A student has a large number of coins of different diameters, all made of the same metal. She wishes to find the density of the metal by a method involving placing the coins in water.

a) State the formula needed to calculate the density.

b) Describe how the measurements of the required quantities are carried out.​

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:

Hey,. its a simple question. hope you learn from the solution. check attached picture

Explanation:

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:

Various uses of water :

1. Water is used for daily purpose like cooking , bathing , cleaning and drinking.

2. Water used as a universal solvent.

3. water maintains the temperature of our body.

4. Water helps in digestion in our body.

5 .water is used in factories and industries.

6. Water is used to grow plants , vegetables and crops.

Answer:

ω₁ = 2.2 rev/s

Explanation:

Conservation of angular momentum

moment of inertia uniform disk is ½mR²

moment of inertia uniform rod about an end mL²/3

We can think of our rod as two rods of mass m/2 and length R

L = ½mR²ω₀

L = (½mR² + 2(m/2)R²/3)ω₁

ω₁ = ω₀(½mR² / (½mR² + mR²/3))

ω₁ = ω₀(½  / (½  + 1/3))

ω₁ = 0.6ω₀

ω₁ = 2.16

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: Wavelength can be defined as the distance between two successive crests or troughs of a wave. It is measured in the direction of the wave.

Explanation:

Wavelength refers to the length or distance between two identical points of neighboring cycles of a wave signal traveling in space or in any physical medium. ... The wavelength of a signal is inversely proportional to its frequency, that is, the higher the frequency, the shorter the wavelength.

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 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:

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 crowning of Charlemagne by Pope Leo III was significant in a number of ways. For Charlemagne, it was necessary because it encouraged to give him higher reliability. It gave him the rank of a dictator, giving him the only ruler in Europe west of the Byzantine emperor in Constantinople.

Answer:

Force,

[tex]F = \frac{kQ_{1} Q_{2} }{ {r}^{2} } \\ F = \frac{(9 \times {10}^{9}) \times (25 \times {10}^{ - 6}) \times (10 \times {10}^{ - 6} ) }{ {(0.85)}^{2} } \\ \\ F = 3.114 \: newtons[/tex]

The magnitude of the force between a 25μC charge exerts on a -10μC charge 8.5cm away would be 311.4 N.

Coulomb's law can be stated as the product of the charges and the square of the distance between them determine the force of attraction or repulsion acting in a straight line between two electric charges.

The math mathematical expression for the coulomb's law given as

F= k Q₁Q₂/r²

where F is the force between two charges

k is the electrostatic constant which is also known as the coulomb constant,it has a value of 9×10⁹

Q₁ and Q₂ are the electric charges

r is the distance between the charges

As given in the problem two charges a 25μC charge exerts on a -10μC charge 8.5cm away

By substituting the respective values in the above formula of Coulomb law

F =9×10⁹×(25×10⁻⁶)×(-10×10⁻⁶)/(8.5×10⁻²)²

F= -311.4 N

A negative sign represents that the force is attractive in nature

Thus, the magnitude of the force is 311.4 N.

Learn more about Coulomb's law from here

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Answer:

The answer is "Option b, c, and a".

Explanation:

Here that the earth pulls on the phone, as it will accelerate towards Earth when we drop it.

We now understand the effects of gravity:

[tex]F \propto M\\\\F\propto \frac{1}{r^2}\\\\or\\\\F \propto \frac{M}{r^2}\\\\Sun (\frac{M}{r^2}) = \frac{10^{28}}{(10^9)^2} = 10^{10}[/tex]

The force of the sun is, therefore, [tex]10^{10}[/tex] times greater and the proper sequence, therefore, option steps are:

b. Pull-on phone from earth

c. Pull-on phone from sun

a. Pull phone from you

Answer:

875 Watts

Explanation:

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

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:

the density of the salt water is 1030 kg/m³

Explanation:

Given;

radius of the cylindrical pool, r = 2 m

depth of the pool, h = 1.3 m

specific gravity of the salt water, γ = 1.03

The atmospheric pressure, P₀ = 1.013 x 10⁵ Pa

Density of fresh water, [tex]\rho _w[/tex] = 1000 kg/m³

The density of the salt water is calculated as;

[tex]Specific \ gravity \ of \ salt\ water \ (\gamma _s_w) = \frac{density \ of \ salt \ water \ (\rho_{sw})}{density \ of \ fresh \ water \ (\rho_{w})} \\\\1.03 = \frac{\rho_{sw}}{1000 \ kg/m^3}\\\\\rho_{sw} = 1.03 \times 1000 \ kg/m^3\\\\\rho_{sw} = 1030 \ kg/m^3[/tex]

Therefore, the density of the salt water is 1030 kg/m³

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.

Answer:

variable

Explanation:

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:

D = .25g/cm³

Explanation:

D = m/V

D = 120g/480cm³

D = .25g/cm³

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:

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

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:

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

Polarized sunglasses creates filter of vertical openings for light. The light rays will reach the eyes of human vertically only.

The sun rays will not reach human eye directly which will create a shield for sun light burden on human eye.

Polarized sunglasses are best used for blocking and eliminating certain wavelengths of light.

Therefore the correct answer is option C. Polarizes Sunglasses are polarized and it filter out certain wavelengths of light.

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Answer:

C . Magnetic dipoles is the correct

Answer:

b). movement of charged particles.

Explanation:

These charges create the nagnetic dipoles.

The vector sum is the algebraic sum if the two vectors have the same direction.

The sum vector is zero if the two vectors have the same magnitude and opposite direction

Vector addition is a process that can be performed graphically using the parallelogram method, see  attached, where the second vector is placed at the tip of the first and the vector sum goes from the origin of the first vector to the tip of the second.

There are two special cases where the vector sum can be reduced to the algebraic sum if the vectors are parallel

case 1. if the two vectors are parallel, the sum vector has the magnitude of the sum of the magnitudes of each vector

case 2. If the two vectors are antiparallel and the magnitude of the two vectors is the same, the sum gives zero.

In summary in the sum of vectors If the vectors are parallel it is reduced to the algebraic sum, also in the case of equal magnitude and opposite direction the sum is the null vector

a) Magnitudes: [tex]\| \vec A\| \ge 0[/tex], [tex]\|\vec B\| \ge 0[/tex], [tex]\|\vec C\| \ge 0[/tex]; Directions: [tex]\theta_{A} \in (-\infty, +\infty)[/tex] for [tex]\|\vec A\|\ne 0[/tex]. Undefined for [tex]\|\vec A\| = 0[/tex], [tex]\theta_{B} \in (-\infty, +\infty)[/tex] for [tex]\|\vec B\|\ne 0[/tex]. Undefined for [tex]\|\vec B\| = 0[/tex], [tex]\theta_{C} \in (-\infty, +\infty)[/tex] for [tex]\|\vec C\|\ne 0[/tex]. Undefined for [tex]\|\vec C\| = 0[/tex].

b) Magnitudes: [tex]\|\vec A\| \ge 0[/tex], [tex]\|\vec B\| \ge 0[/tex], [tex]\|\vec C\| = 0[/tex]; Directions: [tex]|\theta_{A}-\theta_{B}| = 180^{\circ}[/tex], [tex]\theta_{C}[/tex] is undefined.

a) Let suppose that [tex]\vec A \ne \vec O[/tex], [tex]\vec B \ne \vec O[/tex] and [tex]\vec C \ne \vec O[/tex], where [tex]\vec O[/tex] is known as Vector Zero. By definitions of Dot Product and Inverse Trigonometric Functions we derive expression for the magnitude and directions of [tex]\vec A[/tex], [tex]\vec B[/tex] and [tex]\vec C[/tex]:

Magnitude ([tex]\vec A[/tex])

[tex]\|\vec A\| = \sqrt{\vec A\,\bullet\,\vec A}[/tex]

[tex]\| \vec A\| \ge 0[/tex]

Magnitude ([tex]\vec B[/tex])

[tex]\|\vec B\| = \sqrt{\vec B\,\bullet\,\vec B}[/tex]

[tex]\|\vec B\| \ge 0[/tex]

Magnitude ([tex]\vec C[/tex])

[tex]\|\vec C\| = \sqrt{\vec C\,\bullet \,\vec C}[/tex]

[tex]\|\vec C\| \ge 0[/tex]

Direction ([tex]\vec A[/tex])

[tex]\vec A \,\bullet \,\vec u = \|\vec A\|\cdot \|u\|\cdot \cos \theta_{A}[/tex]

[tex]\theta_{A} = \cos^{-1} \frac{\vec A\,\bullet\,\vec u}{\|\vec A\|\cdot \|u\|}[/tex]

[tex]\theta_{A} = \cos^{-1} \frac{\vec A\,\bullet\,\vec u}{\|\vec A\|}[/tex]

[tex]\theta_{A} \in (-\infty, +\infty)[/tex] for [tex]\|\vec A\|\ne 0[/tex]. Undefined for [tex]\|\vec A\| = 0[/tex].

Direction ([tex]\vec B[/tex])

[tex]\vec B\,\bullet \, \vec u = \|\vec B\|\cdot \|\vec u\| \cdot \cos \theta_{B}[/tex]

[tex]\theta_{B} = \cos^{-1} \frac{\vec B\,\bullet\,\vec u}{\|\vec B\|\cdot \|\vec u\|}[/tex]

[tex]\theta_{B} = \cos^{-1} \frac{\vec B\,\bullet\,\vec u}{\|\vec B\|}[/tex]

[tex]\theta_{B} \in (-\infty, +\infty)[/tex] for [tex]\|\vec B\|\ne 0[/tex]. Undefined for [tex]\|\vec B\| = 0[/tex].

Direction ([tex]\vec C[/tex])

[tex]\vec C \,\bullet\,\vec u = \|\vec C\|\cdot\|\vec u\|\cdot \cos \theta_{C}[/tex]

[tex]\theta_{C} = \cos^{-1}\frac{\vec C\,\bullet\,\vec u}{\|\vec C\|\cdot\|\vec u\|}[/tex]

[tex]\theta_{C} = \cos^{-1} \frac{\vec C\,\bullet\,\vec u}{\|\vec C\|}[/tex]

[tex]\theta_{C} \in (-\infty, +\infty)[/tex] for [tex]\|\vec C\|\ne 0[/tex]. Undefined for [tex]\|\vec C\| = 0[/tex].

Please notice that [tex]\vec u[/tex] is the Vector Unit.

b) Let suppose that [tex]\vec A \ne \vec O[/tex] and [tex]\vec B \ne \vec O[/tex] and [tex]\vec C = \vec O[/tex]. Hence, [tex]\vec A = -\vec B[/tex]. In other words, we find that both vectors are antiparallel to each other, that is, that angle between [tex]\vec A[/tex] and [tex]\vec B[/tex] is 180°. From a) we understand that [tex]\|\vec A\| \ge 0[/tex], [tex]\|\vec B\| \ge 0[/tex], but [tex]\|\vec C\| = 0[/tex].

Then, we have the following conclusions:

Magnitude ([tex]\vec A[/tex])

[tex]\|\vec A\| \ge 0[/tex]

Magnitude ([tex]\vec B[/tex])

[tex]\|\vec B\| \ge 0[/tex]

Magnitude ([tex]\vec C[/tex])

[tex]\|\vec C\| = 0[/tex]

Directions ([tex]\vec A[/tex], [tex]\vec B[/tex]):

[tex]|\theta_{A}-\theta_{B}| = 180^{\circ}[/tex]

Direction ([tex]\vec C[/tex]):

Undefined

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.

!! HOPE ITS HELP U !!

Answer:

C. A force must be acting on the object.

Explanation:

This is due to the action of its momentum direction.

[tex].[/tex]

Answer:

Explanation:

That is an amazing fact.

The minus sign is what you have to pay attention to. The earth has a mass of 100 times that of Saturn. As someone on here once noted, Saturn has such a low density that it would float in water.

The answer is D