200, i = 0.5 A, B = 0.050 T

Angle between the plane of loop and magnetic field = 30 Degree

Angle between the normal of loop and the magnetic field = 90 - 30 = 60 degree

θ = 60°

Torque = N i A B Sinθ

Torque = 200 x 0.5 x 0.12 x 0.050 x Sin 60

Torque = 0.52 Nm

4 0

3 years ago

A girl rolls a ball up an incline and allows it to re- turn to her. For the angle and ball involved, the acceleration of the bal

zalisa [80]

Answer:

3.28 m

3.28 s

Explanation:

We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.

Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

V0 = 4 m/s

a = -2.45 m/s^2 (because the acceleration is down slope)

Then:

X(t) = 4*t - 1.22*t^2

And the equation for speed is:

V(t) = V0 + a * t

V(t) = 4 - 2.45 * t

If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:

0 = 4 - 2.45 * t

4 = 2.45 * t

t = 1.63 s

Replacing that time on the position equation:

X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m

To find the time it will take to return we equate the position equation to zero:

0 = 4 * t - 1.22 * t^2

Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:

0 = t * (4 - 1.22*t)

t1 = 0

0 = 4 - 1.22*t2

1.22 * t2 = 4

t2 = 3.28 s

7 0

3 years ago

A carnival Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute. What is the accelerat

Tanzania [10]

If a Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute then the acceleration of a passenger at his lowest point during the ride is 4.11 $\text{m/s}^{2}$ .

Calculation:

Step-1:

It is given that the radius of the Ferris wheel is r=15 m, and the angular speed of the wheel is $\omega$ =5rev/min.

It is required to find the angular acceleration of a passenger at his lowest point during the ride.

The formula required to calculate the angular acceleration is, $a=\omega ^2 r$ .

Step-2:

Now substituting the given values into the equation to get the value of the angular acceleration.

$\begin{aligned}a &=\omega^{2} r \\&=(5 \mathrm{rev} / \mathrm{min})^{2}(15 \mathrm{~m}) \\&=\left(5 \mathrm{rev} / \mathrm{min} \times \frac{\left(\frac{2 \pi}{60}\right) \mathrm{rad} / \mathrm{sec}}{1 \mathrm{rev} / \mathrm{min}}\right)^{2}(15 \mathrm{~m}) \\&=(0.2741 \times 15) \mathrm{m} / \mathrm{s}^{2} \\&=4.11 \mathrm{~m} / \mathrm{s}^{2}\end{aligned}$

The acceleration is towards upwards that means towards the center of the wheel.

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2 years ago