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2 years ago

If an object is placed at the center of carvature of a convance mirror the image formed is called

Physics

1 answer:

Natasha_Volkova [10]2 years ago

3 0

Answer:

When the object is placed between centre of curvature and principal focus of a concave mirror the image formed is beyond C as shown in the figure and it is real, inverted and magnified.

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The train 'A' travelled a distance is f 120 km in 3 hours , Whereas another train 'B' travelled a distance of 180 km in 4 hours

Radda [10]

Answer:

B train!

Explanation:

A train travelled at 40 km/hr. (120/3)

B train traveled at 45 km/hr (180/4)

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3 years ago

Read 2 more answers

Which type of heat transfer causes your face to feel warm when you sit in the sun?

MA_775_DIABLO [31]

Answer:

Radiation

Explanation:

The sun radiates energy to the earth to make it warmer near the equator.

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3 years ago

Is there like an actual definition for momentum?

krok68 [10]

Yes there is.

Momentum of an object is (its mass) times (its speed) .

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3 years ago

What are some indicators of energy transformations?

Annette [7]

Explanation:

Contact, vision, sound, flavor, and smell are all markers of energy transformations. The most basic example would be when we notice something has begun to pass through vision. Whenever an entity accelerates or slows down, energy is constantly transformed.

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3 years ago

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

3 years ago