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nekit [7.7K]
3 years ago
5

jenny's model train is set up on a circular track. There are six telephone poles evenly spaced around the track. It takes the en

gine of her train 10 seconds to go from the first pole to the third pole. How long would it take for the engine to go the entire distance around the track?
Physics
1 answer:
d1i1m1o1n [39]3 years ago
5 0

Answer:

T = 60 s

Explanation:

There are 6 poles on the track which are equally spaced

so the angular separation between the poles is given as

\theta = \frac{2\pi}{6}

\theta = \frac{\pi}{3}

so the angular speed of the train is given as

\omega = \frac{\theta}{t}

\omega = \frac{\pi}{30} rad/s

now we have time period of the train given as

T = \frac{2\pi}{\omega}

T = \frac{2\pi}{\frac{\pi}{30}}

T = 60 s

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We need to directly measure the spectral type in order to determine the surface temperature of a star.

<h3>How do you find the properties of a star?</h3>

Astronomers can determine the temperature of a star by looking at its color and spectrum. The apparent brightness of a star describes how luminous it looks to us. The brightness of a star tells us how bright it really is. The luminance can be determined using both the perceived brightness and the distance.

A star's luminosity, or the total amount of energy it emits each second, is determined by two factors: The stellar photosphere's "Effective Temperature," T. the star's total surface area, which is influenced by its radius, R.

Because it controls how much fuel a star has and how quickly it burns it, a star's mass is its most fundamental characteristic. The majority of a star's life is spent burning hydrogen into helium in its core, which generates energy. The star needs to achieve a balance between gravity and outward pressure in order to continue to be "alive."

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1 year ago
During a race, a sprinter accelerated 1.8 m/s 2 in 2.5 seconds.How many meters per second did the sprint increase with this amou
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During a race, a sprinter accelerated 1.8 m/s 2 in 2.5 seconds. The sprint increase with this amount of acceleration by 4.5 m/s.

<h3>What is acceleration?</h3>

Acceleration is the time rate of change of velocity.

Acceleration a = velocity v / time t

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The sprint increase with this amount of acceleration by 4.5 m/s.

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3 0
2 years ago
A person travelled 350 m east from his home and returns back home an hour has displacement of_?​
Svetradugi [14.3K]

Answer:

vector of zero magnitude

Explanation:

The displacement is a vector magnitude, therefore, in addition to being a module, it has direction and sense.

In this case it moved 350 m and then returned the same 350 m, so the total displacement is zero.

If we draw the vector, one has a directional direction to the right and the other direction to the left, therefore when adding the two vectors gives a vector of zero magnitude

7 0
3 years ago
Very easy physics maths about current and other stuffs only one question please help me
Reil [10]

Answer:

The total voltage of the circuit is 18 volts.

Explanation:

We have, three identical resistors of resistance 3 ohms are connected in series in a circuit.

For a series combination, the equivalent resistance is given by the sum of individual resistances i.e.

R_{eq}=R_1+R_2+R_3\\\\R_{eq}=3+3+3\\\\R_{eq}=9\ \Omega

Let V is the voltage of the battery if the current in the circuit is 2 A. So,

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So, the total voltage of the circuit is 18 volts.

7 0
3 years ago
The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, in
bezimeni [28]

1) Potential difference: 1 V

2) V_b-V_a = -1 V

Explanation:

1)

When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by

\Delta U=q\Delta V

where

q is the charge's magnitude

\Delta V is the potential difference between the initial and final position

In this problem, we have:

q=4.80\cdot 10^{-19}Cis the magnitude of the charge

\Delta U = 4.80\cdot 10^{-19}J is the change in kinetic energy of the particle

Therefore, the potential difference (in magnitude) is

\Delta V=\frac{\Delta U}{q}=\frac{4.80\cdot 10^{-19}}{4.80\cdot 10^{-19}}=1 V

2)

Here we have to evaluate the direction of motion of the particle.

We have the following informations:

- The electric potential increases in the +x direction

- The particle is positively charged and moves from point a to b

Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)

This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative:

V_b-V_a = - 1V

8 0
3 years ago
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