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2 years ago

An experiment is carried out to measure the extension of a rubber band for different loads.

Physics

1 answer:

PtichkaEL [24]2 years ago

3 0

Complete question is;

An experiment is carried out to measure the extension of a rubber band for different loads.

The results are shown in the image attached.

What figure is missing from the table?

Answer:

17.3 cm

Explanation:

The image attached showed values for load, extension and initial length.

Now, the first length there is 15.2 cm and as such it's corresponding extension is 0 because it has no preceding measured length.

The second measured length is 16.2 cm. Since it's initial measured length is 15.2 cm, then the extension has a formula; final length - initial length.

This gives: 16.2 - 15.2 = 1 cm

This corresponds to what is given in the table.

For the next measured length, it is blank but we are given the extension to be 2.1 cm. Now, since the initial measured length is 15.2 cm.

Thus;

2.1 cm = Final length - 15.2 cm

Final length = 15.2 + 2.1

Final length = 17.3 cm

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The magnetic field at the center of a wire loop of radius , which carries current , is 1 mT in the direction (arrows along the w

Citrus2011 [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The magnetic field is $B_{net} = \frac{1}{4} * mT$

And the direction is $-\r k$

Explanation:

From the question we are told that

The magnetic field at the center is $B = 1mT$

Generally magnetic field is mathematically represented as

$B = \frac{\mu_o I}{2R}$

We are told that it is equal to 1mT

$B = \frac{\mu_o I}{2R} = 1mT$

From the first diagram we see that the effect of the current flowing in the circular loop is (i.e the magnetic field generated)

$\frac{\mu_o I}{2R} = 1mT$

This implies that the effect of a current flowing in the smaller semi-circular loop is (i.e the magnetic field generated)

$B_1 = \frac{1}{2} \frac{\mu_o I}{2R}$

and for the larger semi-circular loop is

$B_2 = \frac{1}{2} \frac{\mu_o I}{2 * (2R)}$

Now a closer look at the second diagram will show us that the current in the semi-circular loop are moving in the opposite direction

So the net magnetic field would be

$B_{net} = B_1 - B_2$

$= \frac{1}{2} \frac{\mu_o I}{2R} -\frac{1}{2} \frac{\mu_o I}{2 * (2R)}$

$=\frac{\mu_o I}{4R} -\frac{\mu_o I }{8R}$

$=\frac{\mu_o I}{8R}$

$= \frac{1}{4} \frac{\mu_o I}{2R}$

Recall $\frac{\mu_o I}{2R} = 1mT$

$B_{net} = \frac{1}{4} * mT$

Using the Right-hand rule we see that the direction is into the page which is $-k$

3 0

3 years ago

With the water trap being used, all the possible sources of error would be eliminated.

Phoenix [80]

Answer:

True

Explanation:

7 0

3 years ago

Which elements will bond ionically with barium such that the formula would be written as BaX2? A) Nitrogen, chlorine, and sodium

erma4kov [3.2K]

The answer would be:

B. Chlorine, iodine and Fluorine

Barium has 2 valence electrons. To satisfy the BaX₂ , this would mean that Barium will need to give one of each of its electrons. The elements that need 1 electron would be those that have 7 valence electrons to complete the octet. These elements would fall in group 7 or halogens. Chlorine, iodine and fluorine are all in Group 7, so this would be the best choice.

7 0

3 years ago

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Calculate the elastic potential energy stored in a spring if it has a force constant of 150 N/m. the spring is extended to a len

Alenkinab [10]

Answer:

6.75J

Explanation:

U=1/2KΔx²

U=0.5* 150*0.30^2

4 0

2 years ago

ivann1987 [24]

Answer:

Required energy Q = 231 J

Explanation:

Given:

Specific heat of copper C = 0.385 J/g°C

Mass m = 20 g

ΔT = (50 - 20)°C = 30 °C

Find:

Required energy

Computation:

Q = mCΔT

Q = 20(0.385)(30)

Required energy Q = 231 J

4 0

3 years ago