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3 years ago

I NEED HELP ASAP

Physics

1 answer:

erastovalidia [21]3 years ago

3 0

Answer:

Let the weight of the worker represented by X newton there is a normal force exerted at a distance of 0.67 cm

Explanation:

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Two straight wires separated by a very small distance run parallel to each other, one carrying a current of 3.0 A to the right a

Lyrx [107]

Answer:

Explanation:

I1 = 3 A right

I2 = 6.8 A left

Let the distance is r from both the wires on the same side.

The formula for the magnetic field is given by

$B =\frac{\mu _{0}}{4\pi }\frac{2i}{r}$

As the direction of current in both the wires is opposite to each other

so the net magnetic field is

B = B2 - B1

$B =\frac{\mu _{0}}{4\pi }\frac{2i_{2}}{r}-\frac{\mu _{0}}{4\pi }\frac{2i_{1}}{r}$

$B =\frac{\mu _{0}}{4\pi }\frac{2\times 6.8}{r}-\frac{\mu _{0}}{4\pi }\frac{2\times 3}{r}$

$B =\frac{\mu _{0}}{4\pi }\frac{7.6}{r}$

8 0

3 years ago

If you have a mineral sample with a volume of 4 ml and a mass of 20 grams , what is the density ?

Vaselesa [24]

Since density would be D=M/V, we’d have to literally divide 20 by 4.
20 divided by 4 is 5, so the density of the mineral sample is 5 grams.
Please correct me if I am wrong!

3 0

3 years ago

PLEASE HELLP ILL GIVE BrAINLY AND HEART

ipn [44]

Answer:

Explanation:

5 0

3 years ago

Help please ............

Solnce55 [7]

Answer:

47c

Explanation:

5 0

3 years ago

A cart of mass 6.0 kg moves with a speed of 3.0 m/s towards a second stationary cart with a mass of 3.0 kg. The carts move on a

IgorLugansk [536]

Answer:1.5

Explanation:

Given

mass of first cart $m_1=6 kg$

initial Velocity $u_1=3 m/s$

mass of second cart $m_2=3 kg$

$u_2=0 m/s$

In the absence of External Force we can conserve momentum

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

$v=\frac{m_1u_1+m_2u_2}{m_1+m_2}$

$v=\frac{6\times 3+3\times 0}{6+3}$

$v=2 m/s$

Final kinetic Energy of two masses

$K.E._2=\frac{1}{2}(m_1+m_2)v^2$

$K.E._2=\frac{1}{2}\cdot (3+6)\cdot (2)^2$

$K.E._2=18 J$

Initial Kinetic Energy

$K.E._1=\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2$

$K.E._1=\frac{1}{2}6\times 3^2+0$

$K.E._1=27 J$

$ratio =\frac{K.E._1}{K.E._2}=\frac{27}{18}=1.5$

5 0

4 years ago