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3 years ago

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the light from andromeda galaxy takes about 2.6 million years to reach earth. which of these statements is correct about the and

romeda galaxy? a. it is located at a distance of 2.6 million light years from earth. b. it has to travel a distance of 2.6 million kilometers to reach earth. c. it can be photographed by a space telescope in 2.6 million years. d. it has to move at the rate of 2.6 million kilometers per hour to reach earth.

1 answer:

-Dominant- [34]3 years ago

7 0

A. it is <span>located at a distance of 2.6 million light years from earth</span>

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Two car horns are sounded creating two sound waves with frequencies that differ by a factor of three. How does the speed of the

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Answer:

Remains the same

Explanation:

The speed of waves of higher and lower frequency both will be same.

the speed of sound in a medium is constant and independent of it's frequency. Moreover, when the frequency changes wavelength changes accordingly, such that their product remains constant.

we know that

υ×λ = constant = velocity

υ= frequency

λ= wavelength.

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2 years ago

our friend is constructing a balancing display for an art project. She has one rock on the left (ms=2.25 kgms=2.25 kg) and three

Licemer1 [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The torque produced by the pile of rocks is $\tau = 35.63\ N \cdot m$

b

The distance of the single for equilibrium to occur is $r_s =1.62 \ m$

Explanation:

From the question we are told that

The mass of the left rock is $m_s = 2.25 \ kg$

The mass of the rock on the right $m_p = 10.1 kg$

The distance from fulcrum to the center of the pile of rocks is $r_p = 0.360 \ m$

Generally the torque produced by the pile of rock is mathematically represented as

$\tau = m_p * g * r_p$

Substituting values

$\tau = 10.1 * 9.8 * 0.360$

$\tau = 35.63\ N \cdot m$

Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows

The torque due to the single rock is

$\tau = m_s * g * r_s$

At equilibrium the both torque are equal

$35.63 = m_s * r_s * g$

Making $r_s$ the subject of the formula

$r_s = \frac{35.63 }{m_s * g}$

Substituting values

$r_s = \frac{35.63 }{2.25 * 9.8}$

$r_s =1.62 \ m$

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3 years ago

Mark each of the following statementsas true or false. If a statement refers to "two objects"interacting via some force, you are

slega [8]

Answer:

Explanation:

A,D,E

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2 years ago

A motorcyclist heading east through a small town accelerate at constant 4.0meter per seconds square after he leaves the limits.

SVETLANKA909090 [29]

A) The position at t = 2.0 sec is 43.0 m east

B) The position is 55 m east

Explanation:

A)

In order to solve the problem, we take the east direction as positive direction.

We know that:

- at t = 0, the motorcyclist is at a position of $x_0 = 5.0 m$

- at t = 0, the initial velocity of the motorcyclist is $v_0 = 15.0 m$ east

- The acceleration of the motorcyclist is constant and it is $a=4.0 m/s^2$

Since the motion is a uniformly accelerated motion, the position of the motorcylist is given by the expression

$x(t)=x_0 + v_0t + \frac{1}{2}at^2$

where t is the time.

Substituting t = 2.0 s, we find the position:

$x(2.0)=(5.0)+(15)(2.0)+\frac{1}{2}(4.0)(2.0)^2=43 m$

B)

The velocity of the motoryclist can be found by calculating the derivative of the position. Therefore, it is:

$v(t)=x'(t)=v_0 + at$

where:

$v_0=15.0 m/s$ is the initial velocity

$a=4.0 m/s^2$ is the acceleration

We want to find the time t at which the velocity is

v = 25 m/s

Solving the equation for t,

$t=\frac{v-v_0}{a}=\frac{25-15}{4}=2.5 s$

And therefore, the position at t = 2.5 s is:

$x(2.5s)=5.0+(15.0)(2.5)+\frac{1}{2}(4)(2.5)^2=55 m$

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3 years ago

If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. Wh

In-s [12.5K]

Answer:

v = -v₀ / 2

Explanation:

For this exercise let's use kinematics relations.

Let's use the initial conditions to find the acceleration of the electron

v² = v₀² - 2a y

when the initial velocity is vo it reaches just the negative plate so v = 0

a = v₀² / 2y

now they tell us that the initial velocity is half

v’² = v₀’² - 2 a y’

v₀ ’= v₀ / 2

at the point where turn v = 0

0 = v₀² /4 - 2 a y '

v₀² /4 = 2 (v₀² / 2y) y’

y = 4 y'

y ’= y / 4

We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.

v² = v₀² -2a y’

v² = 0 - 2 (v₀² / 2y) y / 4

v² = -v₀² / 4

v = -v₀ / 2

We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.

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3 years ago