N Rutherford's gold foil experiment the alpha particles pass through which part of the atom?

How do you know the speed of an electromagnetic wave in a vacuum?

Ivenika [448]

The speed of electromagnetic waves in a vacuum is the same as the speed of light. It can be measured by finding the frequency and wavelength of two different waves, and then by that correlation, the speed of the waveform.
Hope this helps you (:

7 0

4 years ago

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

You and a friend are playing with a bowling ball to demonstrate some ideas of Rotational Physics. First, though, you want to cal

RideAnS [48]

Answer:

K_{total} = 19.4 J

Explanation:

The total kinetic energy that is formed by the linear part and the rotational part is requested

$K_{total} = K_{traslation} + K_{rotation}$

let's look for each energy

linear

$K_{traslation}$ = ½ m v²

rotation

$K_{rotation}$ = ½ I w²

the moment of inertia of a solid sphere is

I = 2/5 m r²

we substitute

$K_{total}$ = ½ mv² + ½ I w²

angular and linear velocity are related

v = w r

we substitute

K_{total} = ½ m w² r² + ½ (2/5 m r²) w²

K_{total} = m w² r² (½ + 1/5)

K_{total} = $\frac{7}{10}$ m w² r²

let's calculate

K_{total} = $\frac{7}{10}$ 6.40 16.0² 0.130²

K_{total} = 19.4 J

6 0

3 years ago

Chang sees the following formula on a website about Newton’s second law.

yarga [219]

The correct answer is the correct answer is A

3 0

3 years ago

Read 2 more answers

Please help on this one?

lisov135 [29]

Answer:

The current in a resistor decreases by a factor of 4.

Explanation:

I = $\frac{E}{R}$

Where (I) is the current, (E) is the voltage and (R) is the resistance.

Decreasing the voltage by 4 gives;

I = 0.25 $\frac{E}{R}$

Since you have multiplied $\frac{E}{R}$ by 0.25,

You have to multiply I by 0.25 so the current decreases by a factor of 4.

3 0

3 years ago