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3 years ago

Which galaxy is the most stretched out?

Physics

2 answers:

azamat3 years ago

5 0

The answer is the oddball spiral galaxy. hope this help

RideAnS [48]3 years ago

4 0

Answer:

this is the anwser

Explanation:

The oddball spiral galaxy, called Messier 66, is one-thirdof the Leo Triplet, a group of three interacting galaxies about 35 millionlight-years from Earth (a light-year is the distance light can cover in ayear).

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What is the motion of the object?

aleksley [76]

Answer:

<em>Thus, the object is accelerating to the left</em>

Explanation:

<u>The Net Force</u>

The net force is the result of adding all the forces as vectors acting on a body.

$\vec F=\vec F_1+\vec F_2+...+\vec F_n$

Each vector can be expressed in its rectangular components Fx and Fy, and the sum is the sum of the rectangular components separately.

Second Newton's law gives the relation between the net force and the acceleration of the body:

$\vec F = m.\vec a$

We can see the acceleration is a vector with the same direction as the net force.

The diagram shows two vertical forces and two horizontal forces.

The vertical forces are acting in opposite directions and with the same magnitude, thus they cancel out, leaving zero net force in the y-axis.

The horizontal forces are opposite and with different magnitudes. Since the force acting to the left (F3) has a greater magnitude than the force acting to the right (F4), there is a net force directed to the left with a magnitude of 60 N - 20 N = 40 N

Thus, the object is accelerating to the left

4 0

3 years ago

In a certain region of space, the electric potential is V(x,y,z)=Axy-Bx^2+Cy , where A,B , and C are positive constants.a) Calcu

laiz [17]

Answer:

a) Eₓ = - A y + 2B x , b) Ey = -Ax –C , c) Ez = 0 , d) The correct answer is 3

Explanation:

The electric field and the electric power are related

E = - dV / ds

a) Let's find the electric field on the x axis

Eₓ = - dV / dx

dV / dx = A y - B 2x

Eₓ = - A y + 2B x

b) calculate the electric field on the y-axis

Ey = - dV / dy

dV / dy = A x + C

Ey = -Ax –C

c) the electric field on the z axis

dv / dz = 0

Ez = 0

.d) at which point the electric field is zero

Since the electric field is a vector quantity all components must be zero

X axis

0 = = - A y + 2B x

y = 2B / A x

Axis y

0 = -Ax –C

.x = -C / A

We substitute this value in the previous equation

.y = 2B / A (-C / A)

.y = 2 B C / A2

The correct answer is 3

6 0

3 years ago

Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi

nadya68 [22]

Answer:

$E = 1.85*10^{12}\frac{N}{C}$

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction, and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

$E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}$

Then the electric field at the point of interest is estimated as:

$E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}$

6 0

3 years ago

An ocean wave traveling in one direction has a wavelength of 1.0 m and a frequency of 1.25 Hz. Take the direction of wave propag

aliina [53]

Answer:

a) $v=1*1.25=1.25\: m/s$

b) $y(x,t)=2sin(2\pi( x-1.25 t)$

c) $y(3,10)=1.73 m$

Explanation:

a) The speed of a wave is given by the following equation:

$v=\lambda f$

Where:

λ is the wavelength

f is the frequency

$v=1*1.25=1.25\: m/s$

b) The harmonic wave has the following equation:

$y=Asin(kx-\omega t)$

A is the amplitude (2 m)

k is the wavenumber (2π/λ)

ω is the angular frequency (2πf)

$y(x,t)=2sin(2\pi x-2\pi*1.25 t)$

$y(x,t)=2sin(2\pi( x-1.25 t)$

c) Here we need to find the heigth at x=3 m and t =10 s, so we need to find y(3,10).

$y(3,10)=2sin(2\pi(3-1.25*10)$

$y(3,10)=1.73 m$

I hope it helps you!

6 0

3 years ago

A 800-gram grinding wheel 27.0 cm in diameter is in the shape of a uniform solid disk. (We can ignore the small hole at the cent

SSSSS [86.1K]

Explanation:

d = Diameter of wheel = 27 cm

r = Radius = $\frac{d}{2}=\frac{27}{2}=13.5\ cm$

m = Mass of wheel = 800 g

$\omega_i$ = Initial angular velocity = $245\times \frac{2\pi}{60}\ rad/s$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-245\times \frac{2\pi}{60}}{50}\\\Rightarrow \alpha=-0.51312\ rad/s^2$