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3 years ago

Calculate the force between two small charged spheres having charges of 3×10−7 C and 4×10−7 C placed 20 cm apart.

Physics

1 answer:

xxMikexx [17]3 years ago

3 0

Answer:

27 . 10^-7 or 27/1000

Explanation:

We use the Coulomb Law

k = Coloumb Constant

q1 and q2 are the charges

d is the distance between the spheres

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3 years ago

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3 years ago

An air conditioner running with R-134a on a cycle executed under the saturationdome between the pressure limits of 0.8 MPa and 0

ohaa [14]

Answer:

The COP of the system is = 4.6

Explanation:

Given data

Higher pressure = 1.8 M pa

Lower pressure = 0.12 M pa

Now we have to find out high & ow temperatures at these pressure limits.

Higher temperature corresponding to pressure 1.8 M pa

$T_{H} = 62.9$ °c = 335.9 K

Lower temperature corresponding to pressure 0.2 M pa

$T_{L} = - 10.1$ °c = 262.9 K

COP of the system is given by

$COP = \frac{T_{L} }{T_{H} -T_{L} }$

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3 years ago

A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?

stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, $m_1 = 300\ g = 0.3\ kg$

Initial speed of the cart, $u_1=1.2\ m/s$

Mass of the larger cart, $m_2 = 2\ kg$

Initial speed of the larger cart, $u_2=0$

After the collision,

Final speed of the smaller cart, $v_1=-0.81\ m/s$ (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let $v_2$ is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$m_1u_1+m_2u_2-m_1v_1=m_2v_2$

$v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}$

$v_2=0.301\ m/s$

So, the speed of the large cart after collision is 0.301 m/s.

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vampirchik [111]

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