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3 years ago

If we use 1 millimeter to represent 1 light-year, how large in diameter is the Milky Way Galaxy?

Physics

1 answer:

storchak [24]3 years ago

4 0

Answer:

d.100 meters

Explanation:

The diameter of the Milky Way Galaxy is approximately 100,000 light years.

Here we are using 1 millimiter (1 mm) to represent 1 light-year (1 ly). So, we can set the following proportion:

$1 mm : 1 ly = x : 100,000 ly$

and by finding x, we find the diameter of the Milky Way Galaxy in the scale used:

$x=\frac{(1mm )(100,000 ly)}{1 ly}=100,000 mm = 100 m$

so the correct answer is

d. 100 meters

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How can two tectonic plates move relative to each other

yarga [219]

They are doing the transform type of tectonic plate movement. This is a process where the plates move sideways. But they move at a rate of a few inches a year. Did that help you?

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3 years ago

A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm

Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

$F = - k * \Delta x (1)$

where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
Solving for k:

$k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)$

Assuming no friction present, total mechanical energy mus keep constant.
When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

$U = \frac{1}{2}* k* (\Delta x)^{2} (3)$

Replacing k and Δx by their values, we get:

$U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)$

When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
So, we can write the following expression:

$\frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2} = \frac{1}{2}*k*\Delta x^{2} (5)$

Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

$\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2} = 4J (6)$

⇒ $\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} = 4J - 3.6 J = 0.4 J (7)$

⇒ $\Delta x_{1} = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)$

6 0

3 years ago

The rotational speed of earth is similar to?

m_a_m_a [10]

Actually, the speed of the earth is the same everywhere, taking the angular speed as the valid measure of the speed

8 0

3 years ago

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is

$M_1=m_av_a+m_bv_b$

Since b is initially at rest

$M_1=m_av_a$

After the collision and being $v'_a$ and $v'_b$ the respective velocities, the total momentum is

$M_2=m_av'_a+m_bv'_b$

Both momentums are equal, thus

$m_av_a=m_av'_a+m_bv'_b$

Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

$v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}$

$v_a=0.8176 m/s$

The initial kinetic energy can be found as (provided puck b is at rest)

$K_1=\frac{1}{2}m_av_a^2$

$K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J$

The final kinetic energy is

$K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2$

$K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J$

The change of kinetic energy is

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

3 0

3 years ago

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

Mama L [17]

A) Time needed: 6.24 s

B) Time needed: 2.86 s

Explanation:

In this part, we are told that the power if the engine is constant. The power of the engine is given by

$P=\frac{W}{t}$

where

W is the work done

t is the time

This means that the power of the engine is proportional to the work done, and therefore, to the kinetic energy of the car:

$P=\frac{\frac{1}{2}mv^2}{t}=const.$

where m is the mass of the car and v its velocity.

SInce power is constant, we can write:

$\frac{\frac{1}{2}mv_1^2}{t_1^2}=\frac{\frac{1}{2}mv_2^2}{t_2}$

where:

$t_1=1.40 s$ is the time the car needs to accelerates to $v_1=28.0 mph$

$t_2$ is the time the car needs to accelerate to $v_2=57.0 mph$

Therefore, solving for $t_2$ ,

$t_2 = \frac{v^2}{u^2}t_1=\frac{57^2}{28^2}(1.40)=6.24 s$

First of all, we have to calculate the acceleration of the car. We can do it using the following equation:

$a=\frac{v-u}{t}$

where:

u = 0 is the initial velocity

$v=28.0 mph \cdot \frac{1609 m/mi}{3600 s/h}=12.5 m/s$ is the final velocity

t = 1.40 s is the time elapsed

Substituting, we find the acceleration:

$a=\frac{12.5-0}{1.40}=8.9 m/s^2$

In this part, we are told that the force exerted by the engine is constant: according to Newton's second law, acceleration is proportional to the force,

$F=ma$