A magnetic field is created by a charged particle that is

Physics

2 answers:

Oxana [17]3 years ago

8 0

C is your answer :) !!!!!

romanna [79]3 years ago

4 0

c is the right answer cause i just took this test

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Please help me! It’s due soon! Please help!

Murrr4er [49]

Answer:

do it yourself lazy h.e

Explanation:

kysc lazy btc

4 0

3 years ago

In this lab you will use a cart and a track to explore Newton's second law of motion. You will vary two different variables and

Savatey [412]

Answer:

Newtons II law:

It is defined as "the net force acting on the object is a product of mass and acceleration of the body" . Also it defines that the "acceleration of an object is dependent on net force and mass of the body".

Let us assume that,a string is attached to the cart, which passes over a pulley along the track. At another end of the string a weight is attached which hangs over the pulley. The hanging weight provides tension in the spring, and it helps in accelerating the cart. We assume that the string is massless and no friction between pulley and the string.

Whenever the hanging weight moves downwards, the cart will accelerate to right side.

For the hanging weight/mass

When hanging weight of mass is m₁ and accelerate due to gravitational force g.

Therefore we can write F = m₁ .g

and the tension acts in upward direction T (negetive)

Now, Fnet = m₁ .g - T

= m₁.a

So From Newtons II law F = m.a

3 0

3 years ago

Read 2 more answers

A mercury thermometer is constructed as

Tamiku [17]

Answer:

The change in height of the mercury is approximately 2.981 cm

Explanation:

Recall that the formula for thermal expansion in volume is:

$\frac{\Delta V}{V_0} =\alpha_V\, \Delta\, T\\\Delta V = V_0\,\, \alpha_V\,\,\Delta C$

from which we solved for the change in volume $\Delta V$ due to a given change in temperature $\Delta T$

We can estimate the initial volume of the mercury in the spherical bulb of diameter 0.24 cm ( radius R = 0.12 cm) using the formula for the volume of a sphere:

$V_0=\frac{4}{3} \pi \, R^3\\V_0=\frac{4}{3} \pi \, (0.12\,cm)^3\\V_0=0.007238\,cm^3$