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2 years ago

9

Star X has an apparent magnitude of 1. Star Y has an apparent magnitude of 4. Both stars are in the same star cluster. Which sta

r is more intrinsically luminous and by what factor

1 answer:

sashaice [31]2 years ago

7 0

Answer:

Explanation:

From the given information:

Since both stars are in the same cluster, the magnitude and luminosity relationship can be calculated as:

$m_1 - m_2 = -2.5 log _{10} (\dfrac{L_1}{L_2})$

Given that;

m_1 = 1 and

m_2 = 4

Therefore,

$1 - 4 = -2.5 log _{10} ( \dfrac{L_1}{L_2})$

$3 = -2.5 log _{10} ( \dfrac{L_1}{L_2})$

Making $\dfrac{L_1}{L_2}$ the subject of the formula:

$\implies \dfrac{L_1}{L_2}= 10^{(\dfrac{3}{2.5})}$

=15.84

≅ 16

Hence, we can conclude that star X is more luminous by a factor of 16

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suppose the ball has the smallest possible frequency that allows it to go all the way around the circle. what tension in the str

tatyana61 [14]

The complete question is missing, so i have attached the complete question.

Answer:

A) FBD is attached.

B) The condition that must be satisfied is for ω_min = √(g/r)

C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).

Explanation:

A) I've attached the image of the free body diagram.

B) The formula for the net force is given as;

F_net = mv²/r

We know that angular velocity;ω = v/r

Thus;

F_net = mω²r

Now, the minimum downward force is the weight and so;

mg = m(ω_min)²r

m will cancel out to give;

g = (ω_min)²r

(ω_min)² = g/r

ω_min = √(g/r)

The condition that must be satisfied is for ω_min = √(g/r)

C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).

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3 years ago

tony walks at an average speed of 70m/min from home to school. If the difference between his home and the school is 2100 m, how

finlep [7]

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3 years ago

When you stretch a spring 13 cm past its natural length, it exerts a force of 21

zloy xaker [14]

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3 years ago

Several springs are connected as illustrated below in (a). Knowing the individual springs stiffness k1 = 20 N/m, k2 = 30 N/m, k3

Hatshy [7]

Answer:

The equivalent stiffness of the string is 8.93 N/m.

Explanation:

Given that,

Spring stiffness is

$k_{1}=20\ N/m$

$k_{2}=30\ N/m$

$k_{3}=15\ N/m$

$k_{4}=20\ N/m$

$k_{5}=35\ N/m$

According to figure,

$k_{2}$ and $k_{3}$ is in series

We need to calculate the equivalent

Using formula for series

$\dfrac{1}{k}=\dfrac{1}{k_{2}}+\dfrac{1}{k_{3}}$

$k=\dfrac{k_{2}k_{3}}{k_{2}+k_{3}}$

Put the value into the formula

$k=\dfrac{30\times15}{30+15}$

$k=10\ N/m$

k and $k_{4}$ is in parallel

We need to calculate the k'

Using formula for parallel

$k'=k+k_{4}$

Put the value into the formula

$k'=10+20$

$k'=30\ N/m$

$k_{1}$ ,k' and $k_{5}$ is in series

We need to calculate the equivalent stiffness of the spring

Using formula for series

$k_{eq}=\dfrac{1}{k_{1}}+\dfrac{1}{k'}+\dfrac{1}{k_{5}}$

Put the value into the formula

$k_{eq}=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{35}$

$k_{eq}=8.93\ N/m$

Hence, The equivalent stiffness of the string is 8.93 N/m.

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3 years ago