An atom of tin has an atomic number of 50 and a mass number of 119. How many protons, electrons, and neutrons are found in one neutral atom of tin?
O 50 protons, 69 electrons, 50 neutrons
O 50 protons, 50 electrons, 69 neutrons
69 protons, 50 electrons, 69 neutrons
69 protons, 69 electrons, 50 neutrons

Answer:

Farsighted

Explanation:

Farsightedness also known as hypermetropia is caused by the eye being too short(the eye shortens with advancing age) or the crystalline lines not being sufficiently convergent.

A farsighted person can see far objects but can not see nearby objects. His near point is now farther than the 25 cm near point of a normal eye. Images are formed some distance behind the retina.

This eye defect is corrected by the use of a converging lens to reduce the divergence of the rays entering the eye from an object.

Answer:

Continental polar ( cp):

Explanation:

Cold and dry, originating from high latitudes, typically as air flowing out of the polar highs. This air mass often brings the rattleing cold, dry and clear weather on a perfect winter day and also dry and warm weather on a pleasant day in summer.

Answer:

5.0/s

Explanation:

Answer:

b and a it is this that abewsr

Answer:

According to "http://ww2010.atmos.uiuc.edu"  Diffraction is the slight bending of light as it passes around the edge of an object.

Some examples of Light Defraction would be..

-CD reflecting rainbow colours

-Sun appears red during sunset

-From the shadow of an object

Answer:

x = 0.056 m

ΔKE = 0.489 J

Explanation:

Given that

Angle, θ = 38°

Length, L = 1.7 m

Mass, m = 0.09 kg

Spring constant, K = 590 N/m

If we use the Work-Energy theorem, then we know that Potential Energy, PE = Kinetic Energy, KE

This is mathematically written as

1/2kx² = mgH

The height, H we can get by using the relation

H = L.Sinθ

H = 1.7 * Sin 38

H = 1.7 * 0.6157

H = 1.047 m

Next, we use the Work-Energy theorem

1/2kx² = mgH

1/2 * 590 * x² = 0.09 * 9.8 * 1.047

295 * x² = 0.9234

x² = 0.9235 / 295

x² = 0.00313

x = √0.00313

x = 0.056 m

If the spring loses contact at x = 0.056, definitely, it will also lose contact at x = 0.8

Then we use the formula

ΔKE = mg(H - H1)

ΔKE = mg(xsinθ - x2.sinθ)

Where, x = 1.7 , x2 = 0.8

ΔKE = 0.09 * 9.8 (1.7 * sin 38 - 0.8 * sin 38)

ΔKE = 0.882(1.047 - 0.493)

ΔKE = 0.882 * 0.554

ΔKE = 0.489 J

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Answer:

3 m/s

Explanation:

A= F/m

12,000/ 4000 = 3

Answer:

3 m/s^2

Explanation:

The equation you have to use is F=ma because the problem is a Newton's 2nd law problem.

Our known values are:

F ( Force ) = 12,000 N

m ( mass ) = 4,000 kg

a ( acceleration ) = ?

Now we plug in the known values into the equation and solve

F=ma

12,000=4,000a

We have to divide 4,000 by both sides to isolate the a value

12,000/4,000=4,000/4,000a

The 4,000s on the right of the equation cancel.

And 12,000 divided by 4,000 equals 3

The acceleration (a) is 3 meters per second squared (m/s^2)

Next, check to make sure 3 does work by plugging it back into the equation.

12,000=4,000*3

12,000=12,000 ✔

As you can see, the acceleration will be 3 m/s^2

Answer: B

Explanation:

Answer:

[tex]23.58^{\circ}[/tex] and [tex]53.13^{\circ}[/tex]

[tex]44.43^{\circ}[/tex], second order does not exist

Explanation:

n = Number of lines grating = [tex]1\times10^4\ \text{Lines/cm}[/tex]

[tex]\lambda[/tex] = Wavelength

m = Order

Distance between slits is given by

[tex]d=\dfrac{1}{n}\\\Rightarrow d=\dfrac{1}{1\times 10^4}\\\Rightarrow d=10^{-6}\ \text{m}[/tex]

[tex]\lambda=400\ \text{nm}[/tex]

m = 1

We have the relation

[tex]d\sin\theta=m\lambda\\\Rightarrow \theta=\sin^{-1}\dfrac{m\lambda}{d}\\\Rightarrow \theta=\sin^{-1}\dfrac{1\times 400\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=23.58^{\circ}[/tex]

m = 2

[tex]\theta=\sin^{-1}\dfrac{2\times 400\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=53.13^{\circ}[/tex]

The first and second order angles for light of wavelength 400 nm are [tex]23.58^{\circ}[/tex] and [tex]53.13^{\circ}[/tex].

[tex]\lambda=700\ \text{nm}[/tex]

m = 1

[tex]\theta=\sin^{-1}\dfrac{1\times 700\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=44.43^{\circ}[/tex]

m = 2

[tex]\theta=\sin^{-1}\dfrac{2\times 700\times 10^{-9}}{10^{-6}}[/tex]

Here [tex]\dfrac{2\times 700\times 10^{-9}}{10^{-6}}=1.4>1[/tex] so there is no second order angle for this case.

The first order angle for light of wavelength 700 nm are [tex]44.43^{\circ}[/tex].

Second order angle does not exist.

Answer:

B. in the direction of the force

Explanation:

Sana nakatulong

Answer:

Q = 636.48 J

Explanation:

Given that,

The mass of gold, m = 0.052 kg

The temperature increase from 30°C to 120°C.

The specific heat of gold is 136  J/kg °C.

We need to find the heat needed to warm the gold. The formula for heat needed is given by :

[tex]Q=mc\Delta T\\\\Q=0.052\times 136\times (120-30)\\\\Q=636.48\ J[/tex]

So, 636.48 J of heat is needed to warm gold.

Answer:

A) car - 37500 kg*m/s, minivan - 13332 kg*m/s

B) 50832 kg*m/s

C) 18.83 m/s

Explanation:

Realize that sticky collisions are modeled by: m1v1+m2v2=(m1+m2) vf

conevert to m/s....car going 25 m/s, minivan going 11.11 m/s

A) p=mv

p(car)=(1500)(25)

p(car)=37500 kg*m/s

p(minivan)=(1200)(11.11)

p(minivan)=13332 kg*m/s

B) 37500+13332=50832 kg*m/s

C) 37500+13332=(1500+1200) vf

50832=2700(vf)

18.83 m/s = vf

Answer:

the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

Explanation:

Given;

diameter of the ceiling fan, d = 90 cm = 0.9 m

angular speed of the fan, ω = 64 rpm

time taken for the fan to stop, t = 28 s

The distance traveled by the ceiling fan when it comes to a stop is calculated as;

[tex]d = vt = \omega r\times t= ( \frac{64 \ rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1 \min}{60 \ s} \times 0.9 \ m) \times 28 \ s\\\\d = 168.89 \ m[/tex]

The speed of the tip of a blade 10 s after the fan is turned off is calculated as;

[tex]v = \frac{d}{t} \\\\v = \frac{168.89}{10} \\\\v = 16.889 \ m/s[/tex]

Therefore, the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

Answer: The daughter is named Susie.

Explanation: LIL SUSIE!!!

                      HUH? DIDN'T UNDERSTAND THE QUESTION!

                                        HAVE A GREAT DAY!!!!!

Answer:6/3 Li

Explanation:

I’m not sure what the person under me is talking about but yeah

Answer:

The angular speed is 23.24 rad/s.

Explanation:

Given;

mass of the disk, m = 7 kg

radius of the disk, r = 0.2 m

applied force, F = 42 N

distance moved by disk, d = 0.9 m

The torque experienced by the disk is calculated as follows;

τ = F x d = I x α

where;

I is the moment of inertia of the disk = ¹/₂mr²

α is the angular acceleration

F x r = ¹/₂mr² x α

The angular acceleration is calculated as;

[tex]\alpha = \frac{2Fr}{mr^2} \\\\\ \alpha = \frac{2F}{mr}\\\\\alpha = \frac{2 \times 42 }{7 \times 0.2} \\\\\alpha = 60 \ rad/s^2[/tex]

The angular speed is determined by applying the following kinematic equation;

[tex]\omega _f^2 = \omega_i ^2 + 2\alpha \theta[/tex]

initial angular speed, ωi = 0

angular distance, θ = d/r = 0.9/0.2 = 4.5 rad

[tex]\omega _f^2 = 2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta} \\\\\omega _f = \sqrt{2 \times 60 \times 4.5} \\\\\omega _f = 23.24 \ rad/s[/tex]

Therefore, the angular speed is 23.24 rad/s.

Answer:

320 J

Explanation:

From the question,

KE = 1/2mv².................. Equation 1

Where KE = Kinetic Energy, m = mass of the rock, v = velocity of the rock

Given: m = 40 kg, v = 4m/s

Substitute these values into equation 1

KE = 1/2(40)(4²)

KE = 20×16

KE = 320 J

Hence the kinetic energy of the rock is 320 J

Answer:

W = 1,049 10⁹ J

Explanation:

Work is defined by the relation

         W = F. d = F d cos θ

where tea is the angle between the forces and the displacement.

The total work is the sum of the work of each tug.

Tug 1

       W₁ = F d cos θ₁

the angle measured from the positive side of the x-axis is

       θ₁ = 14 + 90 = 104º

tugboat 2

             W₂ = F d cos θ₂

             θ₂ = 14

we substitute

             W = F d cos θ₁ + F d cos θ₂

             W = F d (cos θ₁ + cos θ₂)

let's calculate

             W = 1.80 10⁶  800 (cos 104 + cos 14)

             W = 1,049 10⁹ J

Answer:

Time = 0.58 seconds

Explanation:

Given the following data;

Initial momentum = 3 kgm/s

Final momentum = 10 kgm/s

Force = 12 N

To find the time required for the change in momentum;

First of all, we would determine the change in momentum.

[tex] Change \; in \; momentum = final \; momentum - initial \; momentum [/tex]

[tex] Change \; in \; momentum = 10 - 3 [/tex]

Change in momentum = 7 kgm/s

Now, we can find the time required;

Note: the impulse of an object is equal to the change in momentum experienced by the object.

Mathematically, impulse (change in momentum) is given by the formula;

[tex] Impulse = force * time [/tex]

Making "time" the subject of formula, we have;

[tex] Time = \frac {impulse}{force} [/tex]

Substituting into the formula, we have;

[tex] Time = \frac {7}{12} [/tex]

Time = 0.58 seconds

Answer:

When R1 = 2.193, R2 = 3.894

When R1 = 3.894, R2 = 2.193

Explanation:

We are told that when R1 and R2 are connected in series, the voltage is 12.6 V and the current is 2.07 A.

Formula for resistance is;

R = V/I

R = 12.6/2.07

R = 6.087 ohms

Since R1 and R2 are connected in series.

Thus; R1 + R2 = 6.087 ohms

R1 = 6.087 - R2

We are also told that when they are connected in parallel, the current is 8.98 A.

Thus, R = 12/8.98

R = 1.403 ohms

Thus;

(1/R1) + (1/R2) = 1/1.403

Let's put 6.087 - R2 for R1;

(1/(6.087 - R2)) + (1/R2) = 1/1.403

Multiply through by 1.403R2(6.087 - R2) to get;

1.403R2 + 1.403(6.087 - R2) = R2(6.087 - R2)

Expanding gives;

1.403R2 + 8.54 - 1.403R2 = 6.087R2 - (R2)²

(R2)² - 6.087R2 + 8.54 = 0

Using quadratic formula, we have;

R2 = 2.193 ohms or 3.894 ohms

Thus,

R1 = 6.087 - 2.193 or R1 = 6.087 - 3.894

R1 = 3.894 or 2.193

When R1 = 2.193, R2 = 3.894

When R1 = 3.894, R2 = 2.193

Answer:

The respiratory system provides oxygen for cells, while the circulatory system transports oxygen to cells.

Explanation:

so the answer is D

Answer:

work done= force × displacement

=800×5

=4000J

Explanation:

The amount of work done is the result of the magnitude of force applied and the displacement of the body due to the force applied. Therefore, work done is defined as the product of the applied force and the displacement of the body.

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

[tex]E_{mid} = E_{right} +E_{left}\\\\E_{right} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt} = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0[/tex]

Therefore, the electric field strength at the mid-point between the two rings is zero.

Your question is a "non sequitur", which means "it doesn't follow".

Your "then" doesn't contradict your "If", so no mystery is implied.

Maybe you're trying to say that matter is somehow not conserved in the equation . . . 4Fe + 302 → 2Fe203 . But it is.  There are 4 Irons and 6 Oxygens on each side, so conservation is not violated here.

I looked up "rust" on Floogle, and got slapped with pages and pages of chemistry that I don't completely understand.  But what it's saying is that rusting is a very complex chemical process, AND it doesn't happen unless there's some water involved.

So the bottom line is that there's a lot more going on than simply

4Fe + 302 → 2Fe203 ,

there's water going in and out of the process at every stage, and when it's all over, you have rusty iron, and mass has been conserved.

Answer:

32.25 s

Explanation:

From the question,

P = W/t.............. Equation 1

Where P = Power, W = work done, t = time.

But

W = F×d................. Equation 2

Where F = force and d = distance

Substitute equation 2 into equation 1

P = F×d/t............... Equation 3

make t the subject of euqation 3

t = (F×d)/P............. Equation 4

Givn: F = 150 N, d = 4.3 m, P = 20 watts.

Substitute these values into equation 4

t = (150×4.3)/20

t = 32.25 s

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Answer:

the minimum wall thickness that will enhance the reflection of light is 146.9 nm

Explanation:

Given the data in the question;

At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.

At the second interface, no shift occurs,

condition for constructive interference;

t = ( m + 1/2) × λ/2n

where m = 0, 1, 2, 3 . . . . . .

now, the condition for the constructive interference;

t = mλ/2n

where t is the thickness of the soap bubble,  λ is the wavelength of light and n is the refractive index of soap bubble.

so the minimum thickness of the film which will enhance reflection of light will be;

t[tex]_{min[/tex] =  ( m + 1/2) × λ/2n

we substitute

t[tex]_{min[/tex] =  ( 0 + 1/2) × 711 /2(1.21)

t[tex]_{min[/tex] = 0.5 × 711/2.42

t[tex]_{min[/tex] = 0.5 × 293.80165

t[tex]_{min[/tex] = 146.9 nm

Therefore,  the minimum wall thickness that will enhance the reflection of light is 146.9 nm

Answer:

a) p₂ = 1.88 kg*m/s

   θ = 273.4 º

b)  Kf = 37% of Ko

Explanation:

a)

       [tex]p_{o1x} = 2.00 kg*m/s (1)[/tex]

       [tex]p_{o1y} = 0 (2)[/tex]

        [tex]p_{o2x} = 0 (3)[/tex]

        [tex]p_{o2y} = 4.00 kg*m/s (4)[/tex]

       [tex]p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)[/tex]

      [tex]p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s (6)[/tex]

       [tex]p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x} (7)[/tex]

       [tex]p_{f2x} = p_{o1x} - p_{f1x} = 2.00kg*m*/s - 2.12 kg*m/s = -0.12 kg*m/s (8)[/tex]

       [tex]p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y} (9)[/tex]

       [tex]p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)[/tex]

       [tex]p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} } = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)[/tex]

       [tex]tg \theta = \frac{p_{2fy} }{p_{2fx} } = \frac{1.88m/s}{(-0.12m/s} = -15.7 (12)[/tex]

b)

       [tex]\frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)[/tex]