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3 years ago

Normally, atoms are electrically neutral because there are__.

Physics

1 answer:

spayn [35]3 years ago

8 0

Answer:

Equal number of protons and electrons

Explanation:

This principle of proton and electron equality was used by J.J. Thompson in the discovery of proton. After he had discovered the electron. He discovered that since atoms are electrically neutral, that means there must be equal number of protons to balance out the electrons present in the atom.

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A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co

Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0

2 years ago

An ac generator with Em = 223 V and operating at 399 Hz causes oscillations in a series RLC circuit having R = 222 Ω, L = 147 mH

Doss [256]

Answer:

Xc= 17.267 Ω, Z= 415.5 Ω, I= 0.537 A

Explanation:

Em = 223 V

f= 300 Hz, R = 222 Ω, L = 147 mH, C = 23.1 μF

Capacitive reactance = Xc=?

Xc= $\frac{1}{2\pi fC}$

Xc=1/2pi *399*23.1*10^-6

Xc= 17.267 Ω

b).

Z= $\sqrt{ R^2 + (Xl - Xc)^2}$

Xl= 2π * f * L

Xl= 2π * 399 * 147 * $10^{-3}$

Xl= 368.5 Ω

Z= $\sqrt{ R^2 + (Xl - Xc)^2}$ = $\sqrt{222^2 + (368.5 - 17.267)^2}$

Z= 415.5 Ω

c).

Current:

I= V / Z= Em / Z

I= 223/415.5

I= 0.537 A

3 0

3 years ago

Cole is riding a sled with initial speed of 5 m/s from west to east. the frictional force of 50 n exists due west. the mass of t

stepan [7]

We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
$F=ma$
where $F=-50 N$ is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
$a= \frac{F}{m}= \frac{-50 N}{100 kg}=-0.5 m/s^2$

Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
$a= \frac{v_f^2-v_i^2}{2d}$
where
$v_f=0$ is the final speed of the sled
$v_i=5 m/s$ is the initial speed
$d$ is the distance covered

By rearranging the equation, we find d:
$d= \frac{v_f^2-v_i^2}{2a}= \frac{-(5 m/s)^2}{2 \cdot (-0.5 m/s^2)}=25 m$

3 0

3 years ago

Can you find the magnetic field based on force? a straight segment of wire 35.0 cm long carrying a current of 1.40 a is in a uni

Lostsunrise [7]

The force exerted by a magnetic field on a wire carrying current is:
$F=ILB \sin \theta$
where I is the current, L the length of the wire, B the magnetic field intensity, and $\theta$ the angle between the wire and the direction of B.

In our problem, the force is F=0.20 N. The current is I=1.40 A, while the length of the wire is L=35.0 cm=0.35 m. The angle between the wire and the magnetic field is $53 ^{\circ}$ , so we can re-arrange the formula and substitute the numbers to find B:
$B= \frac{F}{IL \sin \theta}= \frac{0.20 N}{(1.40 A)(0.35 m)(\sin 53^{\circ})}=0.51 T$

3 0

3 years ago

What is the right answer

DIA [1.3K]

Answer:

Explanation:

6 0

3 years ago