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Vika [28.1K]
3 years ago
14

Sphere A has an excess of 10 ^ 15 electrons and sphere B has an excess of 10 ^ 13 protons. They are separated by 2.0 m. What is

the force between them?
A. -0.58 N
B. 0.58 N
C. 2.30 N
D. -2.30 N
Physics
1 answer:
klasskru [66]3 years ago
4 0
The answer is B
I used these equations then i putted it together.
Charge = number of ( electron or proton ) x charge of ( electron or proton )
Force = k x (q1 q2)/r²

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The magnitude of each force is 208 N the force on the right is applied at an angle 36° and the mass of the block is 17 kg the co
djyliett [7]

Answer:

<em>11.06m/s²</em>

Explanation:

According to Newtons second law of motion

\sm F_x = ma_x\\F_m - F_f = ma_x\\mgsin \theta - \mu R mgcos \theta = ma_x\\

Given

Mass m = 17kg

Fm = 208N

theta = 36 degrees

g = 9.8m/s²

a is the acceleration

Substitute

208 - 0.148(17)(9.8)cos 36 = 17a

208 - 24.6568cos36 = 17a

208 - 19.9478 = 17a

188.05 = 17a

a = 188.05/17

a = 11.06m/s²

<em>Hence the  the magnitude of the resulting acceleration is 11.06m/s²</em>

6 0
2 years ago
So, RCF, or relative centrifugal force should be equal to RCF = <img src="https://tex.z-dn.net/?f=11%2C18%2Ar%2A%28RPM%2F1000%29
zysi [14]

After plugging all the data into the equation, the result of the relative centrifugal force (RCF)  is measured in terms of g.

<h3>What is relative centrifugal force?</h3>

The relative centrifugal force (RCF) or the g force is the radial force generated by the spinning rotor as expressed relative to the earth's gravitational force.

RCF = ac/g

where;

  • ac is centripetal acceleration
  • g is acceleration due to gravity

RCF = \frac{\omega ^2 r}{g} = 1.118\times 10^{-5} \ (RPM)^2 r = 11.18r\ (RPM/1000)^2

where;

  • r is radius in cm

<h3>For example, </h3>

Find the maximum RCF of the JS-4.2 rotor can be obtained from its maximum speed (4200 rpm) and its rmax (250 mm);

RCF = 11.18 \times 25\ cm \times (\frac{4200 \ RPM}{1000} )^2 = 4,930.3 \times g

Thus, after plugging all the data into the equation, the result is measured in terms of g.

Learn more about relative centrifugal force here: brainly.com/question/26887699

#SPJ1

6 0
1 year ago
An automobile moves with a tangential speed of 50.0m/s around a circular track with a 3200.0m diameter. What is the magnitude of
DENIUS [597]

Answer: 0.031rad/s

Explanation:

Tangential speed = 50.0m/s

Diameter = 3200m

Radius = diameter/2

= 3200/2 = 1600m

Angular speed = tangential speed / radius

= 50/1600

= 0.031rad/s

6 0
3 years ago
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression,
eduard

(a) 4.06 cm

In a simple harmonic motion, the displacement is written as

x(t) = A cos (\omega t + \phi) (1)

where

A is the amplitude

\omega is the angular frequency

\phi is the phase

t is the time

The displacement of the piston in the problem is given by

x(t) = (5.00 cm) cos (5t+\frac{\pi}{5}) (2)

By putting t=0 in the formula, we find the position of the piston at t=0:

x(0) = (5.00 cm) cos (0+\frac{\pi}{5})=4.06 cm

(b) -14.69 cm/s

In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:

v(t) = x'(t) = -\omega A sin (\omega t + \phi) (3)

Differentiating eq.(2), we find

v(t) = x'(t) = -(5 rad/s)(5.00 cm) sin (5t+\frac{\pi}{5})=-(25.0 cm/s) sin (5t+\frac{\pi}{5})

And substituting t=0, we find the velocity at time t=0:

v(0)=-(25.00 cm/s) sin (0+\frac{\pi}{5})=-14.69 cm/s

(c) -101.13 cm/s^2

In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:

a(t) = v'(t) = -\omega^2 A cos (\omega t + \phi)

Differentiating eq.(3), we find

a(t) = v'(t) = -(5 rad/s)(25.00 cm/s) cos (5t+\frac{\pi}{5})=-(125.0 cm/s^2) cos (5t+\frac{\pi}{5})

And substituting t=0, we find the acceleration at time t=0:

a(0)=-(125.00 cm/s) cos (0+\frac{\pi}{5})=-101.13 cm/s^2

(d) 5.00 cm, 1.26 s

By comparing eq.(1) and (2), we notice immediately that the amplitude is

A = 5.00 cm

For the period, we have to start from the relationship between angular frequency and period T:

\omega=\frac{2\pi}{T}

Using \omega = 5.0 rad/s and solving for T, we find

T=\frac{2\pi}{5 rad/s}=1.26 s

4 0
3 years ago
A 111 ‑turn circular coil of radius 2.11 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicu
Ira Lisetskai [31]

Answer:

0.0061 J

Explanation:

Parameters given:

Number of turns, N = 111

Radius of turn, r = 2.11 cm = 0.0211 m

Resistance, R = 14.1 ohms

Time taken, t = 0.125 s

Initial magnetic field, Bin = 0.669 T

Final magnetic field, Bfin = 0 T

The energy dissipated in the resistor is given as:

E = P * t

Where P = Power dissipated in the resistor

Power, P, is given as:

P = V² / R

Hence, energy will be:

E = (V² * t) / R

To find the induced voltage (EMF), V:

EMF = [-(Bfin - Bin) * N * A] / t

A is Area of coil

EMF = [-(0 - 0.669) * 111 * pi * 0.0211²] / 0.125

EMF = 0.83 V

Hence, the energy dissipated will be:

E = (0.83² * 0.125) / 14.1

E = 0.0061 J

7 0
3 years ago
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