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3 years ago

A small plastic bead has been charged to -14 nC .

Physics

1 answer:

Studentka2010 [4]3 years ago

3 0

Answer:

The magnitude of the acceleration of a proton at a distance of 1.5cm from the bead is 5.6 x10^13m/s².

The magnitude of the acceleration of a proton at a distance of 1.5cm from the bead is 9.8 x10^16m/s².

Explanation:

Newton's second law states that the total sum of the force acting on a particle in motion is equal to the mass of the particle times the acceleration due to the force. So the electric force between the bead and proton is equal to ma. That is,

Fe = kq1*q2/r² = m*a

The proton had a charge of +1.6x10^-19C and a mass of 1.67×10^-27kg

By substituting these values into the equivalent for a we have that the acceleration of the proton at a distance of 1.5cm form the bead is 5.6×10^13m/s²

The proton had a charge of -1.6x10^-19C and a mass of 9.10×10^-31kg

By substituting these values into the equivalent for a we have that the acceleration of the proton at a distance of 1.5cm form the bead is 9.8×10^16m/s²

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A diver wants to jump from a board, the initial height is 10 meters and he wants to reach a horizontal distance of 2 meters. Wha

bulgar [2K]

Answer:

1.4 m/s

Explanation:

The minimum speed will be when the diver's initial velocity is horizontal.

First, find the time it takes for the diver to fall 10 meters.

Given:

Δy = 10 m

v₀ᵧ = 0 m/s

aᵧ = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

10 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 1.43 s

Now find the initial horizontal velocity.

v = (2 m) / (1.43 s)

v = 1.4 m/s

4 0

3 years ago

A wheel with radius 36 cm is rotating at a rate of 19 rev/s.(a) What is the angular speed in radians per second? rad/s(b) In a t

Sedaia [141]

(a) 119.3 rad/s

The angular speed of the wheel is

$\omega= 19 rev/s$

we need to convert it into radiands per second. We know that

$1 rev = 2 \pi rad$

Therefore, we just need to multiply the angular speed of the wheel by this factor, to get the angular speed in rad/s:

$\omega = 19 rev/s \cdot (2\pi rad/rev))=119.3 rad/s$

(b) 596.5 rad

The angular displacement of the wheel in a time interval t is given by

$\theta= \omega t$

where

$\omega=119.3 rad$

and

t = 5 s is the time interval

Substituting numbers into the equation, we find

$\theta=(119.3 rad/s)(5 s)=596.5 rad$

(c) 127.3 rad/s

At t=10 s, the angular speed begins to increase with an angular acceleration of

$\alpha = 1.6 rad/s^2$

So the final angular speed will be given by

$\omega_f = \omega_i + \alpha \Delta t$

where

$\omega_i = 119.3 rad/s$ is the initial angular speed

$\alpha = 1.6 rad/s^2$ is the angular acceleration

$\Delta t = 15 s - 10 s = 5 s$ is the time interval

Solving the equation,

$\omega_f = (119.3 rad/s) + (1.6 rad/s^2)(5 s)=127.3 rad/s$

(d) 616.5 rad

The angle through which the wheel has rotated during this time interval is given by

$\theta = \omega_i \Delta t + \frac{1}{2} \alpha (\Delta t)^2$

Substituting the numbers into the equation, we find

$\theta = (119.3 rad/s)(5 s) + \frac{1}{2} (1.6 rad/s^2) (5 s)^2=616.5 rad$

(e) 222 m

The instantaneous speed of the center of the wheel is given by

$v_{CM} = \omega R$ (1)

where

$\omega$ is the average angular velocity of the wheel during the time t=10 s and t=15 s, and it is given by

$\omega=\frac{\omega_i + \omega_f}{2}=\frac{127.3 rad/s+119.3 rad/s}{2}=123.3 rad/s$

and

R = 36 cm = 0.36 m is the radius of the wheel

Substituting into (1),

$v_{CM}=(123.3 rad/s)(0.36 m)=44.4 m/s$

And so the displacement of the center of the wheel will be

$d=v_{CM} t = (44.4 m/s)(5 s)=222 m$

8 0

3 years ago

An emf is induced in response to a change in magnetic field inside a loop of wire. Which of the following changes would increase

goldenfox [79]

Answer:

changing the magnetic field more rapidly

Explanation:

According to Faraday's law, whenever there is a change in the magnetic lines of force, it leads the production of induced emf. The magnitude of induced emf is proportional to to the rate of change of flux.

Hence if the magnetic field inside a loop of wire is changed rapidly, the magnitude of induced emf increases in accordance with Faraday's law of electromagnetic induction stated above when the magnetic field is changed more rapidly, hence the answer.

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3 years ago

speed of sound is 343 Ms at 20 degrees Celsius. The frequency heard from the sound is 256 Hz. what is the sounds wavelength?

Lina20 [59]

S= 343m/s
F=256Hz

WL= 343ms/256-1
WL=V/F

= 1.339844m

7 0

3 years ago

A 1000-kg car is moving at 30 m/s around a horizontal unbanked curve whose diameter is 0.20 km. What is the magnitude of the fri

omeli [17]

Answer:

4500 N

Explanation:

When a body is moving in a circular motion it will feel an acceleration directed towards the center of the circle, this acceleration is:

a = v^2/r

where v is the velocity of the body and r is the radius of the circumference:

Therefore, a body with mass m, will feel a force f:

f = m v^2/r

Therefore we need another force to keep the body(car) from sliding, this will be given by friction, remember that friction force is given a the normal times a constant of friction mu, that is:

fs = μN = μmg

The car will not slide if f = fs, i.e.

fs = μmg = m v^2/r

That is, the magnitude of the friction force must be (at least) equal to the force due to the centripetal acceleration

fs = (1000 kg) * (30m/s)^2 / (200 m) = 4500 N

7 0

3 years ago

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