the coefficient of linear expansion of aluminum is 24.0 * 10^-6K^-1 and th density of aluminum at ) C is 2.7 *10^3kg/m^3. what is the density of aluminum at 300 C

Answer:

the  speed of the electron at the given position is 106.2 m/s

Explanation:

Given;

initial position of the electron, r = 9 cm = 0.09 m

final position of the electron, r₂ = 3 cm = 0.03 m

let the speed of the electron at the given position = v

The initial potential energy of the electron is calculated as;

[tex]U_i = Fr = \frac{kq^2}{r^2} \times r = \frac{kq^2}{r} \\\\U_i = \frac{(9\times 10^9)(1.602\times 10^{-19})^2}{0.09} \\\\U_i = 2.566 \times 10^{-27} \ J[/tex]

When the electron is 3 cm from the proton, the final potential energy of the electron is calculated as;

[tex]U_f = \frac{kq^2}{r_2} \\\\U_f = [\frac{(9\times 10^9)\times (1.602 \times 10^{-19})^2}{0.03} ]\\\\U_f = 7.669 \times 10^{-27} \ J \\\\\Delta U = U_f -U_i\\\\\Delta U = (7.699\times 10^{-27} \ J ) - (2.566 \times 10^{-27} \ J)\\\\\Delta U = 5.133 \times 10^{-27} \ J[/tex]

Apply the principle of conservation of energy;

ΔK.E = ΔU

[tex]K.E_f -K.E_i = \Delta U\\\\initial \ velocity \ of \ the \ electron = 0\\\\K.E_f - 0 = \Delta U\\\\K.E_f = \Delta U\\\\\frac{1}{2} mv^2 = \Delta U\\\\where;\\\\m \ is \ the \ mass \ of\ the \ electron = 9.1 1 \times 10^{-31} \ kg\\\\v^2 = \frac{ 2 \Delta U}{m} \\\\v = \sqrt{\frac{ 2 \Delta U}{m}} \\\\v = \sqrt{\frac{ 2 (5.133\times 10^{-27})}{9.11\times 10^{-31}}}\\\\v = \sqrt{11268.935} \\\\v = 106.2 \ m/s[/tex]

Therefore, the  speed of the electron at the given position is 106.2 m/s

Answer:

Volume of a metal block = 24 cm^3

Volume of a block twice as long, wide and high = 192 cm^3

Explanation:

Volume of a block is measured in l*w*h and in the first block, the sides are 3, 2 and 4 and 3*2*4 = 24

Second block, just double each of the lengths to get 6*4*8 = 192

Answer:

C . Magnetic dipoles is the correct

Answer:

b). movement of charged particles.

Explanation:

These charges create the nagnetic dipoles.

After being released, the restoring force exerted by the spring performs

1/2 (5200 N/m) (0.090 m)² = 12.06 J

of work on the block. At the same time, the block's weight performs

- (0.260 kg) g (0.090 m) ≈ -0.229 J

of work. Then the total work done on the block is about

W ≈ 11.83 J

The block accelerates to a speed v such that, by the work-energy theorem,

W = ∆K   ==>   11.83 J = 1/2 (0.260 kg) v ²   ==>   v ≈ 9.54 m/s

Past the equilibrium point, the spring no longer exerts a force on the block, and the only force acting on it is due to its weight, hence it has a downward acceleration of magnitude g. At its highest point, the block has zero velocity, so that

0² - v ² = -2gy

where y is the maximum height. Solving for y gives

y = v ²/(2g) ≈ 4.64 m

Answer:

875 Watts

Explanation:

P = W/t = mgh/t = 700(10)/8 = 875 Watts

Answer:

Their main disadvantage is that they are fairly easy to break and if they do, it results in small splinters of glass and the release of mercury which is quite toxic if absorbed into the body.

ANSWER-:

1/2 mv²

K.E = 1/2 mv²

K.E = 0.01 J.

Hence, the kinetic energy of a body is 0.01 Joule.

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Answer:

The current is more in the parallel combination than in the series combination.

Explanation:

two resistances, R1 and R2 are connected to a battery of voltage V.

When they are in series,

R = R1 + R2

In series combination, the current is same in both the resistors, and it is given by Ohm's law.

V = I (R1 + R2)

[tex]I = \frac{V}{R_1 + R_2}[/tex]..... (1)

When they are connected in parallel.

the voltage is same in each resistor.

The effective resistance is R.

[tex]R = \frac{R_1R_2}{R_1 + R_2}[/tex]

So, the current is

[tex]I = \frac{V(R_1+R_2)}{R_1 R_2}[/tex]..... (2)

So, the current is more is the parallel combination.

Answer:

[tex]T=1.54s[/tex]

Explanation:

From the question we are told that:

Speed of Motor 1 [tex]\omega_1=825rpm=>2 \pi 13.75[/tex]

Speed of Motor 2 [tex]\omega_1=786rpm=>2 \pi 13.1[/tex]

Therefore

Frequency of Motor 1 [tex]f_1=13.75[/tex]

Frequency of Motor 2  [tex]f_2= 13.1[/tex]

Generally the equation for Time Elapsed is mathematically given by

[tex]T=\frac{1}{df}[/tex]

Where

[tex]df=f_1-f_2[/tex]

[tex]df=13.75-13.1[/tex]

[tex]df=0.65Hz[/tex]

Therefore

[tex]T=\frac{1}{65}[/tex]

[tex]T=1.54s[/tex]

The question is incomplete, the complete question is;

how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air, and the other is filled with a dielectric for which k = 3.00; both capacitors have a plate area of 5.00×10 −3  m 2  and a plate separation of 2.00 mm.

The capacitance of the capacitor is the quantity of charge stored by the capacitor.

Given that;

C1= εo k * A/d

εo = permittivity of free space

C1 = 8.85 x 10-12 farad per meter *1 * 5.00×10 −3  m 2/2 * 10^-3

= 2.21 * 10^-11 F

C2 = 8.85 x 10-12 * 3 * 5.00×10 −3  m 2/2 * 10^-3

= 6.63 * 10^-11 F

q1 = C1V1 = 2.21 * 10^-11 C * 12 V

= 2.65 * 10^-10 C

q2 = C2V2 = 6.63 * 10^-11 F * 12 V

= 7.96 * 10^-10 C

qtotal = 2.65 * 10^-10 C + 7.96 * 10^-10 C

qtotal = 1.061 * 10^-9C

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Answer:

366 N

Explanation:

  τ = Iα

FR = ½mR²α

  F = ½mR(Δω/t)

  F = ½(207)(1.50)(0.75)(2π) /2.00

  F = 365.79919...

Answer:

A hypothesis for the period of a pendulum is:

"The period of the pendulum varies with its length"

Explanation:

A hypothesis for the period of a pendulum is:

"The period of the pendulum varies with its length"

To test this hypothesis we can carry out a measurement of a simple pendulum keeping the angle fixed, in general the angle used is about 5º since when placing this value in radiand and the sine of this angle they differ little <5%. therefore measured the time of some oscillations, for example about 10 oscillations, changing the length of the pendulum to test the hypothesis.

If the hypothesis and the model used is correct, the relationship to be tested is

              T² =(4π² /g)   L  

by making a graph of the period squared against the length if obtaining, os a line, the hypothesis is tested.

The work done by [tex]\vec F[/tex] along the given path C from A to B is given by the line integral,

[tex]\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r[/tex]

I assume the path itself is a line segment, which can be parameterized by

[tex]\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath[/tex]

with 0 ≤ t ≤ 1. Then the work performed by F along C is

[tex]\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}[/tex]

Answer:

6000 lines/cm

Explanation:

From the question we are told that:

Grating 1=4000 lines/cm

Grating 2=6000 lines/cm

Generally The Spread of fringes is Larger when the Grating are closer to each other

Therefore

Grating 2 will spread the the light through a larger angle in the first-order pattern because its the closest with 6000 lines/cm

Answer:

D = .25g/cm³

Explanation:

D = m/V

D = 120g/480cm³

D = .25g/cm³

Answer:

The correct solution is "37.5 km".

Explanation:

Given:

Distance between the trains,  

d = 75 km

Speed of each train,

= 15 km/h

The relative speed will be:

= [tex]15 + (-15)[/tex]

= [tex]30 \ km/h[/tex]

The speed of the bird,

V = 15 km/h

Now,

The time taken to meet will be:

[tex]t=\frac{Distance}{Relative \ speed}[/tex]

  [tex]=\frac{75}{30}[/tex]

  [tex]=2.5 \ h[/tex]

hence,

The distance travelled by the bird in 2.5 h will be:

⇒ [tex]D = V t[/tex]

        [tex]=15\times 2.5[/tex]

        [tex]=37.5 \ km[/tex]

Answer:

variable

Explanation:

Answer:

B. is subject to a smaller net force but same acceleration.

Explanation:

F = m*a

So because our force applied is constant from the women pulling on the rope which means the acceleration is the same on both the 4kg create and the 6kg create. The only thing that changes here is the mass of the creates, so there is more tension force between the women and the 6kg create then there is between the 4kg create and the 6kg. It takes less force to move the 4kg create therefore the tension force is less between the two creates.

The net force on both crates is the same and the acceleration of both crates is the same.

The given parameters;

The force on the 4kg crate is calculated as follows;

[tex]F_{4kg } = T + F[/tex]

The force on the 6kg crate is calculated as follows;

[tex]F_{6 kg} = -T + F[/tex]

The net force on both crates is calculated as follows;

[tex]\Sigma F= -T + F - (T + F)\\\\\Sigma F= -2T[/tex]

Thus, we can conclude that the net force on both crates is the same and the acceleration of both crates is the same.

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Answer:

The correct answer is - 8.99N/C

Explanation:

[tex]dE=k\dfrac{dq}{x^2}\\ dq=\lambda{dx}\\ \lambda=2nC/m\\ dq=2dxnC\\ dE=k\dfrac{2dx}{x^2}\\ E=2k\int_1^2\dfrac{dx}{x^2}\\ E=2k(\frac{-1}{x})_1^2=k\times10^{-9}N/C\\ E=8.99\times10^9\times10^{-9}N/C\\ E=8.99N/C\\dE=k[/tex]

THERE ARE COMMONLY THREE SYSTEMS OF UNIT. THEY ARE:-

• CGS System- (Centimeter-Gram-Second system) A metric system of measurement that uses the centimeter, gram and second for length, mass and time.

• FPS System- (Foot–Pound–Second system).

The system of units in which length is measured in foot , mass in pound and time in second is called FPS system. It is also known as British system of units.

• MKS System- (Meter-Kilogram-Second system) A metric system of measurement that uses the meter, kilogram, gram and second for length, mass and time. The units of force and energy are the "newton" and "joule."

Answer:

3.1 kg

Explanation:

Applying,

R = m(g-a)..................... Equation 1

Where R = weight of the scale when the elevator is coming down, a = acceleration of the elevator, g = acceleration due to gravith.

From the question,

Given: m = 5 kg, a = 3.8 m/s²

Constant: g = 9.8 m/s²

Substitute these values into equation 1

R = 5(9.8-3.8)

R = 5(6)

R = 30 N

Hence the spring scale is

m' = R/g

m' = 30/9.8

m' = 3.1 kg

Answer:

endothermic

Explanation:

An endothermic is any process with an increase in the enthalpy H (or internal energy U) of the system. In such a process, a closed system usually absorbs thermal energy from its surroundings, which is heat transfer into the system.

Explanation:

We can write Newton's 2nd law as applied to the sliding mass [tex]m_1[/tex] as

[tex]T - m_1g\sin38 = m_1a\:\:\:\:\:\:\:(1)[/tex]

For the hanging mass [tex]m_2,[/tex] we can write NSL as

[tex]T - m_2g = -m_2a\:\:\:\:\:\:\:(2)[/tex]

We need to solve for a first before we can solve the tension T. So combining Eqns(1) & (2), we get

[tex](m_1 + m_2)a = m_2g - m_1g\sin38[/tex]

or

[tex]a = \left(\dfrac{m_2 - m_1\sin38}{m_1 + m_2}\right)g[/tex]

[tex]\:\:\:\:= 0.30\:\text{m/s}^2[/tex]

Using this value for the acceleration on Eqn(2), we find that the tension T is

[tex]T = m_2(g - a) = (2.6\:\text{kg})(9.51\:\text{m/s}^2)[/tex]

[tex]\:\:\:\:=24.7\:\text{N}[/tex]

Answer:

Here, m=9×10

−31

kg,

q=1.6×10

−19

C,v=3×10

7

ms

−1

,

b=6×10

−4

T

r=

qB

mv

=

(1.6×10

−19

)(6×10

−4

)

(9×10

−31

)×(3×10

7

)

=0.28m

v=

2πr

v

=

2πm

Bq

=

2×(22/7)×9×10

−31

(6×10

−4

)×(1.6×10

−19

)

=1.7×10

7

Hz

Ek=

2

1

mv

2

=

2

1

×(9×10

−31

)×(3×10

7

)

2

J

=40.5×10

−17

J=

1.6×10

−16

40.5×10

−17

keV

=2.53keV

Answer:

the speed of the block at the given position is 21.33 m/s.

Explanation:

Given;

spring constant, k = 3500 N/m

mass of the block, m = 4 kg

extension of the spring, x = 0.2 m

initial velocity of the block, u = 0

displacement of the block, d =1.3 m

The force applied to the block by the spring is calculated as;

F = ma = kx

where;

a is the acceleration of the block

[tex]a = \frac{kx}{m} \\\\a = \frac{(3500) \times (0.2)}{4} \\\\a = 175 \ m/s^2[/tex]

The final velocity of the block at 1.3 m is calculated as;

v² = u² + 2ad

v² = 0 + 2ad

v² = 2ad

v = √2ad

v = √(2 x 175 x 1.3)

v = 21.33 m/s

Therefore, the speed of the block at the given position is 21.33 m/s.

The speed of the block at a height of 1.3 m above the starting position is 21.33 m/s

To solve this question, we'll begin by calculating the acceleration of the block.

F = Ke

Also,

F = ma

Thus,

ma = Ke

Divide both side by m

a = Ke / m

a = (3500 × 0.2) / 4

a = 175 m/s²

v² = u² + 2as

v² = 0² + (2 × 175 × 1.3)

v² = 455

Take the square root of both side

v = √455

v = 21.33 m/s

Learn more about spring constant:

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Answer:

[tex]f=9.55Hz[/tex]

Explanation:

From the question we are told that:

Number of Turns [tex]N=200[/tex]

Length [tex]l=5cm to 10cm[/tex]

Voltage [tex]V=18V[/tex]

Magnetic field [tex]B=300mT[/tex]

Generally, the equation for Frequncy of an amarture is mathematically given by

[tex]f =\frac{ V}{(N B A * 2 pi )}[/tex]

[tex]f =\frac{ 18}{(200 300*10^{-3} (10*10^-2)(5*10^{-2}) * 2 *3.142 )}[/tex]

[tex]f=9.55Hz[/tex]

Answer:

It will travel 350 meters each second.

Explanation:

The unit rate, 350 m/s, tells us that the jet will travel 350 meters per every second elapsed.

Answer:

5.83 seconds

Explanation:

60 seconds in 1 minute

350 meters per second

350/60

=5.83

The masses of the gliders provided in the question differ from the masses mentioned in the "Required" section. I'll use the first masses throughout.

Momentum is conserved, so the total momentum of the system is the same before and after the collision:

m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'

==>

(0.160 kg) (0.710 m/s) + (0.296 kg) (-2.23 m/s) = (0.160 kg) v₁' + (0.296 kg) v₂'

==>

-0.546 kg•m/s ≈ (0.160 kg) v₁' + (0.296 kg) v₂'

where v₁' and v₂' are the gliders' respective final velocities. Notice that we take rightward to be positive and leftward to be negative.

Kinetic energy is also conserved, so that

1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁' )² + 1/2 m₂ (v₂' )²

or

m₁ v₁² + m₂ v₂² = m₁ (v₁' )² + m₂ (v₂' )²

==>

(0.160 kg) (0.710 m/s)² + (0.296 kg) (-2.23 m/s)² = (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²

==>

1.55 kg•m²/s² ≈ (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²

Solve for v₁' and v₂'. Using a calculator, you would find two solutions, one of which we throw out because it corresponds exactly to the initial velocities. The desired solution is

v₁' ≈ -3.11 m/s

v₂' ≈ -0.167 m/s

and take the absolute values to get the magnitudes.

If you want to instead use the masses from the "Required" section, you would end up with

v₁' ≈ -3.18 m/s

v₂' ≈ -0.236 m/s

Answer:

The electrical energy stored in the capacitor will be cut in half.

Explanation:

The energy in a capacitator is given by E=C[tex]V^{2}[/tex]/2 and the formula for the Capacitance in a capacitator is C= [tex]\frac{Q}{V}[/tex] = ε[tex]\frac{A}{d}[/tex] .

So if we replace C = ε[tex]\frac{A}{d}[/tex]  in the first equation we have:

E = ε[tex]\frac{AV^{2} }{2d}[/tex]

Explanation:

the car will be brought back