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3 years ago

Images produced by a cathode ray tube result from induced emf True or false

Physics

1 answer:

3241004551 [841]3 years ago

8 0

I'm not completely sure but i think it's false

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A hard drive disk rotates at 7200 rpm. The disk has a diameter of 5.1 in (13 cm). What is the speed of a point 6.0 cm from the c

ExtremeBDS [4]

Answer:

The speed will be "3.4×10⁴ m/s²".

Explanation:

The given values are:

Angular speed,

w = 7200 rpm

i.e.,

= $7200 \times \frac{2 \pi}{60}$

= $753.6 \ rad/s$

Speed from the center,

r = 6.0 cm

As we know,

⇒ Linear speed, $v=wr$

On putting the estimated values, we get

$=753.6\times 0.06$

$=45.216 \ m$

Now,

Acceleration on disk will be:

⇒ $a=\frac{v^2}{r}$

$=34074 \ m/s^2$

$=3.4\times 10^4 \ m/s^2$

3 0

3 years ago

Write the untis of following physical quantities electric current , force, frequency, density

Shtirlitz [24]

Answer:

1) electric current - ampere

2) force - joule

3 frequency - hertz

4) density - kilogram per cubic metre

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4 0

3 years ago

Read 2 more answers

1.If you are in a car and go around a corner very fast. What happens to you?

andreyandreev [35.5K]

2.low orbit because any satellite in low orbit will encounter some of earths atmosphere it will slow down a bit and fall into a lower orbit

7 0

3 years ago

Explain what will happen if an object has high momentum or low momentum and a force is applied?

Korvikt [17]

-- If the force is applied in the same direction as the object is moving, then the object's momentum in that direction will increase.

-- If the force is applied in the direction OPPOSITE to the way the object is moving, then the object's momentum will decrease.

-- In either case, the CHANGE in the object's momentum will be

(strength of the force) x (length of time the force is applied) .

This quantity is also called "impulse".

5 0

4 years ago

Two rings of radius 5 cm are 20 cm apart and concentric with a common horizontal x-axis. The ring on the left carries a uniforml

Yanka [14]

Answer:

The electric field due to the right ring at a location midway between the two rings is $2.41\times10^{3}\ V/m$

Explanation:

Given that,

Radius of first ring = 5 cm

Radius of second ring = 20 cm

Charge on the left of the ring = +30 nC

Charge on the right of the ring = -30 nC

We need to calculate the electric field due to the right ring at a location midway between the two rings

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{qx}{(x^2+R^2)^{\frac{3}{2}}}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times30\times10^{-9}\times0.1}{((0.1)^2+(0.2)^2)^{\frac{3}{2}}}$

$E=2.41\times10^{3}\ V/m$

Hence, The electric field due to the right ring at a location midway between the two rings is $2.41\times10^{3}\ V/m$

3 0

4 years ago