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3 years ago

Compare the signs of ƒ for lenses and mirrors.

Physics

2 answers:

ELEN [110]3 years ago

4 0

Answer:

A positive focal point for concave mirrors and convex lenses.

A negative focal point for convex mirrors and concave lenses.

Explanation:

STALIN [3.7K]3 years ago

3 0

Answer:

simple

Explanation:

<h3>CONCAVE MIRRORS AND LENSES</h3>

<h3>f= negative</h3>

<h3>CONVEX MIRRORS AND LENSES</h3><h3 /><h3>f= positive</h3>

<h3>PLEASE FOLLOW ME AND MARK IT BRAINLIEST</h3>

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Consider the following situations: i. A ball moving at speed v is brought to rest ii. The same ball starts at rest and is projec

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3 years ago

A square loop of wire is held in a uniform 0.24 T magnetic field directed perpendicular to the plane of the loop. The length of

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Answer:

Explanation:

Given that,

Magnetic field of 0.24T

B = 0.24T

Field perpendicular to plane i.e 90°

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Since it is decreasing

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When L is 14cm, what is the EMF induced?

L = 14cm = 0.14m

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ε = - dΦ/dt

Where flux is given as

Φ = BA

Where A is the area of the square

A = L²

Then, Φ = BL²

Substituting this into the EMF

ε = - dΦ/dt

ε = - d(BL²)/dt

B is constant

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8 0

4 years ago

An object of mass 20 kg is raised vertically through a distance of 8 m above ground level. If g = 10 m/s2, what is the gravitati

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2 years ago

You observe a hockey puck of mass 0.12 kg, traveling across the ice at speed 18.3 m/sec. The interaction of the puck and the ice

Galina-37 [17]

The stopping distance is 143.1 m

Explanation:

First of all, we have to find the acceleration of the hockey puck. This can be done by using Newton's second law of motion:

$\sum F =ma$

where

$\sum F = F_f = -0.14 N$ is the net force acting on the puck (the force of friction, negative because it acts in a direction opposite to the direction of motion)

m = 0.12 kg is the mass of the puck

a is the acceleration

Solving for a,

$a=\frac{\sum F}{m}=\frac{-0.14}{0.12}=-1.17 m/s^2$

The motion of the puck is a uniformly accelerated motion, therefore we can use the following suvat equation:

$v^2-u^2=2as$

where:

v = 0 is the final velocity (the puck comes to a stop)

u = 18.3 m/s is the initial velocity

$a=-1.17 m/s^2$ is the acceleration

s is the stopping distance

And solving for s, we find

$s=\frac{v^2-u^2}{2a}=\frac{0-(18.3)^2}{2(-1.17)}=143.1 m$

Learn more about accelerated motion:

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#LearnwithBrainly

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3 years ago