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3 years ago

Why are we not able to see immediate after we enter a dark hall from a lighter place.

1 answer:

3 0

Technically this is a Biology question;

The 'amount' we can see depends on how much light can get through our pupil to hit our retina.
When there is a lot of light the pupil is small; it doesn't need to be big to let a lot of light in.
When we move to a dark space there is much less light, so the pupil 'dilates' to let enough light so we can see properly.
The period in which one cant see is simply when the pupil hasn't had time to change shape yet so doesn't let in enough light.<span />

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Find initial velocity and time in air

kkurt [141]

Answer:

The answer to your question is: v = 121.46 m/s

Explanation:

Data

h = 30 m

vo = ?

dx = 300 m

Formula

t = $\sqrt{\frac{2h}{g}}$

speed = distance / time

Process

$t = \sqrt{\frac{2(30)}{9.81}} \\$

t= 2.47 s

speed = 300 / 2.47

speed = v = 121.46 m/s

6 0

3 years ago

Choose the correct statement below that accurately describes the shear and normal stresses in a beam. A. Shear stresses are maxi

Scorpion4ik [409]

Answer:

A. Shear stresses are maximum at the neutral axis and normal stresses are maximum furthest from the neutral axis.

Explanation:

Normal stress :

Normal stress is defined as the stress or the restoring force that occurs on the plane when an external axial load is applied on it. For a beam the normal stress is maximum at the point furthest from the neutral axis and is zero at the neutral axis of the beam.

Shear stress :

Shear stress is a stress which occurs when the force acts on the surface of the member in a parallel direction. It changes the shape of the member. For a beam, the shear stress is maximum at the neutral axis.

6 0

3 years ago

A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x

DedPeter [7]

Answer:

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is $\lambda$

Total charge, Q = 7 nC = $7\times 10^{- 9} C$

At x = 2L,

Electric field, $\vec{E_{2L}} = 500N/C$

Coulomb constant, K = $8.99\times 10^{9} N.m^{2}/C^{2}$

Now, we know that:

$\vec{E} = K\frac{Q}{x^{2}}$

Also the line charge density:

$\lambda = \frac{Q}{L}$

Thus

Q = $\lambda L$

Now, for small element:

$d\vec{E} = K\frac{dq}{x^{2}}$

$d\vec{E} = K\frac{\lambda }{x^{2}}dx$

Integrating both the sides from x = L to x = 2L

$\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx$

$\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]$

$\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}$

Similarly,

For the field in between the range 2L< x < 3L:

$\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx$

$\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]$

$\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}$

Now,

If at x = 2L,

$\vec{E_{2L}} = 500 N/C$

Then at x = 3L:

$\frac{\vec{E_{2L}}}{3} = \frac{500}{3} = 166.67 N/C$

4 0

4 years ago

Which of the following factors influence the electrical resistance of a circuit?

stealth61 [152]

-- material and thickness of the interconnecting wires;

-- number, type, and configuration of all the devices
between the in- and out-terminals of the circuit ...
the points that connect to the battery;

-- quality of the connections at the points where
circuit devices connect to wires or to each other.

5 0

3 years ago

Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Us

EleoNora [17]

3) Earth is about 150 million km from the Sun, and the apparent brightness of the Sun in our sky is about 1,300 watts per square meter. Determine the apparent brightness we would measure for the Sun if we were located five times Earth's distance from the Sun. Answer: The Sun would appear 1/25 times as bright.

4 0

4 years ago

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