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2 years ago

Astar begins as a blank a large cloud of gas and dust

1 answer:

3 0

This is a statement but yes a star forms inside nebulae which are gigantic clouds of gas. stars form inside as the gases own gravity pulls it together after which it becomes large enough to perform fusion and become a star.

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During the fission reaction shown, how did the target nucleus change ?

Zarrin [17]

Answer:

A. The target nucleus split into two nuclei, each with fewer nucleons than the original.

Explanation:

5 0

2 years ago

How would the weight of a 7-kilogram concrete block compared to the weight of a 7-kilogram metal sphere?

Hitman42 [59]

Answer:

C. The block and the sphere would have the same weight.

Explanation:

If both the block and the sphere weigh 7 kilograms then they have the same weight. They may look different, but they both weigh 7 kilograms.

3 0

2 years ago

Read 2 more answers

Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the

Over [174]

Answer:

Part a)

$a= 0.32 m/s^2$

Part b)

$F_c = 3.6 N$

Part c)

$F_c = 5.5 N$

Explanation:

Part a)

As we know that the friction force on two boxes is given as

$F_f = \mu m_a g + \mu m_b g$

$F_f = 0.02(10.6 + 7)9.81$

$F_f = 3.45 N$

Now we know by Newton's II law

$F_{net} = ma$

so we have

$F_p - F_f = (m_a + m_b) a$

$9.1 - 3.45 = (10.6 + 7) a$

$a = \frac{5.65}{17.6}$

$a= 0.32 m/s^2$

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

$F_c - F_f = m_b a$

$F_c = \mu m_b g + m_b a$

$F_c = 0.02(7)(9.8) + 7(0.32)$

$F_c = 3.6 N$

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

$F_p - F_f - F_c = m_b a$

$9.1 - 0.02(7)(9.8) - F_c = 7(0.32)$

$F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)$

$F_c = 5.5 N$

3 0

2 years ago

A solid uniform cylinder of mass 4.1 kg and radius 0.057 m rolls without slipping at a speed of 0.79 m/s. What is the cylinder’s

Amiraneli [1.4K]

Answer:

The cylinder’s total kinetic energy is 1.918 J.

Explanation:

Given that,

Mass = 4.1 kg

Radius = 0.057 m

Speed = 0.79 m/s

We need to calculate the linear kinetic energy

Using formula of linear kinetic energy

$K.E_{l}=\dfrac{1}{2}mv^2$

$K.E_{l}=\dfrac{1}{2}\times4.1\times(0.79)^2$

$K.E_{l}=1.279\ J$

We need to calculate the rotational kinetic energy

$K.E_{r}=\dfrac{1}{2}\times I\omega^2$

$K.E_{r}=\dfrac{1}{2}\times\dfrac{1}{2}\times mr^2\times(\dfrac{v}{r})^2$

$K.E_{r}=\dfrac{1}{4}\times m\times v^2$

$K.E_{r}=\dfrac{1}{4}\times4.1\times(0.79)^2$

$K.E_{r}=0.639\ J$

The total kinetic energy is given by

$K.E=K.E_{l}+K.E_{r}$

$K.E=1.279+0.639$

$K.E=1.918\ J$

Hence, The cylinder’s total kinetic energy is 1.918 J.

8 0

2 years ago

Is there ever a situation where an ant will have more momentum than an elephant? Explain why or why not? Question 2 options: Yes

sergiy2304 [10]

An ant can have more momentum than an elephant when the elephant is standing still.

Answer: A

Explanation

The momentum is the quantification of the movement done by an object.

It is found to be dependent on the mass of the object and the velocity with which it is moving.

In the present case, the ant has negligible mass compared to elephant so the momentum can be more for ant only when the velocity with which the elephant is moving tends to be zero.

As the velocity of elephant will be zero, the momentum of elephant will be zero so in this criteria, the moving ant will be having more momentum compared to elephant with zero velocity.

So an elephant with zero velocity means the elephant is standing still.

Thus, the condition in which the ant will be having more momentum compared to elephant is when the elephant stands still.

6 0

2 years ago

Read 2 more answers