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3 years ago

Why are the arms of spiral galaxies typically blue in color?

Physics

1 answer:

Rus_ich [418]3 years ago

3 0

Answer:

Stars are forming in the spiral arms so there are high mass, hot, blue stars in the arms.

Explanation:

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How does using heat as a catalyst affect a chemical reaction?

Goryan [66]

The catalyst lowers the activation energy.

6 0

1 year ago

The maximum mass of a white dwarf is ______ times the mass of the sun.

Savatey [412]

Answer:

(A)

Explanation:

about 1.4 times the mass of our sun.

8 0

2 years ago

Can a goalkeeper at his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will

zmey [24]

The goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

Explanation:

Consider the vertical motion of ball,

We have equation of motion v = u + at

Initial velocity, u = u sin θ

Final velocity, v = 0 m/s

Acceleration = -g

Substituting

v = u + at

0 = u sin θ - g t

$t=\frac{usin\theta }{g}$

This is the time of flight.

Consider the horizontal motion of ball,

Initial velocity, u = u cos θ

Acceleration, a =0 m/s²

Time, $t=\frac{usin\theta }{g}$

Substituting

s = ut + 0.5 at²

$s=ucos\theta \times \frac{usin\theta }{g}+0.5\times 0\times (\frac{usin\theta }{g})^2\\\\s=\frac{u^2sin\theta cos\theta}{g}\\\\s=\frac{u^2sin2\theta}{2g}$

This is the range.

In this problem

u = 30 m/s

g = 9.81 m/s²

θ = 45° - For maximum range

Substituting

$s=\frac{30^2\times sin(2\times 45)}{2\times 9.81}=45.87m$

Maximum horizontal distance traveled by ball without touching ground is 45.87 m, which is less than 95 m.

So the goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

6 0

3 years ago

Starting from rest, a solid sphere rolls without slipping down an incline plane. At the bottom of the incline, what does the ang

Marrrta [24]

Answer:

2/R*sqrt (g*s*sin(θ)) = w

Explanation:

Assume:

- The cylinder with mass m

- The radius of cylinder R

- Distance traveled down the slope is s

- The angular speed at bottom of slope w

- The slope of the plane θ

- Frictionless surface.

Solution:

- Using energy principle at top and bottom of the slope. The exchange of gravitational potential energy at height h, and kinetic energy at the bottom of slope.

ΔPE = ΔKE

- The change in gravitational potential energy is given as m*g*h.

- The kinetic energy of the cylinder at the bottom is given as rotational motion: 0.5*I*w^2

- Where I is the moment of inertia of the cylinder I = 0.5*m*R^2

We have:

m*g*s*sin(θ) = 0.25*m*R^2*w^2

2/R*sqrt (g*s*sin(θ)) = w

- The angular velocity depends on plane geometry θ , distance travelled down slope s, Radius of the cylinder R , and gravitational acceleration g

3 0

3 years ago

a large sphere is on a horizontal field on a sunny day. at a certain time the shadow reaches out a distance of 10 m from the poi

Mars2501 [29]

The radius of the sphere in meters is ,r = $10\sqrt{5} -20$

Think about the angle the ground and the shadow make. Since the sun's beams are parallel, the angle created by the stick's shadow is also equal. Since the stick is 1 m high and its shadow is 2 m long, we know that the stick's angle is arctan 1/2. Therefore, by thinking of a right-angled triangle,

r/10 = tan [arctan(1/2)] = tan (1/2)

Since, tan (θ/2) = 1-cos(θ) / sin(θ)

we find that,

r/10 = $\sqrt{5} -2$

Hence, r = $10\sqrt{5} -20$

So, the radius of the sphere in meters is ,r = $10\sqrt{5} -20$

Learn more about radius (r) of the sphere here;

brainly.com/question/14100787

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5 0

1 year ago