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2 years ago

At which point in the life cycle of a star does nuclear fusion begin?

1 answer:

8 0

I would say B : main sequence is the answer . this is the answer i believe because the star will increase in size and than shine brightly and when it's done , it will get smaller turning into nebula , eventually exploding sometime around the last stage , but not the last stage of b , c, or d

i really hope that this helps you a lot.

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A star is rotating at 5.55x10-5 π rad/s, its velocity decelerates at a constant rate of 0.5x10-9 π rad/s2.?

Basile [38]

<h2>The current rotational period of that star is 10.01 hours.</h2>

Explanation:

Given that,

Initial angular velocity of the star, $\omega=5.55\times 10^{-5}\pi \ rad/s$

It decelerates, final angular speed, $\omega_f=0$

Deceleration, $\alpha =-0.5\times 10^{-9}\pi \ rad/s^2$

It is not required to use the rotational kinematics formula. The angular velocity in terms of time period is given by :

$\omega=\dfrac{2\pi}{T}$

T is current rotational period of that star

$T=\dfrac{2\pi}{\omega}$

$T=\dfrac{2\pi}{5.55\times 10^{-5}\pi \ rad/s}$

T = 36036.03 second

1 hour = 3600 seconds

So, T = 10.01 hours

So, the current rotational period of that star is 10.01 hours. Hence, this is the required solution. Hence, this is the required solution.

4 0

2 years ago

An object has a mass of 7g and a volume of 14cm. What is the density

Andreyy89

Density = mass / volume = 7/14 = 0.5!g/cm^3

8 0

2 years ago

Two samples of water are mixed together.

Anon25 [30]

<h3><u>Answer;</u></h3>

A.75°C

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

Let the change in temp of cold water be x degrees,

while that of hot water be 100 - x degrees.

Heat exchange = mcΔt

Ice

Δt = x

m = 0.50 kg

c = 4.18 kJ/kg*°C

Hot water

Δt = 100 - x

m = 1.5 kg

c = 4.18

But;

Heat lost = heat gained

0.50 * c * x = 1.5 * c * (100 - x)

0.50 *x = 1.5*(100 - x)

0.5x = 150 - 1.5x

0.5x + 1.5x = 150 - 1.5x + 1.5x

2x = 150

x = <u>75° C</u>

Hence; the equilbrium temperature will be 75° C

7 0

2 years ago

An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10

Alexeev081 [22]

Answer:

a) $c=1822.3214\ J.kg^{-1}.K^{-1}$

b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c) The material is peat, possibly.

d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.

Explanation:

Given:

mass of aluminium, $m_a=0.1\ kg$

mass of water, $m_w=0.25\ kg$

initial temperature of the system, $T_i=10^{\circ}C$

mass of copper block, $m_c=0.1\ kg$

temperature of copper block, $T_c=50^{\circ}C$

mass of the other block, $m=0.07\ kg$

temperature of the other block, $T=100^{\circ}C$

final equilibrium temperature, $T_f=20^{\circ}C$

We have,

specific heat of aluminium, $c_a=910\ J.kg^{-1}.K^{-1}$

specific heat of copper, $c_c=390\ J.kg^{-1}.K^{-1}$

specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

Using the heat energy conservation equation.

The heat absorbed by the system of the calorie-meter to reach the final temperature.

$Q_{in}=m_a.c_a.(T_f-T_i)+m_w.c_w.(T_f-T_i)$

$Q_{in}=0.1\times 910\times (20-10)+0.25\times 4186\times (20-10)$

$Q_{in}=11375\ J$

The heat released by the blocks when dipped into water:

$Q_{out}=m_c.c_c.(T_c-T_f)+m.c.(T-T_f)$

where

$c=$ specific heat of the unknown material

For the conservation of energy : $Q_{in}=Q_{out}$

so,

$11375=0.1\times 390\times (50-20)+0.07\times c\times (100-20)$

$c=1822.3214\ J.kg^{-1}.K^{-1}$

This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

The material is peat, possibly.

The material cannot be ice because ice doesn't exists at a temperature of 100°C.

7 0

2 years ago

Is directly proportional to the length and resistivity of the conductor.

aliya0001 [1]

Answer:

B) resistance

Explanation:

the resistance of a wire is proportional to its length, and inversely proportional to its cross-sectional area.

8 0

2 years ago

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