(b) Which statement about beta radiation is true?

2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate

adoni [48]

Answer:

 » Image distance :

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

v is image distance
u is object distance, u is 10 cm
f is focal length, f is 5 cm

${ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}$

 » Magnification :

• Let's derive this formula from the lens formula:

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

» Multiply throughout by fv

${ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}$

• But we know that, v/u is M

${ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = \frac{v}{f} }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = \frac{v}{f} - 1 }}}}$

v is image distance, v is 10 cm
f is focal length, f is 5 cm
M is magnification.

${ \tt{M = \frac{10}{5} - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}$

 » Nature of Image :

Image is magnified
Image is erect or upright
Image is inverted
Image distance is identical to object distance.

4 0

2 years ago

Compare and contrast electric potential energy and electric potential difference? Explain.

forsale [732]

Answer:

Electric Potential Energy:

The energy that is needed to move a charge against an electric firld is called Electric Potential Energy

Electric Potential Difference:

The amount of work done in carrying a unit charge from one point to an other in an electric field is called Electric Potential Difference.

Relation:

Relation between Electric potential and electrical potential energy is given by

$\delta V=\frac{PE}{q}$

Here PE represents Electric potential energy

and $\delta V$ is Electric potential difference

it means electric potential difference is the difference in electric potential energy divided by the charge.

6 0

3 years ago

This speed is measured with respect to the space station the spacecraft was originally launched from. In interstellar space the

valentinak56 [21]

Answer:

15193.62 m/s

Explanation:

t = Time taken = 6.5 hours

u = Initial velocity = 0 (Assumed)

m = Mass of rocket = 1380 kg

F = Thrust force = 896 N

v = Final velocity

a = Acceleration of the rocket

Force

$F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{896}{1380}\\\Rightarrow a=0.6493\ m/s^2$

Equation of motion

$v=u+at\\\Rightarrow v=0+0.6493\times 6.5\times 60\times 60\\\Rightarrow v=15193.62\ m/s$

The velocity of the rocket after 6.5 hours of thrust is 15193.62 m/s

5 0

3 years ago

How to change this to mega and giga prefixes?urgent .thankyou

Akimi4 [234]

$3*10^{8} m = 3*10^{5} km = 3*10^{2} Mm = 3*10^{-1} Gm$

Each increase in the prefix is a division by 1000.

6 0

3 years ago

Before the student releases the cart by cutting the tinsel string, what forces are acting on the cart?

Naya [18.7K]

Answer: TENSION and WEIGHT

Explanation:

Force experienced by the spring is called TENSION while the WEIGHT is the gravitational pull on the body towards the earth surface. Therefore the forces acting on the cart are TENSION and WEIGHT(weight acts downwards (along negative y-axis) while the TENSION upward(along positive y-axis).

5 0

3 years ago