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Anastaziya [24]

3 years ago

12

An object is placed at a certain distance on the principal axis of a concave mirror. The image is formed at a distance of 30cm f

rom the mirror. Find the object distance if radius of curvature R=15cm.

1 answer:

Nikolay [14]3 years ago

6 0

the object distance is 10cm

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At which point does the image appear? <br> W <br> X <br> Y <br> Z

ahrayia [7]

Z I think...........

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3 years ago

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Sladkaya [172]

Answer:

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A two-turn circular wire loop of radius 0.63 m lies in a plane perpendicular to a uniform magnetic field of magnitude 0.219 T. I

Basile [38]

Answer:

The magnitude of the average induced emf in the wire during this time is 9.533 V.

Explanation:

Given that,

Radius r= 0.63 m

Magnetic field B= 0.219 T

Time t= 0.0572 s

We need to calculate the average induce emf in the wire during this time

Using formula of induce emf

$E=-\dfrac{d\phi}{dt}$

$E=-B\dfrac{dA}{dt}$

$E=-B\dfrac{A_{2}-A_{1}}{dt}$

$E=B\dfrac{A_{1}-A_{2}}{dt}$ .....(I)

In reshaping of wire, circumstance must remain same.

We calculate the length when wire is in two loops

$l=2\times 2\pi\times r_{1}$

$l=2\times 2\pi\times 0.63$

$l=7.916\ m$

The length when wire is in one loop

$l=2\pi\times r_{2}$

$7.916=2\times \pi\times r_{2}$

$r_{2}=\dfrac{7.916}{2\times \pi}$

$r_{2}=1.259\ m$

We need to calculate the initial area

$A_{1}=N\times\pi\times r_{1}^2$

Put the value into the formula

$A_{1}=2\times3.14\times(0.63)^2$

$A_{1}=2.49\ m^2$

The final area is

$A_{2}=N\times\pi\times r_{2}^2$

$A_{2}=1\times\pi\times(1.259)^2$

$A_{2}=4.98\ m^2$

Put the value of initial area and final area in the equation (I)

$E=0.219\dfrac{2.49-4.98}{0.0572}$

$E=-9.533\ V$

Negative sign shows the direction of induced emf.

Hence, The magnitude of the average induced emf in the wire during this time is 9.533 V.

6 0

3 years ago

What is a metric unit for energy?

Vilka [71]

Joule (J) is the metric unit for energy

8 0

3 years ago

A student conducts an experiment in which a cart is pulled by a variable applied force during a 2sec time interval. In trial 1,

Harlamova29_29 [7]

Answer:

change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

Explanation:

Let's look for the speed of the car

F = m a

a = F / m

We use kinematics to find lips

v = v₀ + a t

v = v₀ + (F / m) t

The moment is defined by

p = m v

The moment change

Δp = m v - m v₀

Let's replace the speeds in this equation

Δp = m (v₀ + F / m t) - m v₀

Δp = m v₀ + F t - m v₀

Δp = F t

We see that the change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

8 0

3 years ago