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2 years ago

9

When a horse pulls a cart..

1 answer:

elena-s [515]2 years ago

6 0

C, hopefully this helps

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Two particles with masses m and 3m are moving toward each other along the x axis with the same initial speeds Vi. Particle m is

nignag [31]

Answer:

$v1 = \sqrt{(2)}Vi$

$v2 = \sqrt{(2/3)}Vi$

ANGLE is 35.3 degree celcius

Explanation:

Given data:

mass m and 3m

initial speed Vi

particle with mass m is moving toward left while particle with mass 3m is moving toward right

By using conservation of momentum :

$mVi + 3m(-Vi) = mv1 +3mv2$

$-2mVi = m(v1 + 3v2)$

$-2Vi = v1 + 3v2$

conservation of energy :

$m(Vi^2) + 3m(-Vi^2) = mv1^2 + 3mv2^2$

$4mVi^2 = m(v1^2+3v2^2)$

$4Vi^2 = v1^2+3v2^2$

After collision, particle with mass m moves at right angles, thus by considering conservation of momentum in x & y direction,

x direction : $-2mVi = 3m.v2i$

$-2Vi = 3v2i$

y direction : $0 = m(v1)j+3m(v2)j$

$-v1j = 3v2j$

subsitute these value in energy conservation

$v1 = \sqrt{(2)}Vi$

$v2 = \sqrt{(2/3)}Vi$

$angle = tan^{-1}(\frac{\sqrt{(2)}}{2}) =$ 35.3 degree from x-axis

8 0

2 years ago

True or False. You do not need to stretch each side of your body.

Veronika [31]

Answer:

True

Explanation:

5 0

2 years ago

Read 2 more answers

A long, hollow wire with inner radius a and outer radius b carries a uniform current density J. What is the magnetic field as a

sveta [45]

Answer:

The magnetic field in the region a < r < b is $B=\frac{u_{0}I(r^{2}-a^{2}) }{2\pi r(b^{2}-a^{2}) }$

Explanation:

If we have the a < r < b. The formula of current is:

$J=\frac{I_{total} }{A}$

Where:

A = area enclosed by the loop.

Itotal = total current in loop.

$J=\frac{I}{\pi b^{2}-\pi a^{2} }$

$I_{enclosed} =JA_{enclosed}$

$I_{enclosed} =\frac{I(\pi r^{2}- \pi a^{2})}{\pi b^{2}-\pi a^{2} }$

If we have the Ampere`s law:

$\int\limits^a_b {B} \, ds =u_{0} I_{enclosed} \\2B\pi r=u_{0} (\frac{I(\pi r^{2}-\pi ^{2} }{\pi ^{2}-\pi a^{2} } )\\B=\frac{u_{0}I(r^{2}-a^{2}) }{2\pi r(b^{2}-a^{2}) }$

6 0

2 years ago

A positive point charge (q = +8.65 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 2.97 m. A p

ella [17]

Answer:

$r_B = 1.88 m$

Explanation:

As we know that work done by electric force is given as

$W_e = -q\Delta V$

so here we know that charge is moving from

$r_A = 2.97 m$

to another position

so we will have

$W_{AB} = \frac{kq_1q_2}{r_A} - \frac{kq_1q_2}{r_B}$

$-6.95\times 10^{-9} = (9\times 10^9)(8.65\times 10^{-8})(4.56\times 10^{-11})(\frac{1}{2.97} - \frac{1}{r_B})$

$-0.196 = (\frac{1}{2.97} - \frac{1}{r_B})$

$r_B = 1.88 m$

7 0

3 years ago

Explain why the direction of the south equatorial current changes

Allisa [31]

One scientist believe that the ozone layers of the earth had been weakening and waters current changes directly over 5 to 7 months. hope that helped

8 0

3 years ago