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3 years ago

Name the two ends of a magnet

Physics

2 answers:

Alisiya [41]3 years ago

8 0

Answer:

north pole and south pole

hope it is right

astra-53 [7]3 years ago

6 0

The positive and the negative side
I hope this helps :)

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Suppose a new star cluster is born with one o star, 10 a stars, 100 g stars, and 1000 m stars. which stellar type dominates the

uranmaximum [27]

Stars are classified by their spectra (the elements that they absorb) and their temperature. There are seven main types of stars. In order of decreasing temperature, O, B, A, F, G, K, and M. Type O stars are the stars with the highest temperature and biggest luminosity. So, if there is a new born star cluster : 1. with one O star,
2. with 10 A stars,
3. with 10 G stars,
4. with 1000 M stars

the stellar type O will dominate the light output from the cluster.

6 0

3 years ago

A force of 50 N stretches a string by 4 cm,calculate the elastic constant.

murzikaleks [220]

Answer:

50/0.04= answer

8 0

3 years ago

Suppose you increase your walking speed from 6s to 12 m/s in a period of 2 s. What is acceleration

ElenaW [278]

As we know that acceleration is given as

$a = \frac{v_f - v_i}{t}$

here we know that

$v_i = 6 m/s$

$v_f = 12 m/s$

$\Delta t = 2 s$

now we will have

$a = \frac{12 - 6}{2}$

$a = 3 m/s^2$

so acceleration will be 3 m/s^2

5 0

3 years ago

A point charge with a charge q1 = 2.30 μC is held stationary at the origin. A second point charge with a charge q2 = -5.00 μC mo

Alla [95]

Answer:

W = 2.74 J

Explanation:

The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.

This is the electrostatic equivalent of the work-energy theorem.

$W = \Delta U = U_2 - U_1$

where the potential energy is defined as follows

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$

Let's first calculate the distance 'r' for both positions.

$r_1 = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} = \sqrt{(0.170 - 0)^2 + (0 - 0)^2} = 0.170~m\\r_2 = \sqrt{(x_2 - x_0)^2 + (y_2 - y_0)^2} = \sqrt{(0.250 - 0)^2 + (0.250 - 0)^2} = 0.353~m$

Now, we can calculate the potential energies for both positions.

$U_1 = \frac{kq_1q_2}{r_1^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.170)^2} = -3.57~J\\U_2 = \frac{kq_1q_2}{r_2^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.3530)^2} = -0.829~J$

Finally, the total work done on the moving particle can be calculated.

$W = U_2 - U_1 = (-0.829) - (-3.57) = 2.74~J$

4 0

3 years ago

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Was Chantal's snack nutritious? What could she have eaten instead to boost

NeTakaya

Can u go into more detail ?

7 0

3 years ago