All categories

3 years ago

A 2.03 cm high insect is 1.39 m from a 131 mm focal-length lens. Where is the image? How high is it?

Physics

1 answer:

ryzh [129]3 years ago

4 0

Answer:

image is 14.47 cm behind the lens

height is 2.11 mm

Explanation:

Given data

h = 2.03 cm

p = 1.39 m = 139 cm

focal-length f = 131 mm = 13.1 cm

to find out

Where is the image and How high is it

solution

we know focal length formula that is

1/f = 1/p + 1/q

put here value to find q

1/ 13.1 = 1/ 139 + 1 solbe/ q

q = 14.463066 cm

so image is 14.47 cm behind the lens

and

height is calculate

height / h = - q / p

put here all value

height = -14.47 / 139 × 2.03

height = −0.211324 cm

here -ve sign show image is inverted

so height is 2.11 mm

You might be interested in

What would be new resistance if length of conductor is doubled and thickness is halved <br>

nignag [31]

Answer:

The new resistance comes out to be = 4 times of original resistance .

6 0

3 years ago

The direction of an electric field is the direction (5 points)

vlabodo [156]

The direction of an electric field is determined from the behavior of a positive test charge that is set free in the electric field.This charge moves along a distinct vector showing the direction of the electric field Therefore the answer is b. a positive charge will move in the field.

5 0

3 years ago

A 234.0 g piece of lead is heated to 86.0oC and then dropped into a calorimeter containing 611.0 g of water that initally is at

Vaselesa [24]

Answer: $24.70 ^{\circ}C$

Explanation:

Given

mass of lead piece $m_l=234 gm\approx 0.234 kg$

mass of water in calorimeter $m_w=611 gm\approx 0.611 kg$

Initial temperature of water $T_w=24^{\circ}C$

Initial temperature of lead piece $T_l=24^{\circ}C$

we know heat capacity of lead and water are $125.604 J/kg-k$ and $4.184 kJ/kg-k$ respectively

Let us take $T ^{\circ}C$ be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

$m_lc_l(T_l-T)=m_wc_w(T-T_w)$

$0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)$

$86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)$

$86-T=86.97T-2087.49$

$T=\frac{2173.491}{87.97}=24.70^{\circ}C$

3 0

3 years ago

5. This break-dancer's speed is not changing as he spins on his head, but he is

n200080 [17]

Answer:

the velocity is changing therefore the acceleration is changing too.

Explanation:

8 0

3 years ago

Consider the following objects and their locations. Object a 2-kg object 5 cm above the floor Object b 2-kg object 120 cm above

Nataly [62]

Answer:

The gravitational acceleration is same for all objects.

a = b = c = d

Explanation:

Acceleration due to gravity or gravitational acceleration is the force exerted by Earth on unit mass of an object.

Acceleration due to gravity doesn't depend on the height of the object when the height is object is near to the surface of the Earth. Only when the height is comparable to the radius of the Earth, the value of gravitational acceleration changes.

But for the objects here, the gravitational acceleration is independent of the mass or height of the objects and has a constant value of 9.8 m/s².

Therefore, the gravitational acceleration of all the objects is same.

If 'a', 'b', 'c', and 'd' represent gravitational accelerations of objects 'a', 'b', 'c', and 'd' respectively, then a = b = c = d.

5 0

3 years ago