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3 years ago

The solar system is located in the milky way galaxy. what type of galaxy is the milky way?

Physics

2 answers:

Alina [70]3 years ago

7 0

spiral galaxy

It's called a spiral galaxy because if you could view it from the top or bottom, it would look like a spinning pinwheel.

zhuklara [117]3 years ago

4 0

It is a a spiral galaxy !!

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When a ball rolls downhill, the rolling motion results in the ball ____________.

emmainna [20.7K]

I am pretty sure it is A Becoming warm

Since it’s moving and causing friction which makes it warm

Hope this helps

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7 0

3 years ago

Read 2 more answers

A calcium-40 ion has a positive charge that is double the charge of a proton, and a mass of 6.64 ✕ 10−26 kg. At a particular ins

Zina [86]

Answer:

Detailed solution is given below

4 0

3 years ago

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

$E_{pe}\:(elastic\:potential\:energy) = 5184\:J$

$K\:(constant) = 16200\:N/m$

$x\:(displacement) =\:?$

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

$E_{pe} = \dfrac{k*x^2}{2}$

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:

$E_{pe} = \dfrac{k*x^2}{2}$

$5184 = \dfrac{16200*x^2}{2}$

$5184*2 = 16200*x^2$

$10368 = 16200\:x^2$

$16200\:x^2 = 10368$

$x^{2} = \dfrac{10368}{16200}$

$x^{2} = 0.64$

$x = \sqrt{0.64}$

$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

3 0

4 years ago

From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the

Ivahew [28]

Answer:

The launching point is at a distance D = 962.2m and H = 39.2m

Explanation:

It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.

X axis

x = Vox t

t = x / vox

t = 7.1 / 340

t = 2.09 10-2 s

In this same time the height of the window fell

Y = Voy t - ½ g t²

Let's calculate the initial vertical speed, this speed is in the window

Voy = (Y + ½ g t²) / t

Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209

Voy = 27.7 m / s

We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s

Vy = Voy - gt₂

Vy = 0 -g t₂

t₂ = Vy / g

t₂ = 27.7 / 9.8

t₂ = 2.83 s

This is the time it also takes to travel the horizontal and vertical distance

X = Vox t₂

D = 340 2.83

D = 962.2 m

Y = Voy₂– ½ g t₂²

Y = 0 - ½ g t2

H = Y = - ½ 9.8 2.83 2

H = 39.2 m

The launching point is at a distance D = 962.2m and H = 39.2m

6 0

3 years ago

A 50.0-turn circular coil of radius 5.00 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0

Blababa [14]

Answer:

The maximum torque in the coil is $4.9\times 10^{-3}\ N-m$ .

Explanation:

Given that,

Number of turns in the circular coil, N = 50

Radius of coil, r = 5 cm

Magnetic field, B = 0.5 T

Current in coil, I = 25 mA

We need to find the magnitude of the maximum possible torque exerted on the coil. The magnetic torque is given by :

$\tau=NIAB\ \sin\theta$

For maximum torque, $\theta=90^{\circ}$

$\tau=NIAB\\\\\tau=50\times 25\times 10^{-3}\times \pi (0.05)^2\times 0.5\\\\\tau=4.9\times 10^{-3}\ N-m$

So, the maximum torque in the coil is $4.9\times 10^{-3}\ N-m$ .

4 0

3 years ago