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Gnesinka [82]
2 years ago
10

Which of the following statements about globular clusters is true? All stars in the cluster have approximately the same mass. Al

l stars in the cluster are approximately at the same stage in their evolution. Most stars in the cluster are yellow or reddish in color. Most of the stars in the cluster are younger than 10 billion years. There is an approximately equal number of all spectral type stars in the cluster.
Physics
1 answer:
Lilit [14]2 years ago
8 0

Answer:

Most stars in the cluster are yellow or reddish in color. Most

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According to the Law of the Conservation of Matter, if you dissolve 25 grams of sugar into 150 grams of water, the mixture shoul
levacccp [35]
19grams becauce if u
7 0
2 years ago
Which data set has the largest range? A. 55, 57, 59, 60, 61, 49, 48 B. 21, 25, 14, 16, 29, 22, 20 C. 12, 15, 16, 19, 18, 15, 27
Simora [160]
Data D has the largest range.

Data A: 61-48=13
Data B: 29-14=15
Data C:27-12=15
Data D:54-31=23

Therefore, Data D has the largest range.
3 0
3 years ago
An object with total mass mtotal = 14.6 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.9 k
zheka24 [161]

Answer: 1) 0. 2) 4.2 Kg. 3) 15.4 m/s 4) 12.9 m/s 5) 0. 6) 3.62 KJ.

Explanation:

1) Assuming that no external forces act during the collision, total momentum must be conserved. As initially the total mass was at rest, so initial momentum is zero, final momentum of all the system must be 0 also.

2) After the explosion, as mass must be conserved also, the sum of the masses of the three pieces must be equal to the original total mass, so we can write the following:

m₁ + m₂ + m₃ = M = 14.6 Kg = 4.9 Kg + 5.5 Kg + m₃

Solving for m₃, we have:

m₃ = 14.6 Kg - 4.9 Kg -5.5 Kg = 4.2 Kg.

3) and 4)

As momentum is a vector, if it is magnitude must be 0, this means that all his components must be 0 too.

So, we can write two equations, one for the x-component, and other for the y-component, as follows:

pₓ = m₁. v₁ₓ + m₂.v₂ₓ + m₃.v₃ₓ = 0

py = m₁.v₁y + m₂. v₂y + m₃. v₃y =0

Replacing by the values, and solving for v₃ₓ and v₃y, we get:

v₃ₓ = 15.4 m/s

v₃y = 12.9 m/s

v = √(15.4)²+(12.9)² = 20.1 m/s

5) As the center of mass must move as if all the mass were concentrated in this point, and we know that the total momentum must be 0, this tells us that the magnitude of the velocity of the center of mass must be 0 too.

6) As initial kinetic energy is 0, as  the mass was at rest, the increase in the kinetic energy is obtained simply adding the kinetic energy of every piece of mass gained after explosion, as follows:

K = K₁ + K₂ + K₃ = 1/2 (m₁ . v₁² + m₂.v₂² + m₃.v₃²)

Replacing by the values, we get:

K= 3.62 KJ

4 0
3 years ago
a 2000 kg truck is traveling at a velocity of 30 m/s. What velocity must a 1000 kg car have in order to have the same momentum a
Molodets [167]

The car should have a velocity of 60 m/s to attain the same momentum as that of the truck of 2000 kg.

Answer:

Explanation:

Momentum is measured as the product of mass of object with the velocity attained by that object.

Momentum of 2000 kg truck = Mass × Velocity

Momentum of 2000 kg truck = 2000×30 = 60000 N

Similarly, the momentum of 1000 kg car will be 1000× velocity of the 1000 kg car.

Since, it is stated that momentum of 2000 kg truck is equal to the momentum of 1000 kg of car, then the velocity of 1000 kg of car can be determined by equating the momentum of car and truck.

Momentum of 2000 kg truck = Momentum of 1000 kg car

60000=1000×velocity of 1000 kg car

Velocity of 1000 kg car = 60000/1000=60 m/s

So, the car should have a velocity of 60 m/s to attain the same momentum as that of the truck of 2000 kg.

4 0
3 years ago
A cyclotron is to accelerate protons to an energy of 5.4 MeV. The superconduction electromagnet of the cyclotron produces a 2.9T
mart [117]
<h3><u>Answer;</u></h3>

Radius = 0.0818 m

Angular velocity = 2.775 × 10^7 rad/sec

<h3><u>Explanation;</u></h3>

The mass of proton m=1.6748 × 10^-27 kg;  

Charge of electron e= 1.602 × 10^-19 C;  

kinetic energy E= 2.7 MeV

                          = 2.7 × 10^6 × 1.602 × 10^-19 J;

                          = 4.32 × 10^-13 Joules

But; K.E =0.5m*v^2,

Hence v=√(2K.E/m)

Velocity = 2.27 × 10^7 m/s

Angular velocity, ω = v/r

Therefore; V = ωr

Hence; V = √(2K.E/m) = ωr

r= √(2E/m)/w = √E*√(2*m)/(eB)

  = √E * √(2×1.6748×10^-27)/(1.602×10^-19 ×2.9)

but E =  4.32 × 10^-13 Joules

  r = 0.0818 m

Angular speed

Angular velocity, ω = v/r , where r is the radius and v is the velocity

Therefore;

Angular velocity = 2.27 × 10^7 / 0.0818 m

                            = 2.775 × 10^7 rad /sec

3 0
3 years ago
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