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kolezko [41]
3 years ago
12

How was hubble important to our understanding of galaxies?

Physics
1 answer:
dusya [7]3 years ago
4 0

Astronomer Hubble was the one that discovered that the universe has many galaxies and that the universe is expanding.  That's why they name the Hubble Telescope under his name.


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A 150.0-kg crate rests in the bed of a truck that slows from 50.0 km/h to a stop in 12.0 s. The coefficient of static friction b
Leokris [45]

By Newton's second law,

• the net force acting vertically on the crate is 0, and

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0   ==>   <em>n</em> = <em>mg</em> = 1470 N

where <em>n</em> is the magnitude of the normal force; and

• the net force acting in the horizontal direction on the crate is also 0, with

∑ <em>F</em> = <em>f</em> - <em>b</em> = 0   ==>   <em>b</em> = <em>f</em> = <em>µn</em> = 0.645 (1470 N) = 948.15 N

where <em>b</em> is the magnitude of the braking force, <em>f</em> is (the maximum) static friction, and <em>µ</em> is the coefficient of static friction. This is to say that static friction has a maximum magnitude of 948.15 N. If the brakes apply a larger force than this, then the crate will begin to slide.

Note that we are taking the direction of the truck's motion as it slows down to be the positive horizontal direction. The brakes apply a force in the negative direction to slow down the truck-crate system, and static friction keeps the crate from sliding off the truck bed so that the frictional force points in the positive direction.

Let <em>a</em> be the acceleration felt by the crate due to either the brakes or friction. Use Newton's second law again to solve for <em>a</em> :

<em>f</em> = <em>ma</em>   ==>   <em>a</em> = (948.15 N) / (150.0 kg) = 6.321 m/s²

With this acceleration, the truck will come to a stop after time <em>t</em> such that

0 = 50.0 km/h - (6.321 m/s²) <em>t</em>   ==>   <em>t</em> ≈ (13.9 m/s) / (6.321 m/s²) ≈ 2.197 s

and this is the smallest stopping time possible.

8 0
2 years ago
A 32.5 g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final temperature of b
lbvjy [14]

Answer:

T = 17.26 ^oC

Explanation:

At thermal equilibrium we have heat given by aluminium must be equal to the heat absorbed by the water

so we will have

Q_1 = Q_2

m_1s_1\Delta T_1 = m_2s_2\Delta T_2

so we will have

32.5(900)(45.8 - T) = 105.3(4186)(T - 15.4)

so we have

(45.8 - T) = 15.1(T - 15.4)

so we have

16.1 T = 277.87

T = 17.26 ^oC

3 0
3 years ago
What determines the quality of a conductor
aksik [14]
<span>number of free electrons present.</span>
8 0
3 years ago
A .422 kg coffee mug rests on a table top 0.63 m above the floor. What is the potential energy of the mug with respect to the fl
kondaur [170]

Answer:

2.61 J

Explanation:

Since potential energy U = mgy where m = mass of object, g = acceleration due to gravity = 9.8 m/s² and y = height of object above the ground.

Now for the coffee mug, m= 0.422 kg and it is 0.63 m on a table, so it is 0.63 m above the ground. Thus, y = 0.63 m.

We compute U

U = mgy

= 0.422 kg × 9.8 m/s² × 0.63 m

= 2.605 J

≅ 2.61 J

So, the potential energy of the mug with respect to the floor is 2.61 J

4 0
3 years ago
A disk shaped grindstone of mass 3.0 kg and radius 8 cm is spinning at 600 rpm. After the power is shut off the frictional torqu
seropon [69]

Answer:

\theta=50\ revolution

Explanation:

It is given that,

Mass of the grindstone, m = 3 kg

Radius of the grindstone, r = 8 cm = 0.08 m

Initial speed of the grindstone, \omega_i=600\ rpm=62.83\ rad/s

Finally it shuts off, \omega_f=0

Time taken, t = 10 s

Let \alpha is the angular acceleration of the grindstone. Using the formula of rotational kinematics as :

\alpha =\dfrac{\omega_f-\omega_i}{t}

\alpha =\dfrac{0-62.83}{10}

\alpha =-6.283\ rad/s^2

Let \theta is the number of revolutions of the grindstone after the power is shut off. Now using the third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta

\theta=\dfrac{\omega_f^2-\omega_i^2}{2\alpha }

\theta=\dfrac{-62.83^2}{2\times -6.283}

\theta=314.15\ radian

\theta=49.99\ revolution

or

\theta=50\ revolution

So, the number of revolutions of the grindstone after the power is shut off is 50.

7 0
3 years ago
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