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3 years ago

What type of radiation is given off in the nuclear reaction 210 84po → 206 82pb + ?

Physics

1 answer:

PIT_PIT [208]3 years ago

8 0

That is the addition of 4 2He
that's an alpha particle Emmision

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A 90 kg painter is standing on a horizontal wooden scaffolding of length 10 m, which is supported on each end by a rope. The pai

RSB [31]

Explanation:

For equilibrium, $\sum M = 0$ .

So, $8 m \times mg - (10 m) T_{1}$ = 0

$T_{1} = \frac{8 \times mg}{10}$

= $\frac{8 \times 90 \times 9.8}{10}$

= 705.6 N

Also, for equilibrium $\sum F_{y}$ = 0

$T_{1} + T_{2} - mg$ = 0

or, $T_{2} = mg - T_{1}$

= $90 \times 9.8 - 705.6$

= 176.4 N

Thus, we can conclude that the tension in the first rope is 176.4 N.

8 0

4 years ago

A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  r

Hatshy [7]

Answer:

The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

Explanation:

Given that,

Angular acceleration $\alpha=\pi\ rad/s^2$

Angular displacement $\theta=\pi\ rad$

Angular velocity $\omega =2\pi\ rad/s$

We need to calculate the angular velocity at the beginning

Using formula of angular velocity

$\alpha =\dfrac{\omega^2-\omega_{0}^2}{2\theta}$

$\omega_{0}^2=\omega^2-2\alpha\theta$

Where, $\alpha$ = angular acceleration

$\omega$ = angular velocity

Put the value into the formula

$\omega_{0}^2=(2\pi)^2-2\times\pi\times\pi$

$\omega=\sqrt{2\pi^2}$

$\omega_{0}=\pi\sqrt{2}\ rad/s$

Hence, The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

3 0

3 years ago

As a gaseous element condenses, the atoms become ________ and they have ________ attraction for one another.

romanna [79]

Answer:

blank 1: close together

blank 2: more

3 0

1 year ago

A disk-shaped merry-go-round of radius 2.13 m and mass 175 kg rotates freely with an angular speed of 0.651 rev/s. A 55.4 kg per

GREYUIT [131]

Answer:

Explanation:

To find out the angular velocity of merry-go-round after person jumps on it , we shall apply law of conservation of ANGULAR momentum

I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω

I₁ is moment of inertia of disk , I₂ moment of inertia of running person , I is the moment of inertia of disk -man system , ω₁ and ω₂ are angular velocity of disc and man .

I₁ = 1/2 mr²

= .5 x 175 x 2.13²

= 396.97 kgm²

I₂ = m r²

= 55.4 x 2.13²

= 251.34 mgm²

ω₁ = .651 rev /s

= .651 x 2π rad /s

ω₂ = tangential velocity of man / radius of disc

= 3.51 / 2.13

= 1.65 rad/s

I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω

396.97 x .651 x 2π + 251.34 x 1.65 = ( 396.97 + 251.34 ) ω

ω = 3.14 rad /s

kinetic energy = 1/2 I ω²

= 3196 J

8 0

3 years ago

If you were stuck in the center of the bridge when it was swaying, where would be the safest place to walk back to land?

Ludmilka [50]

Straight
You already have to momentum of walking forward, and going back and forth are the same distance. If you go back then you would have to stop, turn and walk, but if you go forward you just have to walk.

4 0

3 years ago