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melamori03 [73]
3 years ago
5

Suppose that you measure a galaxy's redshift, and from the redshift you determine that its recession velocity is 30,000 (3×10^4)

kilometers per second. According to Hubble's law, approximately how far away is the galaxy?
Physics
1 answer:
anyanavicka [17]3 years ago
5 0

Answer:

1.4 billion light years away

Explanation:

v = Recessional velocity = 30000 km/s[/tex]

H_0 = Hubble constant = \frac{65}{3.2\times 10^6}\ ly

D = Distance to the galaxy

According to Hubble's law

v=H_0D\\\Rightarrow D=\frac{v}{H_0}\\\Rightarrow D=\frac{3\times 10^{4}}{65}\times 3.2\times 10^6\\\Rightarrow D=1476923076.92307\ ly\\\Rightarrow D=1.4\times 10^9\ ly

The galaxy is 1.4 billion light years away

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When the starter motor on a car is engaged, there is a 320 A current in the wires between the battery and the motor. Suppose the
Salsk061 [2.6K]

Answer:

0.5

Explanation:

because it is V uwbsusvegwjosnfvehdbuxbdusndgdghqbwbwbbeuehdbdhdhdhdudhdhdudhdnskowoqllqlqlqnebe

8 0
2 years ago
Describe each of Newton’s Laws of Motion in ice skating. What can you design/develop to improve ice skating?
denis23 [38]

Newton's three laws of motion can be used to describe the motion of the ice skating.

<h3>Newton's first law of motion</h3>

Newton's first law of motion states that an object at rest or uniform motion in a straight line will continue in that state unless it is acted upon by an external force.

  • Based on this law, once the ice skating starts, it will continue endlessly unless external force stops it.

<h3>Newton's second law of motion</h3>

Newton's second law of motion states that the force applied to an object is directly proportional to the product of mass and acceleration of an object.

  • Based on this law, the force applied to the ice skating is equal to the product of mass and acceleration of the ice skating.

<h3>Newton's third law of motion</h3>

This law states that action and reaction are equal and opposite.

  • Based on this law, the force applied to the ice skating is equal in magnitude to the reaction of ice.

Learn more about Newton's law here: brainly.com/question/3999427

3 0
2 years ago
Read 2 more answers
A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 34 m/s when a 60
german

Answer:

34 m/s

Explanation:

m = Mass of glider with person = 680 kg

v = Velocity of glider with person = 34 m/s

m_1 = Mass of glider without person = 680-60 kg

v_1 = Gliders speed just after the skydiver lets go

m_2 = Mass of person = 60 kg

v_2 = Velcotiy of person = 34 m/s

As the linear momentum of the system is conserved

m_1v_1+m_2v_2=mv\\\Rightarrow v_1=\dfrac{mv-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{680\times 34-60\times 34}{680-60}\\\Rightarrow v_1=34\ m/s

The gliders speed just after the skydiver lets go is 34 m/s

3 0
3 years ago
A bicycle wheel has a diameter of 63.0 cm and a mass of 1.75 kg. Assume that the wheel is a hoop with all of the mass concentrat
Masteriza [31]

Answer:

F2 = 834 N

Explanation:

We are given the following for the bicycle;

Diameter; d1 = 63 cm = 0.63 m

Mass; m = 1.75 kg

Resistive force; F1 = 121 N

For the sprocket, we are given;

Diameter; d2 = 8.96 cm = 0.0896 m

Radius; r2 = 0.0896/2 = 0.0448 m

Radial acceleration; α = 4.4 rad/s²

Now moment of inertia of the wheel which is assumed to be a hoop is given by; I = m(r1)²

Where r1 = (d1)/2 = 0.63/2

r1 = 0.315 m

Thus, I = 1.75 × 0.315²

I = 0.1736 Kg.m²

The torque is given by the relation;

I•α = F1•r1 - F2•r2

Where F2 is the force that must be applied by the chain to give the wheel an acceleration of 4.40 rad/s².

Thus;

0.1736 × 4.4 = (121 × 0.315) - (0.0448F2)

>> 0.76384 = 38.115 - (0.0448F2)

>> 0.0448F2 = 38.115 - 0.76384

>> F2 = (38.115 - 0.76384)/0.0448

>> F2 = 833.73 N

Approximately; F2 = 834 N

7 0
2 years ago
A ball of 10kg falls from rest from a height of 150m, Neglating air resistance, calculate its kinetic energy after falling a dis
GaryK [48]

Answer: 3920\ J

Explanation:

Given

mass of ball m=10 kg

It is placed at a height of 150 m

It is dropped from the height and allowed to free fall for 40 m

Velocity acquired by the ball during this fall is given by v^2-u^2=2as

Insert u=0, a=g

\Rightarrow v^2-0=2\times 9.8\times 40\\\Rightarrow v=\sqrt{784}\\\Rightarrow v=28\ m/s

Kinetic energy at this instant

K.E.=\dfrac{1}{2}\times 10\times 28^2\\\\\Rightarrow K.E.=3920\ J

3 0
2 years ago
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