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3 years ago

Which BEST compares the motions of dots C and D?

Physics

2 answers:

Alex787 [66]3 years ago

8 0

Option C is correct as c is speeding up but d moving with same or constant velocity

melisa1 [442]3 years ago

3 0

As per the graph there is a motion map which denotes the motion of two dots C and D respectively.

As per the graph the motion of dot C is non uniform .It is so because it is covering unequal distance in equal interval of time.Its speed is increasing at every instant .Hence it is speeding up.

The motion of dot D is uniform .It covers equal distance in equal interval of time.Hence the dot D is moving at a constant velocity.

Hence out of four options given in the question,the third option i.e C is the correct answer .

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Phobos's Orbit. Phobos orbits Mars at a distance of 9,380 km from the center of the planet and has a period of 0.3189 days. Assu

Elenna [48]

Answer:

Explanation:

The relation between time period of moon in the orbit around a planet can be given by the following relation .

T² = 4 π² R³ / GM

G is gravitational constant , M is mass of the planet , R is radius of the orbit and T is time period of the moon .

Substituting the values in the equation

(.3189 x 24 x 60 x 60 s)² = 4 x 3.14² x ( 9380 x 10³)³ / (6.67 x 10⁻¹¹ x M)

759.167 x 10⁶ = 8.25 x 10²⁰ x 39.43 / (6.67 x 10⁻¹¹ x M )

M = .06424 x 10²⁵

= 6.4 x 10²³ kg .

4 0

3 years ago

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.

Read more about current

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4 0

2 years ago

:) What is practical machine? what is the reda<br>tion between MA and VR in a practical<br>machine?

denis-greek [22]

Answer: For ideal machine efficiency = 1. Hence M.A = V. R. The V. R of an ideal machine and the practical machine is a constant or is the same for both

3 0

3 years ago

What is the momentum of a 1, 500 kg car traveling at 3m/s?

ruslelena [56]

4,500 kgm/s bc 1,500x3

6 0

3 years ago

Read 2 more answers

Please Help!

irga5000 [103]

The maximum force that the tires can exert on the road before slipping is 16200 N.

From the information in the question;

The coefficient of static friction = 0.9

The mass of the car = 1800 kg

Using the formula;

μ = F/R

μ = coefficient of static friction

F = force on the tires

R = the reaction force

But recall that the reaction is equal in magnitude to the weight of the car.

W=R

Hence; R = 1800 kg × 10 ms-2 = 18000 N

Making F the subject of the formula;

F = μR

Substituting values;

F = 18000 N × 0.9

F = 16200 N

Hence, the maximum force that the tires can exert on the road before slipping is 16200 N.

Learn more: brainly.com/question/18754989

6 0

3 years ago