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2 years ago

6

A slide with an image 4cm×2cm is placed at a distance of 10 cm behind a converging lens and a clear image is formed on a screen

1.1 m from the slide. the size of the image on the screen is

1 answer:

pav-90 [236]2 years ago

7 0

Answer:

44cm x 22cm

Explanation:

u= 10 cm

v= 1.1 cm

m=v/u= 1.1/10

m=11

hence the size of the image.

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A car is rounding a 100-m-radius curve at 25 m/s.What is the minimum possible coefficient of static friction between the tires a

Crazy boy [7]

Answer:

The minimum possible coefficient of static friction between the tires and the ground is 0.64.

Explanation:

if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :

Fc = f

m×(v^2)/(R) = μ×m×g

(v^2)/(R) = g×μ

μ = (v^2)/(R×g)

= ((25)^2)/((100)×(9.8))

= 0.64

Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.

4 0

3 years ago

A boat cruises 60 meters west in 10 seconds. What is its instantaneous velocity?

kari74 [83]

A - 6 meters a second ( 6m/s)

3 0

3 years ago

Read 2 more answers

A. What is the RMS speed of Helium atoms when the temperature of the Helium gas is 343.0 K? (Possibly useful constants: the atom

kkurt [141]

Answer:

(a) 1462.38 m/s

(b) 2068.13 m/s

Explanation:

(a)

The Kinetic energy of the atom can be given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 10⁻²³ J/k

K.E = Kinetic Energy of atoms = 343 K

T = absolute temperature of atoms

The K.E is also given as:

K.E = (1/2)mv²

Comparing both equations:

(1/2)mv² = (3/2)KT

v² = 3KT/m

v = √[3KT/m]

where,

m = mass of Helium = (4 A.M.U)(1.66 X 10⁻²⁷ kg/ A.M.U) = 6.64 x 10⁻²⁷ kg

v = RMS Speed of Helium Atoms = ?

Therefore,

v = √[(3)(1.38 x 10⁻²³ J/K)(343 K)/(6.64 x 10⁻²⁷ kg)]

<u>v = 1462.38 m/s</u>

(b)

For double temperature:

T = 2 x 343 K = 686 K

all other data remains same:

v = √[(3)(1.38 x 10⁻²³ J/K)(686 K)/(6.64 x 10⁻²⁷ kg)]

<u>v = 2068.13 m/s</u>

8 0

3 years ago

Which statements describe what is occurring at t=5 seconds? Check all that apply

Alex73 [517]

Answer: The object changed directions

The object sped up

Explanation:

7 0

3 years ago

Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated

Artemon [7]

Answer: Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN

Explanation: To find the answer we need to know more about the Newton's law of gravitation.

<h3>What is Newton's law of gravitation?</h3>

Gravitation is the force of attraction between any two bodies.
Every body in the universe attracts every other body with a force.
This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between these two masses.
Mathematically we can expressed it as,

$F=\frac{GMm}{r^2} \\where, G=6.67*10^-11Nm^2kg^-2$

<h3>How to solve the problem?</h3>

Here, we have given with the data's,

$M=8.22*10^9kg\\m=1.38*10^8 kg\\r=1.43*10^3m$

Thus, the force of attraction between these two bodies will be,

$F=6.67*10^-11*\frac{8.22*10^9*1.38*10^8}{1.43*10^3} =52.9kN$

Thus, if two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg and, If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN.

Learn more about the Newton's law of gravitation here:

brainly.com/question/28045318

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6 0

2 years ago