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4 years ago

What are the main activities involved in studying physics?

1 answer:

7 0

The main activity that is involved in studying of physics is the study of natural laws. The study of physics has to do with many aspects of the universe. Physics majorly looks into the natural laws that operate in the universe and describe how they affect matter in relation to time.

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An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

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

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0

3 years ago

Question 1 of 10

Novay_Z [31]

Answer:

C. 5.6 × 10^11 N/C

Explanation:

The electric field $E$ at a distance $R$ from a charge $Q$ is given by

$E = k\dfrac{Q}{R^2}$

where $k = 9*10^9Nm/C$ is the coulomb's constant.

Now, in our case

$R = 0.0075m$

$Q = 0.0035C$ ;

therefore,

$E = (9*10^9)\dfrac{0.0035C}{(0.0075m)^2}$

$\boxed{E = 5.6*10^{11}N/C.}$

which is choice C from the options given<em> (at least it resembles it).</em>

6 0

3 years ago

100 Points!!!

Kaylis [27]

(1) The image of an object placed further from the lens than the focal point will be upside down and smaller than the object.

(2) When light rays reflect, they bounce back.

(3) Images formed by a concave lens will look magnified.

(4) When light rays enter a different medium, they bend.

<h3>1.0 Object placed further from the lens than the focal point</h3>

The image of an object placed further from the lens than the focal point will be diminished and inverted.

Thus, the correct answer will be "upside down and smaller than the object".

<h3>2.0 What is reflection of light?</h3>

The ability of light to bounce back when it strike a hard surface is known as refection.

<h3>3.0 Image formed by concave lens</h3>

A concave lens is diverging lens is usually virtual, erect and magnified.

<h3>4.0 Refraction of light</h3>

The change in speed of light when it travels from medium to another medium is known as refraction. Refraction is also, the ability of light to bend around obstacles.

Learn more about reflection and refraction of light here: brainly.com/question/1191238

4 0

2 years ago

When a balloon is deflating, why does air leave the balloon?

Gwar [14]

Answer:

<em>When a balloon deflates air moves out of the balloon </em><em>because the pressure inside the balloon is higher than the pressure outside the balloon.</em>

Explanation:

An inflated balloon has a high pressure region on its inside. Gases always move from a region of high pressure to a region of low pressure. When a balloon is inflated its membrane stretches making it even more porous.

The gas molecules inside the balloon easily diffuse out through this membrane. The diffusion rate may differ depending on the type of gas filled inside the balloon and the material of the balloon. For example helium balloon deflates faster than common air balloon.

This is because helium is a light element and can escape easier than gases like nitrogen and oxygen through the porous membrane of the balloon.

6 0

3 years ago

A 90 kg astronaut Travis is stranded in space at a point 12 m from his spaceship. In order to get back to his ship, Travis throw

insens350 [35]

Answer:

Explanation:

This is a recoil problem, which is just another application of the Law of Momentum Conservation. The equation for us is:

$[m_av_a+m_ev_e]_b=[m_av_a+m_ev_e]_a$ which, in words, is

The momentum of the astronaut plus the momentum of the piece of equipment before the equipment is thrown has to be equal to the momentum of all that same stuff after the equipment is thrown. Filling in:

$[(90.0)(0)+(.50)(0)]_b=[(90.0)(v)+(.50)(-4.0)]_a$

Obviously, on the left side of the equation, nothing is moving so the whole left side equals 0. Doing the math on the right and paying specific attention to the sig fig's here (notice, I added a 0 after the 4 in the velocity value so our sig fig's are 2 instead of just 1. 1 is useless in most applications).

0 = 90.0v - 2.0 and

2.0 = 90.0v so

v = .022 m/s This is the rate at which he is moving TOWARDS the ship (negative was moving away from the ship, as indicated by the - in the problem). Now we can use the d = rt equation to find out how long this process will take him if he wants to reach his ship before he dies.

12 = .022t and

t = 550 seconds, which is the same thing as 9.2 minutes

7 0

3 years ago