All categories

4 years ago

What is meant by power of accommodation??

Physics

1 answer:

musickatia [10]4 years ago

6 0

The ability of the lens of the eye to self adjust its focal length.

You might be interested in

Find the final velocity of a car that accelerates at +2 m/s2 for +4m from an

Stolb23 [73]

Answer:

Explanation:

according to third equation of motion

2as=vf²-vi²

vf²=2as+vi²

vf=√2as+vi²

vf=√2as+vi

vf=√2*2*4+3

vf=√16+3

vf=4+3=7

so final velocity is 7 m/s

5 0

3 years ago

For a particular scientific experiment, it is important to be completely isolated from any magnetic field, including the earth's

Alexus [3.1K]

Answer:

50000 μT

Explanation:

From the given information:

the diameter of the loop = 1.0 mm = 0.001 m

no of turns (N) = 200

current (I) = 0.199 A

radius = d/2 = 0.001/2

= 5 × 10⁻⁴ m

Recall that;

the magnetic field at the centre of circular wire is:

$= \dfrac{\mu I N}{2R}$

$= \dfrac{4 \pi \times 10^{-7} \times 200 \times0.199}{2\times 5\times 10^{-4}}$

= 0.05 T

= 50000 μT

Since the centre of the earth's magnetic field is given to be equal to the magnetic field produced by the wire, then:

the earth's magnetic field = 50000 μT

5 0

3 years ago

The ball will oscillate along the z axis between z=d and z=−d in simple harmonic motion. What will be the angular frequency ω of

Alinara [238K]

Any kind of frequency, including the angular kind, is closely involved with
time. Still, for some unknown reason,you've given us no time information
whatsoever ... a peculiar decision on your part, since we can be sure that
it's right there, inexorably intertwined with the part of the question that you
DID copy and share with us.

Furthermore and moreover, for one with no prior experience with simple
harmonic motion, the many symbols in this question such as ' d ', ' a ',
' << ', ' d₂ ', and ' a₂ ' would be of no help at all to guide him toward a
solution. On the contrary, he would conclude that the question itself
had been posted by some alien life form.

To sum up: Come back and post the drawing that goes along with the
question, make sure you have presented all of the information that the
question includes, and then we'll talk.

5 0

3 years ago

An irregular object of mass 3 kg rotates about an axis, about which it has a radius of gyration of 0.2 m, with an angular accele

Artemon [7]

Answer:

0.06 Nm

Explanation:

mass of object, m = 3 kg

radius of gyration, k = 0.2 m

angular acceleration, α = 0.5 rad/s^2

Moment of inertia of the object

$I = mK^{2}$

I = 3 x 0.2 x 0.2 = 0.12 kg m^2

The relaton between the torque and teh moment off inertia is

τ = I α

Wheree, τ is torque and α be the angular acceleration and I be the moemnt of inertia

τ = 0.12 x 0.5 = 0.06 Nm

6 0

3 years ago

Identical twins, each with mass 61.0 kg, are on ice skates and at rest on a frozen lake, which may be taken as frictionless. Twi

Anna007 [38]

Answer:

Immediately after throwing the backpack away, twin A would be moving away from twin B at approximately $0.630\; \rm m \cdot s^{-1}$ .

Initially, twin B would not immediately be moving. However, after the backpack hits her, she would move away from twin A at approximately $0.526\; \rm m \cdot s^{-1}$ if she held onto the backpack.

Explanation:

Consider this scenario in three steps:

Step one: twin A is carrying the backpack.
Step two: twin A throws the backpack away; the backpack is en route to twin B;
Step three: twin B starts to move after the backpack hits her.

Since all external forces are ignored, momentum should be conserved when changing from step one to step two, and from step two to step three.

In step one, neither twin A nor the backpack is moving. Their initial momentum would be zero. That is:

$p(\text{twin A, step one}) = 0$ .
$p(\text{backpack, step one}) = 0$ .

Therefore:

$p(\text{backpack, step one}) +p(\text{twin A, step one}) = 0$ .

In step two, the backpack is moving towards twin B at $3.20\; \rm m \cdot s^{-1}$ . Since the mass of the backpack is $12.0\; \rm kg$ , its momentum at that point would be:

$\begin{aligned}p(\text{backpack, step two}) &= m \cdot v \\ &= 12.0\;\rm kg \times 3.20\; \rm m \cdot s^{-1} = 38.4\; \rm kg\cdot m \cdot s^{-1} \end{aligned}$ .

Momentum is conserved when twin A throws the backpack away. Hence:

$\begin{aligned}&p(\text{backpack, step two}) +p(\text{twin A, step two}) \\ &= p(\text{backpack, step one}) +p(\text{twin A, step one})\end{aligned}$ .

Therefore:

$p(\text{twin A, step two}) \\ &= p(\text{backpack, step one}) +p(\text{twin A, step one}) - p(\text{backpack, step two}) \\ &= -38.4\; \rm kg \cdot m \cdot s^{-1}\end{aligned}$ .

The mass of twin A (without the backpack) is $61.0\; \rm kg$ . Therefore, her velocity in step two would be:

$\begin{aligned} v(\text{twin A, step two}) &= \frac{p}{m} \\ &= \frac{-38.4\; \rm kg \cdot m \cdot s^{-1}}{61.0\; \rm kg} \approx -0.630\; \rm m \cdot s^{-1}\end{aligned}$ .

Note that while the velocity of the backpack is assumed to be greater than zero, the velocity of twin A here is less than zero. Since the backpack is moving towards twin B, it can be concluded that twin A is moving in the opposite direction away from twin B.

<h3>From step two to step three</h3>

In step two:

$p(\text{twin B, step two}) = 0$ since twin B is not yet moving.
$p(\text{backpack, step two}) = 38.4\; \rm kg \cdot m\cdot s^{-1}$ from previous calculations.

Assume that twin B holds onto the incoming backpack. Thus, the velocity of the backpack and twin B in step three will be the same. Let $v(\text{twin B and backpack, step three})$ denote that velocity.

In step three, the sum of the momentum of twin B and the backpack would thus be:

$\begin{aligned}& m(\text{twin B}) \cdot v(\text{twin B and backpack, step three}) \\ &+ m(\text{backpack}) \cdot v(\text{twin B and backpack, step three})\end{aligned}$ .

Simplify to obtain:

$(m(\text{twin B}) + m(\text{backpack})) \cdot v(\text{twin B and backpack, step three})$ .

Momentum is conserved when twin B receives the backpack. Therefore:

$\begin{aligned}& (m(\text{twin B}) + m(\text{backpack})) \cdot v(\text{twin B and backpack, step three})\\ =&p(\text{twin B, step two}) +p(\text{backpack, step two})\\ =& 38.4\; \rm kg \cdot m\cdot s^{-1} \end{aligned}$ .

Therefore:

$\begin{aligned}& v(\text{twin B and backpack, step three})\\ =&\frac{p(\text{twin B, step two}) +p(\text{backpack, step two})}{m(\text{twin B}) + m(\text{backpack})}\\ =& \frac{38.4\; \rm kg \cdot m\cdot s^{-1}}{61.0\; \rm kg + 12.0\; \rm kg} \approx 0.526\;\rm m\cdot s^{-1} \rm \end{aligned}$ .

In other words, if twin B holds onto the backpack, then (after doing so) she would be moving away from twin A at approximately $0.526\; \rm m \cdot s^{-1}$ .

6 0

3 years ago

What is meant by power of accommodation??​

What is meant by power of accommodation??