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3 years ago

What do neutrons and protons have in common? How are they different?

Physics

2 answers:

alexandr1967 [171]3 years ago

8 0

They both can be found in an atom, but nuetrons have no charge, while protons have a positive charge

NARA [144]3 years ago

4 0

this is easy, i've just done this like last year... they are similar because the are both in the nucleus. they are different because the protons have a positive charge and the neutrons have no charge at all.

HOPE THIS HELPS!!

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If a sinusoidal electromagnetic wave with intensity 18 W/m2 has an electric field of amplitude E, then a 36 W/m2 wave of the sam

Neporo4naja [7]

Answer:

The correct option is D

Explanation:

From the question we are told that

The intensity of the first electromagnetic wave is $I = 18 \ W/m^2$

The amplitude of the electric field is $E_{max}_1 =A$

The intensity of the second electromagnetic wave is $I = 36 \ W/m^2$

Generally the an electromagnetic wave intensity is mathematically represented as

$I = \frac{1}{2} * \epsilon_o * c * E_{max}^2$

Looking at this equation we see that

$I \ \ \alpha \ \ E^2_{max}$

=> $\frac{I_1}{I_2} = [ \frac{ E_{max}_1}{ E_{max}_2} ] ^2$

=> $E_{max}_2 = \sqrt{\frac{x}{y} } * E_{max}_1$

=> $E_{max}_2 = \sqrt{\frac{36}{18} } * E$

=> $E_{max}_2 = \sqrt{2 } E$

5 0

3 years ago

You are conducting a cross using Drosophila melanogaster. The results of your cross indicate that the recombination frequency is

Lapatulllka [165]

Answer: Explained below

Explanation: The calculations are not very accurate. The distance would be underestimated because double crossovers are not observed.

4 0

4 years ago

A 0.50-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.9 m on a frictionless horizontal su

Sedaia [141]

Answer:

$\boxed {\boxed {\sf 18 \ m/s}}$

Explanation:

The ball is moving in a circle, so the force is centripetal.

One formula for calculating centripetal force is:

$F_c= \frac{mv^2}r}$

The mass of the ball is 0.5 kilograms. The radius is 1.9 meters. The centripetal force is 85 Newtons or 85 kg*m/s².

$F_c$ = 85 kg*m/s²
m= 0.5 kg
r= 1.9 m

Substitute the values into the formula.

$85 \ kg*m/s^2 = \frac{0.5 \ kg *v^2}{1.9 \ m}$

Isolate the variable v. First, multiply both sides by 1.9 meters.

$(1.9 \ m)(85 \ kg*m/s^2) = \frac{0.5 \ kg *v^2}{1.9 \ m}*1.9 \ m$

$(1.9 \ m)(85 \ kg*m/s^2) = {0.5 \ kg *v^2}$

$161.5 \ kg*m^2/s^2 = 0.5 \ kg*v^2$

Divide both sides by 0.5 kilograms.

$\frac {161.5 \ kg*m^2/s^2}{0.5 \ kg} = \frac{0.5 \ kg*v^2}{0.5 \ kg}$

$\frac {161.5 \ kg*m^2/s^2}{0.5 \ kg} =v^2$

$323 \ m^2/s^2 = v^2$

Take the square root of both sides of the equation.

$\sqrt {323 \ m^2/s^2} =\sqrt{ v^2$

$\sqrt {323 \ m^2/s^2} =v$

$17.9722007556 \ m/s =v$

The original measurements have 2 significant figures, so our answer must have the same.

For the number we found, 2 sig fig is the ones place. The 9 in the tenth place tells us to round the 7 to an 8.

$18 \ m/s =v$

The maximum speed is approximately <u>18 meters per second.</u>

8 0

3 years ago

A luggage handler pulls a suitcase of mass 19.6 kg up a ramp inclined at an angle 24.0 ∘ above the horizontal by a force F⃗ of m

Dvinal [7]

(a) 638.4 J

The work done by a force is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement

Here we want to calculate the work done by the force F, of magnitude

F = 152 N

The displacement of the suitcase is

d = 4.20 m along the ramp

And the force is parallel to the displacement, so $\theta=0^{\circ}$ . Therefore, the work done by this force is

$W_F=(152)(4.2)(cos 0)=638.4 J$

b) -328.2 J

The magnitude of the gravitational force is

W = mg

where

m = 19.6 kg is the mass of the suitcase

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$W=(19.6)(9.8)=192.1 N$

Again, the displacement is

d = 4.20 m

The gravitational force acts vertically downward, so the angle between the displacement and the force is

$\theta= 90^{\circ} - \alpha = 90+24=114^{\circ}$

Where $\alpha = 24^{\circ}$ is the angle between the incline and the horizontal.

Therefore, the work done by gravity is

$W_g=(192.1)(4.20)(cos 114^{\circ})=-328.2 J$

c) 0

The magnitude of the normal force is equal to the component of the weight perpendicular to the ramp, therefore:

$R=mg cos \alpha$

And substituting

m = 19.6 kg

g = 9.8 m/s^2

$\alpha=24^{\circ}$

We find

$R=(19.6)(9.8)(cos 24)=175.5 N$

Now: the angle between the direction of the normal force and the displacement of the suitcase is 90 degrees:

$\theta=90^{\circ}$

Therefore, the work done by the normal force is

$W_R=R d cos \theta =(175.4)(4.20)(cos 90)=0$

d) -194.5 J

The magnitude of the force of friction is

$F_f = \mu R$

where

$\mu = 0.264$ is the coefficient of kinetic friction

R = 175.5 N is the normal force

Substituting,

$F_f = (0.264)(175.5)=46.3 N$

The displacement is still

d = 4.20 m

And the friction force points down along the slope, so the angle between the friction and the displacement is

$\theta=180^{\circ}$

Therefore, the work done by friction is

$W_f = F_f d cos \theta =(46.3)(4.20)(cos 180)=-194.5 J$

e) 115.7 J

The total work done on the suitcase is simply equal to the sum of the work done by each force,therefore:

$W=W_F + W_g + W_R +W_f = 638.4 +(-328.2)+0+(-194.5)=115.7 J$

f) 3.3 m/s

First of all, we have to find the work done by each force on the suitcase while it has travelled a distance of

d = 3.80 m

Using the same procedure as in part a-d, we find:

$W_F=(152)(3.80)(cos 0)=577.6 J$

$W_g=(192.1)(3.80)(cos 114^{\circ})=-296.9 J$

$W_R=(175.4)(3.80)(cos 90)=0$

$W_f =(46.3)(3.80)(cos 180)=-175.9 J$

So the total work done is

$W=577.6+(-296.9)+0+(-175.9)=104.8 J$

Now we can use the work-energy theorem to find the final speed of the suitcase: in fact, the total work done is equal to the gain in kinetic energy of the suitcase, therefore

$W=\Delta K = K_f - K_i\\W=\frac{1}{2}mv^2\\v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(104.8)}{19.6}}=3.3 m/s$

6 0

3 years ago

Give one real-life example of the Doppler Effect and explain what causes this phenomenon. Please write at least 3 complete sente

Elanso [62]

Dipper effect of an oncoming train get louder as it approaches and sound diminishes as it goes away sound traveling

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3 years ago