If the magnitude of b⃗ increases while its direction remains unchanged, how will the magnetic flux through the coil change?

How does the thickness of the lens affects its focal length plz ans this i will mark as brainliest

Oksanka [162]

Hello! :)

The focal length of the lens tells you how far away from the lens a focused image is created, if light rays approaching the lens are parallel. A lens with more “bending power” has a shorter focal length, because it alters the path of the light rays more effectively than a weaker lens. Most of the time, you can treat a lens as being thin and ignore any effects from the thickness, because the thickness of the lens is much less than the focal length. But for thicker lenses, how thick they are does make a difference, and in general, results in a shorter focal length.

Hope I helped and didn’t answer too late!

Good luck and stay COOL!

~ Destiny ^_^

5 0

3 years ago

Krishna wants to measure the mass and volume of a key. Which tools should she use?

Andrei [34K]

Answer:

a balance and a beaker of water

Explanation:

The mass can easily be measured with balance. If you drop the key in a beaker of water, the water inside the beaker will increase and this increases the volume of water that will be equal to volume of key.

8 0

3 years ago

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mixer [17]

Answer:

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5 0

3 years ago

Read 2 more answers

Person is lifting a 250 N dumbbell. The weight is 30 cm from the pivot point of the elbow. What force must be exerted five from

qwelly [4]

Refer to the diagram shown below.

The force, F, is applied at 5 cm from the elbow.

For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N

Answer: 1500 N

4 0

3 years ago

A wheel of radius 25cm has eight spokes. It is mounted on a fixed axle and is rotating at a constant angular speed w. You shoot

spayn [35]

Explanation:

We will assume that the rim of the wheel is also very thin, like the spokes. The distance s between the spokes along the rim is

$s = \frac{1}{8}C = \frac{1}{8}(2\pi)(0.25\:\text{m}) = 0.196\:\text{m}$

The 20-cm arrow, traveling at 6 m/s, will travel its length in

$t = \dfrac{0.2\:\text{m}}{6\:\text{m/s}} = \dfrac{1}{30}\:\text{s}$

The fastest speed that the wheel can spin without clipping the arrow is

$v = \dfrac{s}{t} = \dfrac{0.196\:\text{m}}{\left(\dfrac{1}{30\:\text{s}}\right)} = 5.9\:\text{m/s}$

The angular velocity $\omega$ of the wheel is given by

$\omega = \dfrac{v}{r} = \dfrac{5.9\:\text{m/s}}{0.25\:\text{m}} = 23.6\:\text{rad/s}$

In terms of rev/s, we can convert the answer above as follows:

$23.6\:\dfrac{\text{rad}}{\text{s}}×\dfrac{1\:\text{rev}}{2\pi\:\text{rad}} = 3.8\:\text{rev/s}$

As you probably noticed, I did the calculations based on the assumption that I'm aiming for the edge of the wheel because this is the part of the wheel where a point travels a longer linear distance compared to ones closer to the axle, thus giving the arrow a better chance to pass through the wheel without getting clipped by the spokes. If you aim closer to the axle, then the wheel needs to spin slower to allow the arrow to get through without hitting the spokes.

3 0

2 years ago