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

A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an ang

koban [17]

Answer:

magnitude of the magnetic field 0.692 T

Explanation:

given data

rectangular dimensions = 2.80 cm by 3.20 cm

angle of 30.0°

produce a flux Ф = 3.10 × $10^{-4}$ Wb

solution

we take here rectangular side a and b as a = 2.80 cm and b = 3.20 cm

and here angle between magnitude field and area will be ∅ = 90 - 30

∅ = 60°

and flux is express as

flux Ф = $\int \vec{B}.d\vec{A}$ .................1

and Ф = BA cos∅ ............2

so B = $\frac{\phi }{Acos\theta }$

and we know

A = ab

B = $\frac{\phi }{abcos\theta }$ ..............3

put here value

B = $\frac{3.10\times 10^{-4} }{2.80 \times 10^{-2}\times 3.20 \times 10^{-2}\times cos60}$

solve we get

B = 0.692 T

8 0

3 years ago

If the coefficient of kinetic friction between tires and dry pavement is 0.84, what is the shortest distance in which you can st

liberstina [14]

Answer:

The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

Explanation:

Given;

coefficient of kinetic friction, μ = 0.84

speed of the automobile, u = 29.0 m/s

To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;

v² = u² + 2ax

where;

v is the final velocity

u is the initial velocity

a is the acceleration

x is the shortest distance

First we determine a;

From Newton's second law of motion

∑F = ma

F is the kinetic friction that opposes the motion of the car

-Fk = ma

but, -Fk = -μN

-μN = ma

-μmg = ma

-μg = a

- 0.8 x 9.8 = a

-7.84 m/s² = a

Now, substitute in the value of a in the equation above

v² = u² + 2ax

when the automobile stops, the final velocity, v = 0

0 = 29² + 2(-7.84)x

0 = 841 - 15.68x

15.68x = 841

x = 841 / 15.68

x = 53.64 m

Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

4 0

3 years ago

Someone please help its a simple power problem.

SOVA2 [1]

Well 200 doubled or (x2)=400 if that’s what it means

7 0

3 years ago

It took David 2 seconds to lift a 5 Newton bag of toys from the floor to the top of the top shelf, which is 3 meters tall. How m

just olya [345]

So power is equal to work over time and work is force times distance, you do 5 times 3 and get 15 dividing by 2 gives us 7.5 W answer c

7 0

3 years ago

A first order reaction, A -> products, has a rate reaction of .00250 Ms-1 when [A] = . 484 M. (a) What is the rate constant,

tamaranim1 [39]

Answer: a) The rate constant, k, for this reaction is $0.00516s^{-1}$

b) No $t_{\frac{1}{2}}$ does not depend on concentration.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$A\rightarrow products$

Given: Order with respect to $A$ = 1

Thus rate law is:

a) $Rate=k[A]^1$

k= rate constant

$0.00250=k[0.484]^1$

$k=0.00516s^{-1}$

The rate constant, k, for this reaction is $0.00516s^{-1}$

b) Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{2.303}{k}\log\frac{100}{50}$

$t_{\frac{1}{2}}=\frac{0.69}{k}$

Thus $t_{\frac{1}{2}}$ does not depend on concentration.

8 0

3 years ago