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sweet [91]
3 years ago
6

The unit of the electric resistance is ???​

Physics
1 answer:
vaieri [72.5K]3 years ago
7 0

Answer:

<h2>Ohms</h2>

Step-by-step-explanation :

Electric resistance

Dimension length :

Squared mass per time cubed , that is , electric current squared.

The SI derived unit of electric resistance is the ohm,(volt per ampere).

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What is the chief benefit of anaerobic exercise?
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A proton and an alpha particle are momentarily at rest at adistance r from each other. They then begin to move apart.Find the sp
Arte-miy333 [17]

Answer:

The unknown quantities are:

E and F

The final velocity of the proton is:

√(8/3) k e^2/(m*r)

Explanation:

Hello!

We can solve this problem using conservation of energy and momentum.

Since both particles are at rest at the beginning, the initial energy and momentum are:

Ei = k (q1q2)/r

pi = 0

where k is the coulomb constant (= 8.987×10⁹ N·m²/C²)

and q1 = e and q2 = 2e

When the distance between the particles doubles, the energy and momentum are:

Ef = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

pf = m1v1 + m2v2

with m1 = m,   m2 = 4m,    v1=vf_p,    v2 = vf_alpha

The conservation momentum states that:

pi = pf      

Therefore:

m1v1 + m2v2 = 0

That is:

v2 = (1/4) v1

The conservation of energy states that:

Ei = Ef

Therefore:

k (q1q2)/r = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

Replacing

      m1 =  m, m2 = 4m, q1 = e, q2 = 2e

      and   v2 = (1/4)v1

We get:

(1/2)mv1^2 = k e^2/r + (1/2)4m(v1/4)^2 =  k e^2/r + (1/8)mv1^2

(3/8) mv1^2 = k e^2/r

v1^2 = (8/3) k e^2/(m*r)

3 0
3 years ago
A diverging lens (f1 = -12.0cm) is located 50.0 cm to the left of a converging lens (f2 = 34.0 cm). A 2.0 cm-tall object stands
pishuonlain [190]

Answer:

The final image relative to the converging lens is 34 cm.

Explanation:

Given that,

Focal length of diverging lens = -12.0 cm

Focal length of converging lens = 34.0 cm

Height of object = 2.0 cm

Distance of object = 12 cm

Because object at focal point

We need to calculate the image distance of diverging lens

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}

\dfrac{1}{v}=\dfrac{1}{-12}-\dfrac{1}{-12}

v=\infty

The rays are parallel to the principle axis after passing from the diverging lens.

We need to calculate the image distance of converging lens

Now, object distance is ∞

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{34}-\dfrac{1}{\infty}

v=34

The image distance is 34 cm right to the converging lens.

Hence, The final image relative to the converging lens is 34 cm.

5 0
3 years ago
An athlete practicing for a track meet ran 400 meter in 50 seconds. What was his average speed?
sattari [20]

Speed= Distance ÷ time

Speed= 400 ÷ 50 =8 m/s

7 0
3 years ago
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