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2 years ago

11

Planet X is in a stable circular orbit around a star, as shown in the figure. Which of the following graphs best predicts the an

gular momentum of the planet as a function of its horizontal position from point A to point B if the planet is moving counterclockwise as viewed in the figure?

1 answer:

damaskus [11]2 years ago

5 0

Answer:

C

Explanation:

Angular momentum is the product of moment of inertia and angular velocity.

L = I × ω

Since the planet follows a stable circular orbit, I and ω are constant and non-zero. Therefore, the angular momentum is constant and non-zero.

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A fault in which the hanging wall moves down relative to the footwall is a _____.

LuckyWell [14K]

<span>Normal fault. Normal (extensional ) fault is a displacement of a rock as a result of rock-mass movement and occurs when the crust is stretching. Because of the stretching the thickness of the crust is reduced and the crust or horizontally extended. </span>

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3 years ago

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A solenoid of length 18 cm consists of closely spaced coils of wire wrapped tightly around a wooden core. The magnetic field str

Kisachek [45]

Answer:

$B_2 = 1.71 mT$

Explanation:

As we know that the magnetic field near the center of solenoid is given as

$B = \frac{\mu_0 N i}{L}$

now we know that initially the length of the solenoid is L = 18 cm and N number of turns are wounded on it

So the magnetic field at the center of the solenoid is 2 mT

now we pulled the coils apart and the length of solenoid is increased as L = 21 cm

so we have

$\frac{B_1}{B_2} = \frac{L_2}{L_1}$

now plug in all values in it

$\frac{2.0 mT}{B_2} = \frac{21}{18}$

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2 years ago

Net force needed to accelerate a 1000-kg car at 0.5g

just olya [345]

It would be 2000N ( newtons )

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3 years ago

Compute the dot product of the vectors u and v, and find the angle between the vectors. Bold v equals 7 Bold i minus Bold j and

OLga [1]

Answer:

$\theta = 106.3 degree$

Explanation:

As we know that

$\vec w = -\hat i + 7\hat j$

$\vec v = 7\hat i - \hat j$

also we know that

$\vec v. \vec w = -14$

it is given as

$\vec v. \vec w = (-\hat i + 7\hat j).(7\hat i - \hat j)$

$\vec v. \vec w = - 7 - 7 = -14$

also we can find the magnitude of two vectors as

$|v| = \sqrt{(-1)^2 + (7)^2}$

$|v| = \sqrt{50}$

similarly we have

$|w| = \sqrt{(7^2) + (-1)^2}$

$|w| = \sqrt{50}$

now we know the formula of dot product as

$\vec v. \vec w = |v||w| cos\theta$

$-14 = (\sqrt{50})^2cos\theta$

$\theta = cos^{-1}(\frac{-14}{50})$

$\theta = 106.3 degree$

3 0

3 years ago

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

givi [52]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

$f = f_0 (\frac{v_0}{v_0-v})$

Here,

$f_0$ = Frequency of Source

$v_s$ = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

$f = f_0 (\frac{v_0}{v_0-v})$

$300 = f_0 (\frac{343}{343-v})$

$(300*343) - 300v = 343f_0$

Now the second expression will be,

$f' = f_0 (\frac{v_0}{v_0-v/2})$

$290 = (343)(\frac{v_0}{343-v/2})$

$290*343-145v = 343f_0$

Dividing the two expression we have,

$\frac{(300*343) - 300v}{290*343-145v} = 1$

Solving for v, we have,

$v = 22.12m/s$

Therefore the speed of the train before and after slowing down is 22.12m/s

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3 years ago