What energy store is in the hammer
BEFORE it hits the nail?​

Answer:

The new force becomes 4 times the initial force.

Explanation:

The force of attraction or repulsion is given by the relation as follows :

[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]

Where

d is the distance between the interacting charges

F is inversely proportional to the distance between charges.

If the distance is halved, d'=(d/2), new force is given by :

[tex]F'=k\dfrac{q_1q_2}{d'^2}\\\\=k\dfrac{q_1q_2}{(\dfrac{d}{2})^2}\\\\=k\dfrac{q_1q_2}{\dfrac{d^2}{4}}\\\\=4\times \dfrac{kq_1q_2}{d^2}\\\\F'=4F[/tex]

So, the new force becomes 4 times the initial force.

Answer:

While Bohr's atomic model hypothesizes that electrons move in particular energy levels around the nucleus, the electron cloud model suggests that electrons move in an unpredictable pattern but are more likely to be in certain regions than others.

Explanation:

Answer:

i dont know i am right but here Luminous intensity is defined as dI=dΨλ / dΩ, where dΨλ is the luminous flux (light energy flux in watts per m2) emitted within a solid angle dΩ. The light energy flux may be expressed in terms of the incident x-ray energy flux and the x-ray absorption and conversion properties of the scintillator(7,8,9).

Explanation:

Answer:

9.3m/s

Explanation:

Based on the law of conservation of momentum

Sum of momentum before collision = sum of momentum after collision

m1u1 +m2u2 = m1v1+m2v2

m1 = 8kg

u1 = 15.4m/s

m2 = 10kg

u2 = 0m/s(at rest)

v1 = 3.9m/s

Required

v2.

Substitute

8(15.4)+10(0) = 8(3.9)+10v2

123.2=31.2+10v2

123.2-31.2 = 10v2

92 = 10v2

v2 = 92/10

v2 = 9.2m/s

Hence the velocity of the 10.0 kg object after the collision is 9.2m/s

Answer:

a) if the two trains are going to collide

b) x = 538 m

Explanation:

To solve this exercise we use the kinematics relations in one dimension.

We set a reference system at the starting point of the passenger train

Freight train, we use index 1 for this train

          x₁ = x₀ + v₁ t

passenger train

          x₂ = v₂ t - ½ a₂ t²

as we are using the same system to measure the position of the two trains, at the meeting point the position must be the same

           x₁ = x₂

we substitute

          x₀ + v₁ t = v₂ t - ½ a₂ t²

         ½ a₂ t² + t (v₁ -v₂) + x₀ = 0

we subjugate the values

         ½ 0.1 t² + t (15-25) + 200 = 0

         0.05 t² - 10 t + 200 = 0

          t² - 200 t + 4000 = 0

we solve the second degree system

         t = [200 ±[tex]\sqrt{200^2 - 4 \ 4000}[/tex] ]/ 2

         t = [200 ± 154.9] / 2

         t₁ = 177.45 s

         t₂ = 22.55 s

therefore for the smallest time  the two trains must meet  t₂ = 22.55 s

merchandise train

            x = 200+ 15  22.55

            x = 538 m

passenger train

           x = 25 22.55 -1/2  0.100  22.55²

           x = 538 m

we see that the trains meet for this distance

a) if the two trains are going to collide

b) x = 538 m

c) see attachment for schematic graphics

Answer:

Explanation:

For resistance of a wire , the formula is as follows .

R = ρ L/S

where ρ is specific resistance , L is length and S is cross sectional area of wire .

for first wire resistance

R₁ =  ρ 3L/3a = ρ L/a

for second wire , resistance

R₂ = ρ 3L/6a

= .5 ρ L/a

For 3 rd wire resistance

R₃ = ρ 6L/3a

= 2ρ L/a

For fourth wire , resistance

R₄ = ρ 6L/6a

=  ρ L/a

So the smallest resistance is of second wire .

Its resistance is .5 ρ L/a

Answer:

The answer is below

Explanation:

We are going to use Gauss’ law to find the electric field equation. Since electric field is coming from an infinite line of charge, hence it is going out in a radial direction.  

Therefore we use the area of the electric field which passes through, forming a Gaussian cylinder. We neglect the ends of the area.

Hence:

[tex]\int\limits {E} \, dA=\frac{Q_{enc}}{\epsilon_o}\\\\E(2\pi rL)= \frac{\lambda L}{\epsilon_o}\\\\E=\frac{\lambda}{2\pi r\epsilon_o} \\\\Given \ that:\\\\r=a=8.7\ cm=0.087\ m, \lambda=-2.3 \mu C/cm=-2.3*10^{-4}\ C/m,\epsilon_o=8.85*10^{-12}F/m.\\\\Hence:\\\\E=\frac{-2.3*10^{-4}}{2\pi *0.087*8.85*10^{-12}}=-4.75*10^7\ N/C[/tex]

The value of the x-component of the electric field is -475213.968 newtons per coulomb.

In this question we shall apply Gauss' Law to determine the magnitude of the electric field ([tex]E_{x}[/tex]), in newtons per coulomb, rapidly and based on the assumptions of uniform charge distribution and cylindrical symmetry.

[tex]\frac{Q_{enc}}{\epsilon_{o}} = \oint\,\vec E\,\bullet d\vec A[/tex] (1)

Where:

Based on all assumptions, we simplify (1) as follows:

[tex]\frac{\lambda\cdot l}{\epsilon_{o}} = E \cdot (2\pi\cdot r\cdot l)[/tex]

And the equation of the x-component of the electric field is:

[tex]E = \frac{\lambda}{2\pi\cdot \epsilon_{o}\cdot r}[/tex] (2)

Where [tex]\lambda[/tex] is the linear charge density, in coulomb per meter.

If we know that [tex]\lambda = -2.3\times 10^{-6}\,\frac{C}{m}[/tex] and [tex]a = 0.087\,m[/tex], then the electric field produced by the line of charge at point P is:

[tex]E = \frac{\left(-2.3\times 10^{-6}\,\frac{C}{m} \right)}{2\pi\cdot \left(8.854\times 10^{-12}\,\frac{s^{4}\cdot A^{2}}{kg\cdot m^{3}} \right)\cdot (0.087\,m)}[/tex]

[tex]E_{x} = -475213.968 \,\frac{N}{C}[/tex]

The value of the x-component of the electric field is -475213.968 newtons per coulomb. [tex]\blacksquare[/tex]

The figure is missing, we present the corresponding image in the file attached below.

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Answer: The correct statement is that the Last Quarter Moon (rises near midnight and sets near midday).

Explanation:

Phases of the moon also called the LUNAR PHASE can be defined as the different shades of illumination on the moon as seen from the earth. The moon is the natural satellite of the earth that illuminates upon reflection of light from the sun. This means it doesn't have power to shine on its own. When carefully observed, there are times it gets dark and beings to glow brighter over a period of time. This occurs because as the moon completes its four weeks lunar cycle round the earth, how much of its face we see illuminated by sunlight depends on the angle the Sun makes with the Moon.

There are 8 main types of the moon phases these includes:

--> New moon: This is when the moon is not visible to the earth because it's between the earth and the sun. It rises at sunrise and sets at sunset.

--> The waxing crescent: At this phase the moon gets brighter and illuminated from the sun that a crescent shape is seen.

--> First quarter: this occurs one week after the new moon. The moon rises at noon and sets at midnight.

--> The waxing gibbous: This occurs after the first quarter phase where more than half of the lit part of the moon is seen.

--> Full moon: This occurs when the moon and the sun are opposite each other. That is way it is said to rise at sunset and sets at sun rise.

--> The waning gibbous: this occurs when more than half of the lit part of the moon gradually becomes darker

--> Third quarter ( Last Quarter): The moon rises at midnight and sets at noon. This occurs a week after the full moon.

--> The waning crescent: This occurs after the last quarter phase where a very thin fading crescent shaped moon is seen, just before the Moon is invisible again at the start of the cycle, the new moon.

Answer and Explanation: A charge exerts a force over another charge even if they are very far apart. This force is called Electrostatic Force.

If the two charges have the same sign, e.g. both aare positive, the force between them is opposite. If they have opposite sign, the force is towards each other. In other words, for electrostatic force, equal charges repel and different charges attract.

So,

1. If Q2 and Q3 have opposite signs, it is TRUE force in Q2 will go the left;

2. If the 2 are negative, they have the same sign, so it's FALSE force is to the right;

Sentences 3 and 4 are also TRUE due to the reasons described above;

5. If the charges have opposite signs, it means force is towards each other, or, to the right, so the sentence is TRUE;

1. Force is directly proportional to charges in Coulomb [C] and inversely proportional to distance squared in [m]:

[tex]F=\frac{k.q.Q}{r^{2}}[/tex]

where k is a constant that equals 9 x 10⁹ N.m²/C²

Calculating force between 1 and 2:

[tex]F_{12}=\frac{9.10^{9}(1.9.10^{-6})(2.84.10^{-6})}{(0.301)^{2}}[/tex]

[tex]F_{12}=536.02.10^{-3}[/tex] N

Force between 2 and 3:

[tex]F_{23}=\frac{9.10^{9}(2.84.10^{-6})(3.03.10^{-6})}{(0.169)^{2}}[/tex]

[tex]F_{23}=2711.63.10^{-3}[/tex] N

Total force is the net force. Since Q2 is negative and the others are positive, force of 2 related to 1 is to left and related to 3 is to the right. Therefore, total force is the difference between those two forces:

[tex]F_{T}=2711.63.10^{-3}-536.02.10^{-3}[/tex]

[tex]F_{T}=2175.61.10^{-3}[/tex] N

The total force on Q2 is 2175.61 x 10⁻³ N

2. For net force to be 0, [tex]F_{13}=F_{23}[/tex]. Suppose distance from 1 to 3 is x, then from 2 to 3 is [tex]x-0.301[/tex]

Calculating:

[tex]\frac{k(1.90.10^{-6})(3.03.10^{-6})}{x^{2}}=\frac{k(2.84.10^{-6})(3.03.10^{-6})}{(x-0.301)^{2}}[/tex]

[tex]\frac{5.757.10^{-12}}{x^{2}} =\frac{8.6052.10^{-12}}{x^{2}-0.602x+0.090601}[/tex]

[tex]\frac{5.757.10^{-12}}{8.6052.10^{-12}}=\frac{x^{2}}{x^{2}-0.602x+0.090601}[/tex]

[tex]x^{2}=0.67x^{2}-0.40x+0.061[/tex]

[tex]0.33x^{2}+0.40x-0.061=0[/tex]

roots = 0.14 or -1.35

Solving quadratic equation gives 2 roots, but one of the roots is negative. As distance is a measure that cannot be negative, the solution is x = 0.14.

The distance of Q3 relative to Q1 is 0.14 m

To Find :

Which is larger: 65 mph (miles per hour) or 120 kph (kilometers per hour).

Solution :

We know, 1 mph = 1.61 kph

So, 65 mph = 1.61 × 65 kph

65 mph = 104.65 kph

Since, 65 mph is 104.65 kph which is smaller than 120 kph.

Therefore, 120 kph is faster than 65 mph by ( 120 - 104.65 ) = 15.35 kph.

Answer:

The automobile is running at speed of 23.806 meters per second.

Explanation:

From Kinematic we remember that acceleration ([tex]a[/tex]) can be defined by this ordinary differential equation in terms of distance:

[tex]a = v\cdot \frac{dv}{ds}[/tex] (1)

Where:

[tex]v[/tex] - Speed of the automobile, measured in meters per second.

[tex]s[/tex] - Distance travelled by the automobile, measured in meters.

If we know that [tex]v = 0.5\cdot s - 0.0025\cdot s^{2}[/tex], then the equation of acceleration is:

[tex]a = (0.5\cdot s - 0.0025\cdot s^{2})\cdot \left(0.5-0.0050\cdot s\right)[/tex]

[tex]a = s\cdot (0.5-0.0025\cdot s)\cdot (0.5-0.0050\cdot s)[/tex]

[tex]a = s\cdot (0.0025\cdot s - 0.5)\cdot (0.0050\cdot s-0.5)[/tex]

But distance covered by the vehicle is defined by the following formula:

[tex]s = \theta \cdot r[/tex] (2)

Where:

[tex]\theta[/tex] - Arc angle, measured in radians.

[tex]r[/tex] - Radius, measured in radians.

Then, we expand (1) by means of this result:

[tex]a = \theta\cdot r \cdot (0.0025\cdot \theta\cdot r -0.5)\cdot (0.0050\cdot \theta \cdot r-0.5)[/tex]

[tex]a = \theta\cdot r \cdot (1.25\times 10^{-5}\cdot \theta^{2}\cdot r^{2}-3.75\times 10^{-3}\cdot \theta\cdot r +0.25)[/tex]

[tex]a = 1.25\times 10^{-5}\cdot \theta^{3}\cdot r^{3}-3.75\times 10^{-3}\cdot \theta^{2}\cdot r^{2}+0.25\cdot \theta \cdot r[/tex]

And finally we get the following third order polynomial:

[tex]1.25\times 10^{-5}\cdot \theta^{3}\cdot r^{3}-3.75\times 10^{-3}\cdot \theta^{2}\cdot r^{2}+0.25\cdot \theta \cdot r - a = 0[/tex] (3)

If we know that [tex]r = 50\,m[/tex], [tex]a = 0.6\cdot g[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the polynomial becomes into this:

[tex]1.5625\cdot \theta^{3}-9.375\cdot \theta^{2} +12.5\cdot \theta - 5.886 = 0[/tex] (3b)

This polynomial can be solved analytically by Cardano's Method or by numerical methods. The roots of the polynomial are, respectivelly:

[tex]\theta_{1} \approx 4.365\,rad[/tex], [tex]\theta_{2} \approx 0.818+i\,0.441\,rad[/tex], [tex]\theta_{3}\approx 0.818 -i\,0.441\,rad[/tex], [tex]\theta_{4} \approx 1.563\,rad[/tex]

Both first and fourth roots are physically reasonable solution, but the latter represents the angle where automobile begins to skid first. Then, the automobile begins to skid at an angle of 1.563 radians relative to x axis.

The distance travelled by the automobile is: ([tex]r = 50\,m[/tex], [tex]\theta \approx 1.563\,rad[/tex])

[tex]s = (1.563\,rad)\cdot (50\,m)[/tex]

[tex]s = 78.15\,m[/tex]

Lastly, the speed of the automobile at this location is: ([tex]s = 78.15\,m[/tex])

[tex]v = 0.5\cdot s - 0.0025\cdot s^{2}[/tex] (4)

[tex]v = 0.5\cdot (78.15)-0.0025\cdot (78.15)^{2}[/tex]

[tex]v = 23.806\,\frac{m}{s}[/tex]

The automobile is running at speed of 23.806 meters per second.

Shivering is a response of an organism when the temperature drops. In shivering,  muscles contract and expand quickly and the body temperature rises​.

Warm-blooded animals shiver (also known as shudder) as a physiological reaction to cold and acute fear. To preserve homeostasis, the shivering reaction is activated when the body's core temperature lowers.

Small skeletal muscle movements start, producing warmth as the muscles use up their energy. Another sign of a fever is shivering because the person may feel cold.

The hypothalamic set point for temperature rises during a fever. Up until the new set point is reached, the patient experiences cold until the increased set point raises body temperature (pyrexia).

Rigors are the medical term for extremely cold symptoms accompanied by ferocious shivering. Rigors happen as a result of the patient's body shivering in an effort to physiologically raise body temperature to the new set point.

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Answer:

Statement A is greater than Statement B.

Explanation:

Statement A: 2.567 km, to two significant figures..

To 2 sig figures means only 2 whole numbers should be left after approximation. Thus, 2.567 to 2 significant figures is 2.6 km

Statement B: 2.567 km, to three significant figures. To 3 sig figures means only 3 whole numbers should be left after approximation. Thus, 2.567 to 3 significant figures is 2.57 km

Comparing both values, statement A is obviously greater than Statement B

Answer:

220 km

Explanation:

3 hours at 40 km/hr

40 km = 1 hour

40 km times 3 = 120 km

2 hours at 50 km/hr

50 km = 1 hour

50 times 2 = 100 km

Total distance

120 km + 100 km = 220 km

Answer:

1/9

Explanation:

Let A denote the bigger piece and let B denote the smaller piece.

We are told that one with three times the mass of the other.

Therefore, we have;

M_a = 3M_b

Firecracker is placed in the block and it explodes and thus, momentum is conserved.

Thus;

V_ai = V_bi = 0

Where V_ai is initial velocity of piece A and V_bi is initial velocity of piece B.

Since initial momentum equals final momentum, we have;

P_i = P_f

Thus;

0 = (M_a × V_af) + (M_b × V_bf)

Since M_a = 3M_b, we have;

(3M_b × V_af) + (M_b × Vbf) = 0

Making V_af the subject, we have;

V_af = -⅓V_bf

The kinetic energy gained by each block during the explosion will later be lost due to the negative work done by friction. Thus;

W_f = -½M_b•(v_bf)²

Now, let's express the work is in terms of the force and the distance.

Thus;

W_f = F_f × Δx × cos 180°

Frictional force is also expressed as μmg

Thus;

W_f = -μM_b × g × Δx

Earlier, we saw that;

W_f = -½M_b•(v_bf)²

Thus;

-½M_b•(v_bf)²= -μM_b × g × Δx

Δx = (v_bf)²/2μg

Let the distance travelled by block A be Δx_a and that travelled by B be Δx_b

Thus;

Δx_a/Δx_b = ((v_ba)²/2μg)/((v_bf)²/2μg)

Δx_a/Δx_b = ((v_af)²/((v_bf)²)

Δx_a/Δx_b = (-⅓V_bf)²/(V_bf)²

Δx_a/Δx_b = 1/9

Answer:

c

Explanation:

a vector quantity has both magnitude and direction

The value of  20m/s, east is a vector quantity is Hence, option (C) is correct.

A physical quantity that has both directions and magnitude is referred to as a vector quantity.

A lowercase letter with a "hat" circumflex, such as "û," is used to denote a vector with a magnitude equal to one. This type of vector is known as a unit vector.

Given values  200V, 100kg/m, 50J/(kg°C) are denoting magnitude of different physical quantity. Hence, they and scalar quantity ( Physical quantities with merely magnitude and no direction are referred to as scalar quantities. These physical quantities can be explained just by their numerical value without any further guidance.).

But The value of  20m/s, east has a magnitude of 20 m/s and a direction along east. Hence, 20m/s, east denotes a  vector quantity is Hence, option (C) is correct.

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Answer:

If the speed does not change at all, the object would be moving at a constant speed.

Answer:

Option A

Explanation:

To solve this problem we need to apply the momentum conservation, and analyze the data.

For this problem, I will call the initial velocities as V₁ and V₂, while the final velocities will be V₃ and V₄.

According to the momentum principle, this states the following:

m₁V₁ + m₂V₂ = m₁V₃ + m₂V₄   (1)

From this equation we can write an expression in function of V₃ and V₄. We also know that coefficient of restitution is 0.6. Knowing this, we can write the expression that will help us to solve for the final velocities:

e = V₄ - V₃ / 2    (2)

With both expressions we can solve for the final velocities. Let's use (1) first and see what we can simplify first by replacing the given data:

(3*4) + (4*2) = 3V₃ + 4V₄

12 + 8 = 3V₃ + 4V₄

20 = 3V₃ + 4V₄   (3)

This is all we can do for now. Let's use (2) now:

0.6 = V₄ - V₃ / 2

1.2 = V₄ - V₃

V₄ = 1.2 + V₃   (4)

Now, we can replace (4) into (3), and then, solve for V₃:

20 = 3V₃ + 4(1.2 + V₃)

20 = 3V₃ + 4.8 + 4V₃

15.2 = 7V₃

V₃ = 15.2 / 7

We have the value of one final velocity, let's see the other one.

V₄ = 1.2 + V₃

V₄ = 1.2 + 2.17

The closest values to these results are in option A, so this will be the correct option.

Hope this helps

Answer:

a.-7.5 kg m/s

b.-375 N

Explanation:

We are given that

Mass of football, m=0.5 kg

Initial velocity ,u=15m/s

Final speed ,v=0

Time, t=0.02 s

a. We have to find the impulse delivered to the ball.

We know that

Impulse=Change in momentum

I=m(v-u)

Using the formula

[tex]I=0.5(0-15)=-7.5 kg m/s[/tex]

Hence,  the impulse delivered to the ball=-7.5 kg m/s

(b)

We know that

Force,[tex]F=\frac{|mpulse}{time}[/tex]

Using the formula

[tex]F=\frac{-7.5}{0.02}=-375 N[/tex]

Answer:

the signs on the plates are      A +    B-    C +

Explanation:

The electric field for an infinite plate is perpendicular to it, this field is outgoing if the charge on the plate is positive and incoming if the charge on the plate is negative.

Let's analyze the situation presented, we have two infinite plates A and B with the elective field directed to the right, therefore the charge of the plate must be

            plate A    positive charge

           plate B     Negative charge

We also have a third plate C and they indicate that the field between B and C is directed to the left, therefore

           plate B     negative charged

            plate C   positive charged

in short the signs on the plates are

    A +    B-    C +

Answer:

276.12 pC

Explanation:

We are given that

Area,A=[tex]2.6 cm^2=2.6\times 10^{-4}m^2[/tex]

Where [tex]1 cm^2=10^{-4} m^2[/tex]

[tex]d=0.25 mm=0.25\times 10^{-3} m[/tex]

[tex]1mm=10^{-3} m[/tex]

Potential difference, V=30  V

We have to find the charge on the capacitor when the plates are pulled to a separation of 0.55 mm.

We know that

Charge ,[tex]Q=\frac{\epsilon_0 A V}{d}[/tex]

Where [tex]\epsilon_0=8.85\times 10^{-12}[/tex]

Using the formula

[tex]Q=\frac{8.85\times 10^{-12}\times 2.6\times 10^{-4}\times 30}{0.25\times 10^{-3}}[/tex]

[tex]Q=2.7612\times 10^{-10} C[/tex]

[tex]Q=276.12p C[/tex]

[tex]1 pC=10^{-12} C[/tex]

When the plates are now pulled to a separation of 0.55 mm.Then, the charge on the plates remain same because the  battery has been disconnected.

Therefore, charge on the capacitor=276.12 pC

Answer:

Explanation:

Properties of the sun

Introduction to Stars A. The Sun 1. ...

Layers of the sun (Atmosphere) a. Photosphere: Layer that emits the radiation we see. ...

Layers of the sun (Interior) a. ...

Solar Wind: a. ...

The sun in X-Rays: a. ...

Luminosity: The measure of the energy put out by the sun. ...

Luminosity continued: c. ...

Luminosity continued: e.

The heat of the Sun's surface is so large that neither solid nor liquid can reside there. The heat is the physical property of the sun.

The Sun is a billion star at the core of our universe, a hot. It is an incandescent ball of hydrogen and helium.

The Sun is millions of kilometers away from the land, and life is not possible without its energy.

The major properties of the sun are as follows;

1. The mass of the sun is 1.98892 x 10³⁰ kg.

2 The density of the sun is 1.622 x 10⁵ kg/m³

3. The surface gravity of the sun is 27.94 g.

4. The diameter is 1,391,000 kilometers.

5. The volume of the Sun is 1.412 x 1018 km³

The heat of the Sun's surface is so hot that neither solid nor liquid can reside there instead, the component materials are mostly gaseous atoms with a few molecules. As a result, no permanent surface exists.

Hence heat, density, and mass are the major properties of the sun.

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Answer:

84 cm

Explanation:

The resolving power of the eye = 1.22 λ / D where λ is wavelength of light and D is diameter of the eye .

Putting the values in the formula

The resolving power of the eye = 1.22 λ / 2.9 x 10⁻³

Let the minimum distance you need to stand from one of Seurat's paintings to see the dots blended together  be D .

d = distance between two dots = diameter of dot = 2 x 10⁻³ m .

Given ,

λ D / d =  1.22 λ / 2.9 x 10⁻³

D /2 x 10⁻³   =  1.22  / 2.9 x 10⁻³

D = 2 x 1.22 / 2.9 m

= .84 m

= 84 cm .

minimum distance you need to stand from one of Seurat's paintings to see the dots blended together is 84 cm .

Answer:

Explanation:

Mechanical weathering types

Mechanical weathering is the breaking down of rocks into smaller pieces without changing the composition of the minerals in the rock. This can be divided into four basic types – abrasion, pressure release, thermal expansion and contraction, and crystal growth.

Answer:

joule is the answer

Explanation:

I hope it helps you

Answer:

True.

Explanation:

Welding requires extensive training because welding involves fire and we need to use fire safety measurements. A normal man can't just simply go and weld so a person must require extensive training for welding.