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V125BC [204]
3 years ago
7

2. The nucleus holds two particles:

Physics
1 answer:
noname [10]3 years ago
4 0
Proton and neutrons is the answer
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A computer hard disk starts from rest, then speeds up with an angular acceleration of 190 rad/s2rad/s2 until it reaches its fina
Otrada [13]

Answer:

962 rpm.

Explanation:

given,

angular acceleration = 190 rad/s²

initial angular speed = 0 rad/s

final angular speed = 7200 rpm

                                 =7200\times\dfrac{2\pi}{60}

                                 =754\ rad/s

we need to calculate the revolution of disk after 10 s.

time taken to reach the final angular velocity

    using equation of angular motion

 \omega_f - \omega_i = \alpha t

 754 - 0 =190\times t

    t = 4 s

rotation of wheel in 4 s

\theta =\omega_i t+  \dfrac{1}{2}\alpha t

\theta = \dfrac{1}{2}\alpha t^2

\theta = \dfrac{1}{2}\times 190 \times 4^2

 θ = 1520 rad

 \theta = \dfrac{1520}{2\pi}

 \theta =242\ rev

now, revolution of the disk in next 6 s

angular velocity is constant

\omega_f = \dfrac{\theta_f-\theta_i}{t_f-t_i}

754 = \dfrac{\theta_f-1520}{10-4}

θ_f = 6044 rad

θ_f = \dfrac{6044}{2\pi}

revolution of the computer hard disk

θ_f =  962 rpm.

total revolution of the computer disk after 10 s is equal to 962 rpm.

3 0
3 years ago
A 6 kg table has a net force of 15 N acting on it. What is the magnitude of the acceleration of the table?
beks73 [17]

Answer:

2.5m/s2

Explanation:

F=ma

a=F/m

a=15/6

a=2.5m/s2

8 0
4 years ago
Light with a wavelength of λ = 689 nm. is incident on a single slit of width w = 3.75 micrometers. A screen is located L = 0.65
rodikova [14]

Answer:

119.42 x 10⁻³ m

Explanation:

The distance of first dark fringe from central bright spot will be equal to the width of fringe .

Fringe width = λL /w

where λ is wavelength of light, w is slit width , L is distance of screen .

So required distance

= \frac{689\times10^{-9}\times0.65}{3.75\times10^{-6}}

= 119.42 x 10⁻³ m

8 0
3 years ago
A satellite is spinning at 0.01 rev/s. The moment of inertia of the satellite about the spin axis is 2000 kg · m2. Paired thrust
enyata [817]

Answer:

Explanation:

Angular velocity of satellite

= 2π x .01

= .02 π rad /s

Initial angular momentum

Moment of inertia x angular velocity

= 2000 x .02 π

= 125.6 unit

Linear impulse produced by each thruster

= 15 N.s

Angular impulse

= 15 x 1.5 = 22.5 unit

Total angular impulse in 30 pulses

= 22.5 x 2 x 30

1350

This angular impulse will add total angular momentum of

1350 unit

So total angular momentum after 30 pulses

= 1350 + 22.5

= 1372.5 unit

So  final angular  velocity

= final  angular momentum / moment of inertia

= 1372.5 / 2000

=   0 .686 rad /s

3 0
3 years ago
Use the chart to determine the half-life of carbon-14
lubasha [3.4K]

The correct answer to the question is B) 5,700 years.

EXPLANATION:

Before going to answer this question, first we have to understand half life period of a radio active substance.

The half life period [\ T_{1/2}\ ] of a radio active substance is defined as the time in which half of the radio active specimen has undergone decay.

As per the question, the initial concentration of the radioactive carbon-14 is given as 10,000.

Hence, initial concentration N_{0} =\ 10,000.

After 5,700 years, the remaining radio active specimen is 5000. It means that half of the radio active specimen has undergone decay in 5,700 years.

Hence,  half life period of the radio active carbon-14 specimen is 5,700 years.

7 0
3 years ago
Read 2 more answers
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