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3 years ago

Please help me this is a test and it needs to be done in a few

Physics

2 answers:

Amanda [17]3 years ago

7 0

Answer:

I think A because astroids are stronger than a crash. Hope I helped! :)

Explanation:

Lelu [443]3 years ago

5 0

Answer:

aaaaa

Explanation:

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How do sex cells differ from other human cells?

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Sex cells have half as many chromosomes as other cells. Sex cells are therefore called haploid.

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3 years ago

If earth squeezed to the size of moon.Calculate the weight of 100kg mass in the squeezed earth.

Ugo [173]

Answer:

833.9kg that's the answer

5 0

3 years ago

QuestionDetails:

sleet_krkn [62]

To solve the two parts of this problem, we will begin by considering the expressions given for gravitational potential energy and finally kinetic energy (to find velocity). From the potential energy we will obtain its derivative that is equivalent to the Force of gravitational attraction. We will start considering that all the points on the ring are same distance:

$r = \sqrt{x^2+R^2}$

Then the potential energy is

$U = \frac{-GMm}{\sqrt{x^2+R^2}}$

PART A) The force is excepted to be along x-axis.

Therefore we take a derivative of U with respect to x.

$F = -\frac{dU}{dx}$

$F = -\frac{d}{dx}(GMm(\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}}))$

$F = \frac{GMmx}{(x^2+R^2)^{3/2}}$

This expression is the resultant magnitude of the Force F.

PART B) The magnitude of loss in potential energy as the particle falls to the center

$U = GMm(\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}})$

According to conservation of energy,

$\frac{1}{2}mv^2 = GMm (\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}})$

$\therefore v = \sqrt{2GM(\frac{1}{R}-\frac{1}{x^2+R^2})}$

7 0

3 years ago

Waves of wavelength 24 cm and amplitude 3.0 em are used to create a standing wave. The distance from one node tothe next in the

NikAS [45]

Answer:

D. 12 cm

Explanation:

A node is a point on a standing wave that does not vibrate.

The nodes of a standing wave are shown in the following sketch.

The red dots are the nodes of the standing wave.

It is observed that the distance between two adjacent nodes is half the wavelength of the wave.

Therefore, if the wavelength of the wave is 24 cm, then the distance from one node to the net must be 24 / 2 = 12 cm.

Hence, choice D is the correct answer.

7 0

1 year ago

A 0.050-kg lump of clay moving horizontally at 12 m/s strikes and sticks to a stationary 0.15-kg cart that can move on a frictio

valina [46]

Answer:

Explanation:

It is given that,

Mass of lump, m₁ = 0.05 kg

Initial speed of lump, u₁ = 12 m/s

Mass of the cart, m₂ = 0.15 kg

Initial speed of the cart, u₂ = 0

The lump of clay sticks to the cart as it is a case of inelastic collision. Let v is the speed of the cart and the clay after the collision. As the momentum is conserved in inelastic collision. So,

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$v=\dfrac{0.05\ kg\times 12\ m/s+0}{0.05\ kg+0.15\ kg}$

v = 3 m/s

So, the speed of the cart and the clay after the collision is 3 m/s. Hence, this is the required solution.

3 0

3 years ago

Please help me this is a test and it needs to be done in a few​

Please help me this is a test and it needs to be done in a few