What is the minimum work needed to push a 925-kg car 207 m up along a 14.5° incline? Ignore friction.

Answer:

Option C is the correct answer.

Explanation:

By Charles's law we have

        V ∝ T

That is

       [tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

Here given that

      V₁ = 0.20 cubic meter

      T₁ = 333 K

      T₂ = 533 K

Substituting

      [tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}\\\\\frac{0.20}{333}=\frac{V_2}{533}\\\\V_2=\frac{0.20}{333}\times 533=0.3198m^3[/tex]

New volume of the gas  = 0.3198 m³

Option C is the correct answer.

Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

[tex]\Delta P.E=mg(h_{2}-h_{1})[/tex]

[tex]\Delta P.E=10000\times9.8\times(10000-0)[/tex]

[tex]\Delta P.E=10000\times9.8\times10000[/tex]

[tex]\Delta P.E=980000000\ J[/tex]

[tex]\Delta P.E=980\ MJ[/tex]

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

[tex]\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)[/tex]

[tex]\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)[/tex]

[tex]\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)[/tex]

[tex]\Delta K.E=148298642\ J[/tex]

[tex]\Delta K.E=148.3\ MJ[/tex]

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

The problem involves applying principles of soil mechanics to a direct shear test on a sand specimen. The calculations involve converting the specimen's physical dimensions to an area and applying formulas related to shear stress and friction angle.

The problem given is an application of soil mechanics principles in civil engineering. The situation is a direct shear test on a dry sand specimen. To answer this, we need to understand principles related to shearing force, shearing stress and their relationship with angle of internal friction (φ‘) and normal stress.

Shear stress can be calculated using the formula τ = F/A, where F is the force and A is the area over which the force is distributed. The area can be calculated based on the specimen's dimensions using A = πr² (where r is the radius of the specimen, which is half of the diameter). Given the normal stress and the shear stress, the angle of friction φ’ can be calculated using the formula tan(φ‘) = τ / σ, where σ is the normal stress.

To calculate the shear force required to cause failure under a different normal stress, we use the above formula in reverse, solving for τ (which represents the shear stress under the new normal stress), then multiply by the area to obtain the force. In other words, F = τA.

Please note that this is a simplified calculation ignoring potential complexities of real-world soil behavior.

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The angle of friction φ' is approximately 30.50°. For a normal stress of 200 kN/m², the required shear force to cause failure is approximately 367.97 N.

To determine the angle of friction (φ') and the required shear force at a different normal stress, we will conduct the following calculations:

a. Determining the angle of friction, φ’

Determine the cross-sectional area (A) of the specimen:

Diameter (d) = 63 mm = 0.063 m

Area (A) = π/4 × d² = (3.1416/4) × (0.063²) ≈ 0.003117 m²

Calculate the shear stress (τ) at failure:

τ = Shear Force (F) / Area (A)

F = 276 N

τ = 276 N / 0.003117 m² ≈ 88555.19 N/m² = 88.56 kN/m²

Calculate the angle of friction (φ'):

Normal stress (σ) = 150 kN/m²

The relationship between shear stress and normal stress in terms of the angle of friction (φ') is given by:

τ = σ × tan(φ')

88.56 kN/m² = 150 kN/m² × tan(φ')

tan(φ') = 88.56 / 150 ≈ 0.5904

φ' = atan(0.5904) ≈ 30.50°

b. For a normal stress of 200 kN/m², Shear Force required to cause failure:

Calculate the shear stress (τ) using the angle of friction:

σ = 200 kN/m²

τ = σ × tan(φ')

τ = 200 kN/m² × tan(30.50°) ≈ 118.08 kN/m²

Determine the required shear force (F):

τ = F / A

F = τ × A

F = 118.08 kN/m² × 0.003117 m² ≈ 367.97 N

Conclusion:

The angle of friction (φ') is approximately 30.50°.

The shear force required to cause failure at a normal stress of 200 kN/m² is approximately 367.97 N.

Answer:

T=78.48 N

Explanation:

We know that the moment developed at a hinge equals zero

Thus summing moments about hinge we have

(See attached figure)

[tex]2.0\times g\times \frac{1.5}{2}+1.50\times 5\times g-T\times \frac{3}{4}1.5=0\\\\Solving\\\\T=\frac{1}{1.125}(88.29)\\\\T=78.48N[/tex]

Answer:

The speed of the skier after moving 100 m up the slope are of V= 25.23 m/s.

Explanation:

F= 280 N

m= 80 kg

α= 12º

μ= 0.15

d= 100m

g= 9,8 m/s²

N= m*g*sin(α)

N= 163 Newtons

Fr= μ * N

Fr= 24.45 Newtons

∑F= m*a

a= (280N - 24.5N) / 80kg

a= 3.19 m/s²

d= a * t² / 2

t=√(2*d/a)

t= 7.91 sec

V= a* t

V= 3.19 m/s² * 7.91 s

V= 25.23 m/s

Answer:

The pressure that the skater exerts on ice if they are balanced on a single skate is P= 18,312 * 10⁶ Pa = 180.72 atm.

Explanation:

width= 0.025cm = 2.5 * 10⁻⁴ m

lenght= 15 cm = 0.15m

m= 70 kg

g= 9.81 m/s²

S= width * lenght

S= 3.75 * 10⁻⁵ m²

F= m*g

F= 686.7 N

P= F/S

P= 18.312 * 10⁶ Pa= 180.72 atm

Answer:

- 7.56 C

Explanation:

E = 105 N/C downwards

m = 81 kg

Let the charge on the man is q.

To lose te contact with the ground, the electrostatic foece should be balanced by the weight of the person.

The charge should be negative in nature so that the direction of electrostatic force is upwards and weight is downwards.

q E = m g

q = (81 x 9.8) / 105 = 7.56 C

Answers:

(a) [tex]0.0073kg[/tex]

(b) Volume gold: [tex]3.79(10)^{-7}m^{3}[/tex], Volume cupper: [tex]7.6(10)^{-8}m^{3}[/tex]

(c) [tex]17633.554kg/m^{3}[/tex]

Explanation:

We are told the total mass [tex]M[/tex] of the coin, which is an alloy  of gold and copper is:

[tex]M=m_{gold}+m_{copper}=7.988g=0.007988kg[/tex]   (1)

Where  [tex]m_{gold}[/tex] is the mass of gold and [tex]m_{copper}[/tex] is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats [tex]K[/tex] and mass is:

[tex]K=24\frac{m_{gold}}{M}[/tex]   (2)

Finding [tex]{m_{gold}[/tex]:

[tex]m_{gold}=\frac{22}{24}M[/tex]   (3)

[tex]m_{gold}=\frac{22}{24}(0.007988kg)[/tex]   (4)

[tex]m_{gold}=0.0073kg[/tex]   (5)  This is the mass of gold in the coin

The density [tex]\rho[/tex] of an object is given by:

[tex]\rho=\frac{mass}{volume}[/tex]

If we want to find the volume, this expression changes to: [tex]volume=\frac{mass}{\rho}[/tex]

For gold, its volume [tex]V_{gold}[/tex] will be a relation between its mass [tex]m_{gold}[/tex]  (found in (5)) and its density [tex]\rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}[/tex]:

[tex]V_{gold}=\frac{m_{gold}}{\rho_{gold}}[/tex]   (6)

[tex]V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}[/tex]   (7)

[tex]V_{gold}=3.79(10)^{-7}m^{3}[/tex]   (8)  Volume of gold in the coin

For copper, its volume [tex]V_{copper}[/tex] will be a relation between its mass [tex]m_{copper}[/tex]  and its density [tex]\rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}[/tex]:

[tex]V_{copper}=\frac{m_{copper}}{\rho_{copper}}[/tex]   (9)

The mass of copper can be found by isolating [tex]m_{copper}[/tex] from (1):

[tex]M=m_{gold}+m_{copper}[/tex]  

[tex]m_{copper}=M-m_{gold}[/tex]  (10)

Knowing the mass of gold found in (5):

[tex]m_{copper}=0.007988kg-0.0073kg=0.000688kg[/tex]  (11)

Now we can find the volume of copper:

[tex]V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}[/tex]   (12)

[tex]V_{copper}=7.6(10)^{-8}m^{3}[/tex]   (13)  Volume of copper in the coin

Remembering density is a relation between mass and volume, in the case of the coin the density [tex]\rho_{coin[/tex] will be a relation between its total mass [tex]M[/tex] and its total volume [tex]V[/tex]:

[tex]\rho_{coin}=\frac{M}{V}[/tex] (14)

Knowing the total volume of the coin is:

[tex]V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3}[/tex] (15)

[tex]\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}}[/tex] (16)

Finally:

[tex]\rho_{coin}=17633.554kg/m^{3}}[/tex] (17)  This is the total density of the British sovereign coin

Answer:

The electric field at x = 0.500 m is 0.02 N/C.

Explanation:

Given that,

Point charge at the origin[tex]q_{1} = 7.00\ nC[/tex]

Second point charge[tex]q_{2}=-250\ nC[/tex]  at x = +0.800 m

We calculate the electric field at x = 0.500 m

Using formula of electric field

[tex]E=\dfrac{kq}{r^2}[/tex]

The electric field at x = 0.500 m

[tex]E=\dfrac{k\times7\times10^{-9}}{(5)^2}+\dfrac{k\times(-2.50)\times10^{-9}}{(3)^2}[/tex]

[tex]E=9\times10^{9}(\dfrac{7\times10^{-9}}{25}-\dfrac{2.50\times10^{-9}}{9})[/tex]

[tex]E = 0.02\ N/C[/tex]

Hence, The electric field at x = 0.500 m is 0.02 N/C.

Answer: The electric field at x = 0.5 m is equal to 1.96 N/C, and the direction is in the postive x-axis (to the rigth)

Explanation:

I will use the notations (x, y, z)

the first particle is located at the point (0m, 0m, 0m) and has a charge q1 = 7.00 nC

the second particle is located at the point (0.8m, 0m, 0m) and has a charge q2 =  -2.50 nC

Now, we want to find the electric field at the point (0.5m, 0m, 0m)

First, we can see that we only work on the x-axis, so we can think about this problem as one-dimensional.

First, the electric field done by a charge located in the point x0 is equal to:

E(x) = Kc*q/(x - x0)^2

where Kc is a constant, and it is Kc = 8.9*10^9 N*m^2/C^2

then, the total magnetic field will be equal to the addition of the magnetic fields generated by the two charges:

E(0.5m) = Kc*q1/0.5m^2 + Kc*q2/(0.5m - 0.8m)

E(0.5m) = Kc*(7.0nC/(0.5m)^2 - 2.5nC/(0.3m)^2)

E(0.5m) = Kc*(0.22nC/m^2)

now, remember that Kc is in coulombs, so we must change the units from nC to C

where 1nC = 1*10^-9 C

E(0.5m) = (8.9*10^9 N*m^2/C^2)*(0.22x10^-9C/m^2) = 1.96 N/C

the fact that is positive means that it points in the positve side of the x-axis.

Answer:

D). [tex]327 ^0 C[/tex]

Explanation:

As we know that temperature scale is linear so we will have

[tex]\frac{^0C - 0}{100 - 0} = \frac{K - 273}{373 - 273}[/tex]

now we have

[tex]\frac{^0 C - 0}{100} = \frac{K - 273}{100}[/tex]

so the relation between two scales is given as

[tex]^0 C = K - 273[/tex]

now we know that in kelvin scale the absolute temperature is 600 K

so now we have

[tex]T = 600 - 273 = 327 ^0 C[/tex]

so correct answer is

D). [tex]327 ^0 C[/tex]

Answer:

The angular position of the wheel after 2.7 seconds is θf= 11.35 rad.

Explanation:

θi= 8.8 rad

ωi= 0 rad/seg

α= 0.7 rad/seg²

θf= θi + ωi * t + α * t² / 2

θf= 11.35 rad

In physics, inertia refers to an object's resistance to a change in motion, and it is directly proportional to the object's mass. This means the object with the greater mass will have more inertia. Given the provided options, the 7-kilogram object at rest has the greatest inertia because it has the most mass.

The subject in question relates to inertia, a concept in physics that describes an object's resistance to a change in motion. Inertia is directly proportional to an object's mass, meaning an object with more mass exhibits greater inertia. Hence, considering the options a. A 2-kilogram object moving at 5 m/s, b. A 5-kilogram object moving at 3 m/s, c. A 7-kilogram object at rest, and d. A 3-kilogram object moving at 4 m/s, the object that has the greatest inertia would be c. A 7-kilogram object at rest. This is because it has the greatest mass out of all the options.

Inertia is associated with Newton's first law of motion, which states that an object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force. This can be seen in daily life - for example, it's more difficult to push a heavy truck into motion than a small toy because the truck has a greater mass and hence more inertia.

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Answer:

2.00 m/s²

Explanation:

Given

The Mass of the metal safe, M = 108kg

Pushing force applied by the burglar,  F = 534 N

Co-efficient of kinetic friction, [tex]\mu_k[/tex] = 0.3

Now,

The force against the kinetic friction is given as:

[tex]f = \mu_k N = u_k Mg[/tex]

Where,

N = Normal reaction

g= acceleration due to the gravity

Substituting the values in the above equation, we get

[tex]f = 0.3\times108\times9.8[/tex]

or

[tex]f = 317.52N[/tex]

Now, the net force on to the metal safe is

[tex]F_{Net}= F-f[/tex]

Substituting the values in the equation we get

 [tex]F_{Net}= 534N-317.52N[/tex]

or

[tex]F_{Net}= 216.48[/tex]

also,

[tex]F_{Net}= M\times [/tex]acceleration of the safe

Therefore, the acceleration of the metal safe will be

acceleration of the safe=[tex] \frac{F_{Net}}{M} [/tex]

or

 acceleration of the safe=[tex] \frac{216.48}{108} [/tex]

or

acceleration of the safe=[tex] 2.00 m/s^2 [/tex]

Hence, the acceleration of the metal safe will be  2.00 m/s²

The acceleration of the safe is determined by factoring in the force exerted by the burglar, the static friction that initiates movement, and the kinetic friction that must be overcome when the safe is in motion. When these factors are calculated, the acceleration comes out to be approximately 2 m/s².

The subject in question deals with two types of force: the force applied by the burglar and the frictional force which acts against the direction of the motion. The gravitational force acting on the safe, also known as its weight, can be calculated by multiplying the safe's mass (108 kg) with the acceleration due to gravity (approx. 9.80 m/s²), which gives us a value of 1058.4 N. This weight also represents the normal force, as the safe is on a horizontal plane.

The maximum force of static friction, calculated using the formula ƒs_max = μsN (where μs is the coefficient of static friction and N is the normal force), turns out to be 0.4 * 1058.4 N = 423.36 N. This implies the burglar needs to exert a force greater than this to overcome the static friction and set the safe in motion.

Given that the burglar can apply a maximum force of 534 N, this is significantly greater than the static friction, inducing motion in the safe. Once the safe is moving, it's the force of kinetic friction that matters. Calculating this force gives us 0.3 * 1058.4 N = 317.52 N. This is the force that has to be overcome to maintain the safe in motion.

Using Newton's second law (F = ma), we can determine the acceleration by subtracting kinetic friction from the applied force and dividing it by the mass of the safe. This gives us an acceleration of (534N - 317.52N) / 108kg = 2 m/s². Therefore, the safe would indeed move, and its acceleration would be 2 m/s².

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Answer:

14.7 Volt

Explanation:

C1 = 2.74 F, C2 = 7.46 F, C3 = 10.1 F

Here C1 and C3 are in parallel

So, Cp = C1 + C3 = 2.74 + 10.1 = 12.84 F

Now Cp and C2 are in series

1 / C = 1 / Cp + 1 / C2

1 / C = 1 / 12.84 + 1 / 7.46

C = 4.72 F

Let q be the total charge

q = C V = 4.72 x 40 = 188.8 C

Voltage across C2

V2 = q / C2 = 188.8 / 7.46 = 25.3 V

Voltage across C2 or c3

V' = V - V2 = 40 - 25.3

V' = 14.69

V' = 14.7 Volt

Answer:

1.586m

Explanation:

let 'u' be the velocity of the roof tile when it reaches the window

now using the equation of motion

[tex]s=ut+\frac{1}{2}at^{2}[/tex]

where s= distance travelled

u=velocity

a=acceleration of the object

t=time taken to travel the distance 's'

given:

s=1.7m (distance covered to pass the window)

t=0.25s (Time taken to pass the window)

a=g=9.8m/s^2 (since the roof tile is moving under the action of gravity)

thus, substituting the values in the above equation we get

[tex]1.7=u\times 0.25+\frac{1}{2}\times 9.8\times0.25^{2}[/tex]

u=5.575m/s

This is the velocity when the tile  touches the window top.

Let's take this in second scenario as the tile's final velocity(v).

Now we have another equation of motion as

[tex]v^{2}-u^{2}=2as[/tex]

initial speed (when starts to fall) will be zero.

So the distance travelled (h) i.e the height from which the tile falls from the top of the window is given by,

substituting the values in the above equation, we get

[tex]5.575^{2}-0^{2}=2\times 9.8\times h[/tex]

[tex]h=\frac{5.575^{2}-0^{2}}{2\times 9.8}[/tex]

h=1.586m

Hence, the window roof is 1.586m far away from the roof

Final answer:

To determine the height above the window where the roof tile fell, we use the kinematic equations with the given time and the height of the window. After calculating the velocity at the bottom of the window and the time taken to pass, we find the total distance fallen from the roof and subtract the window height to get the height above the window.

To determine how far above the top of the window the roof is where the roof tile fell, we will apply kinematic equations for uniformly accelerated motion, which in this case, is due to gravity. The tile falls past the window in 0.25 s, covering a distance of 1.7 m.

First, we calculate the velocity of the tile at the bottom of the window using the equation v = v_0 + at, where v_0 is the initial velocity (0 m/s since it falls from rest), a is the acceleration due to gravity (9.81 m/s2), and t is the time it takes to pass the window. We assume the tile's speed at the top of the window is approximately the same as at the bottom since the window height is relatively small. This gives us a calculation to find the speed at the bottom: v = 0 m/s + (9.81 m/s2)(0.25 s). We'll use this speed as an average speed to simplify the calculation.

Using d = vt, where d is the distance covered (1.7 m) and v is the average velocity, we solve for the time it takes to pass the window. This provides half the time of passage, thus t = d/v. Then we use the time to find the total distance fallen from the roof, d_total, using the equation d_total = [tex]0.5at^2[/tex].  Finally, we subtract the window height from d_total to find the height above the window where the tile fell, which is the answer to the question.

Answer:

8.6 kg

Explanation:

According to the question,

1 Kg = 2.21 pound

2.21 pound = 1 kg

1 pound = 1 / 2.21 kg

19 pound = 19 / 2.21 kg = 8.59 kg

By rounding to nearest tenth, it is equal to 8.6 kg.

Answer:

a. 76.92 MW

b. 26.92 MW

Explanation:

η = Efficiency of the power plant = 0.35

Q₂ = rate at which heat rejected to surrounding = 50 MW

Q₁ = Rate of Input heat = ?

Efficiency of the power plant is given as

[tex]\eta =1-\frac{Q_{2}}{Q_{1}}[/tex]

[tex]0.35 =1-\frac{50}{Q_{1}}[/tex]

Q₁ = 76.92 MW

Net Rate of work is given as

Q = Q₁ - Q₂

Q = 76.92 - 50

Q = 26.92 MW

Answer:

Option (e)

Explanation:

A = 45 cm^2 = 0.0045 m^2, d = 0.080 mm = 0.080 x 10^-3 m,

Energy density = 100 J/m

Let Q be the charge on the plates.

Energy density = 1/2 x ε0 x E^2

100 = 0.5 x 8.854 x 10^-12 x E^2

E = 4.75 x 10^6 V/m

V = E x d

V = 4.75 x 10^6 x 0.080 x 10^-3 = 380.22 V

C = ε0 A / d

C = 8.854 x 10^-12 x 45 x 10^-4 / (0.080 x 10^-3) = 4.98 x 10^-10 F

Q = C x V = 4.98 x 10^-10 x 380.22 = 1.9 x 10^-7 C

Q =  190 nC

Answer:

The magnitude of the angular momentum of the cylinder and the rotational kinetic energy of the cylinder are 0.0205 Kgm²/s and 0.01317 J

Explanation:

Given that,

Moment of inertia = 0.016 kg m²

Radius = 6.0

Linear speed = 7.7 m/s

We need to calculate the angular momentum

Using formula of angular momentum

[tex]L=I\omega[/tex]

Where, L = angular momentum

I = moment of inertia

[tex]\omega[/tex] =angular velocity

Put the value into the formula

[tex]L=0.016\times\dfrac{7.7}{6.0}[/tex]

[tex]L=0.0205\ Kg m^2/s[/tex]

We need to calculate the rotational kinetic energy of the cylinder

Using formula of Rotational kinetic energy

[tex]K.E=\dfrac{1}{2}\times I\omega^2[/tex]

[tex]K.E= \dfrac{1}{2}\times I\times(\dfrac{v}{r})^2[/tex]

[tex]K.E= \dfrac{1}{2}\times0.016\times(\dfrac{7.7}{6.0})^2[/tex]

[tex]K.E=0.01317\ J[/tex]

Hence, The magnitude of the angular momentum of the cylinder and the rotational kinetic energy of the cylinder are 0.0205 Kg m²/s and 0.01317 J

Answer:

The pressure will be of 399.17 mmHg.

Explanation:

p1= 754 mmHg

V1= 4.5 L

p2= ?

V2= 8.5 L

p1*V1 = p2*V2

p2= (p1*V1)/V2

p2= 399.17 mmHg

Answer:

[tex]r= 2.03*10^7m[/tex]

V = 1.45x10^3 m/s

Explanation:

number of days in sec = 1.02days * 86400s = 88128 s              

Mass of Mars is 6.42*10^23 Kg

gravitational constant G = 6.674*10^{-11}

[tex]T^{2} = \frac {4pi^2}{GM} *r^3[/tex]

[tex]\frac {88128^2}{(9.22*10^{-13})} = r^3[/tex]

[tex]r= 2.03*10^7m[/tex]

(b) [tex]V=\frac{2 \pi*r}{T}[/tex]

[tex]V=\frac{(2 \pi(2.03*10^7))}{(88128)} = 1.45x10^3 m/s[/tex]

V

Explanation:

It is given that,

Let q₁ and q₂ are two small positively charged spheres such that,

[tex]q_1+q_2=5\times 10^{-5}\ C[/tex].............(1)

Force of repulsion between the spheres, F = 1 N

Distance between spheres, d = 2 m

We need to find the charge on the sphere with the smaller charge. The force is given by :

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

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

[tex]q_1q_2=\dfrac{1\ N\times (2\ m)^2}{9\times 10^9}[/tex]

[tex]q_1q_2=4.45\times 10^{-10}\ C[/tex]............(2)

On solving the system of equation (1) and (2) using graph we get,

[tex]q_1=0.0000384\ C=38.4\ \mu C[/tex]

[tex]q_2=0.0000116\ C=11.6\ \mu C[/tex]

So, the charge on the smaller sphere is 11.6 micro coulombs. Hence, this is the required solution.

Combined charge = 5.0 x 10-5 C
Distance between spheres = 2.0 m
Force = 1.0 N

Formula:
Coulomb's Law: F = k * (q1 * q2) / r2

Calculations:
q1 + q2 = 5.0 x 10-5 C (Given)
q1 = q2 - x (Assuming q2 is larger charge)
Substitute and solve for x

Answer:
The charge on the sphere with the smaller charge is 2.5 x 10-5 C or 25 micro-coulombs.

Answer:

convection C

Explanation:

Answer:

C. convection

Answer:

The value of D is [tex]7.753\times10^{-14}\ m^2/s[/tex]

Explanation:

Given that,

Temperature = 705°C

Maximum diffusion [tex]D_{0}=4.5\times10^{-5}\ m^2/s[/tex]

Activation energy [tex]Q_{d} = 164 kJ/mol[/tex]

We need to calculate the value of D

Using formula of diffusion coefficient

[tex]D=D_{0}\ exp\ (\dfrac{-Q_{d}}{RT})[/tex]

Where, D = diffusion coefficient

[tex]D_{0}[/tex] = Maximum diffusion coefficient

[tex]Q_{d}[/tex] = Activation energy

T = temperature

R = Gas constant

Put the value into the formula

[tex]D=4.5\times10^{-5}\ exp\ (\dfrac{-164\times10^{3}}{8.31\times705+273})[/tex]

[tex]D=7.753\times10^{-14}\ m^2/s[/tex]

Hence, The value of D is [tex]7.753\times10^{-14}\ m^2/s[/tex]

Final answer:

To find the diffusion coefficient D at 705°C, we use the Arrhenius equation, convert the activation energy to eV, and then calculate D using the given values of D0, Qd, Boltzmann's constant, and the temperature in Kelvin. After performing the calculations, we will obtain the required value for D at 705°C.

Explanation:

To calculate the value of D (diffusion coefficient) at 705°C for the diffusion of a species in a metal, we use the Arrhenius type equation for diffusion: D = D0 × exp(-Qd / (k × T)), where D0 is the pre-exponential factor, Qd is the activation energy for diffusion, k is Boltzmann's constant (8.617 x 10-5 eV/K), and T is the absolute temperature in Kelvin (K).

First convert the temperature from Celsius to Kelvin: T = 705°C + 273.15 = 978.15 K.

Then plug in the given values: D0 = 4.5 x 10-5 m2/s and Qd = 164 kJ/mol (which is 164000 J/mol).

Using these values, calculate D at 705°C:

D = 4.5 x 10-5 m2/s × exp(-164000 J/mol / (8.617 x 10-5 eV/K × 978.15 K))

Since 1 eV = 1.602 x 10-19 J, we can convert the activation energy to eV by dividing by this conversion factor:

Qd in eV = 164000 J/mol / (1.602 x 10-19 J/eV) = 1023629.84 eV/mol

Now insert the activation energy in eV into the equation:

D = 4.5 x 10-5 m2/s × exp(-1023629.84 eV/mol / (8.617 x 10-5 eV/K × 978.15 K))

After performing the calculations, we will obtain the required value for D at 705°C.

Answer:

Option (D)

Explanation:

The definition of specific heat is given below

The amount of heat required to raise the temperature of 1 kg substance by 1 degree Celsius.

Q = m c (T2 - T1)

c = Q / m × (T2 - T2)

Answer:

Charge, [tex]q=-2.14\times 10^{-5}\ C[/tex]

Explanation:

It is given that,

Mass of the charged particle, m = 1.44 g = 0.00144 kg

Electric field, E = 660 N/C

We need to find the charge of that particle to remain stationary when placed in a downward-directed in the given electric field such that its weight is balanced by the electrostatic force i.e.

[tex]mg=qE[/tex]

[tex]q=\dfrac{mg}{E}[/tex]

[tex]q=\dfrac{0.00144\ kg\times 9.81\ m/s^2}{660\ N/C}[/tex]

q = 0.0000214 C

[tex]q=2.14\times 10^{-5}\ C[/tex]

Since, the electric field is acting in downward direction, so the electric force will act in opposite direction such that they are in balanced position. Hence, the charge must be negative.

i.e. [tex]q=-2.14\times 10^{-5}\ C[/tex]

The magnetic force acting on a charged particle moving perpendicular to the field is:

[tex]F_{b}[/tex] = qvB

[tex]F_{b}[/tex] is the magnetic force, q is the particle charge, v is the particle velocity, and B is the magnetic field strength.

The centripetal force acting on a particle moving in a circular path is:

[tex]F_{c}[/tex] = mv²/r

[tex]F_{c}[/tex] is the centripetal force, m is the mass, v is the particle velocity, and r is the radius of the circular path.

If the magnetic force is acting as the centripetal force, set [tex]F_{b}[/tex] equal to [tex]F_{c}[/tex] and solve for v:

qvB = mv²/r

v = qBr/m

Due to the work-energy theorem, the work done on the proton by the potential difference V becomes the proton's kinetic energy:

W = KE

W is work, KE is kinetic energy

W = Vq

KE = 0.5mv²

Therefore:

Vq = 0.5mv²

Substitute v = qBr/m and solve for V:

V = 0.5qB²r²/m

Given values:

m = 1.67×10⁻²⁷kg (proton mass)

B = 0.750T

q = 1.60×10⁻¹⁹C (proton charge)

r = 1.84×10⁻²m

Plug in the values and solve for V:

V = (0.5)(1.60×10⁻¹⁹)(0.750)²(1.84×10⁻²)²/1.67×10⁻²⁷

V = 9120V

The proton, initially accelerated through a potential difference V and then making a circular path in a magnetic field, allows us to calculate that V is approximately 8.74 x 10^5 volts.

The question involves the concept of a proton moving in a magnetic field after being accelerated through a voltage V. As a proton enters a magnetic field perpendicular to its path, it follows a circular arc. Let's use the known concepts of physics to derive the required voltage.

The radius of the proton's path can be calculated using the Lorentz force formula: F = qvB = mv^2/r, where q is the charge of the proton, v is the velocity, B is the magnetic field, m is the proton's mass, and r is the radius. From this equation, we can express velocity as v = qBr/m.

Next, we know that the kinetic energy of the proton (K.E.) equals the work done on it, which is the voltage times the charge of the proton: K.E. = qV. Also, the kinetic energy can be expressed as K.E. = 1/2 mv^2.

Equating these two forms of kinetic energy, we get 1/2 mv^2 = qV. Substituting our expression for velocity from above, we find V = ( q^2 B^2 r^2) / (2 m).

Plug in the known values: q = 1.60 x 10^-19 C, B = 0.750 T, r = 1.84 x 10^{-2} m (converted from cm to m), and m = 1.67 x 10^-27 kg (mass of a proton), we find that V is approximately 8.74 x 10^5 volts.

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Using the given intensity of the electromagnetic wave and the fundamental constants, we substitute into the formula B = √(2I/μoc²). The resulting amplitude of the magnetic field is approximately 7.98 × 10-⁶ Tesla.

The student is asking for the calculation of the amplitude of a magnetic field given the intensity of an electromagnetic wave. This belongs to the realm of Physics, specifically electromagnetism. We can use the formula I = 1/2μoc²B² to solve for this, where I is the intensity, μo is the permeability of free space, c is the speed of light, and B is the maximum strength of the magnetic field.

First, we rearrange the formula to solve for B, yielding B = √(2I/μoc²). Substituting the given values, we get B = √(2*80x10⁶ W/m²/(4π×10−7 T m/A * (3.0×10⁸ m/s)²). Calculating this gives us a magnetic field amplitude of approximately 7.98 × 10-⁶ T.

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Final answer:

To find the amplitude of the magnetic field for an electromagnetic wave with a given intensity, use the formula I = cε0B2. For an intensity of 80 MW/m2, the calculated magnetic field amplitude is approximately 1.69×10−3 T (1.69 mT).

Explanation:

The intensity I of an electromagnetic wave can be related to its magnetic field B using the relationship I = cε0B2, where c is the speed of light in a vacuum, and ε0 is the permittivity of free space. Given that the intensity I is 80 MW/m2, we can use the given values for c (3.0×108 m/s) and ε0 (8.85×10−12C2/N·m2) to find the amplitude of the magnetic field.

First, rearrange the formula to solve for B:

B = √(I / (cε0))

Substitute the given values:

B = √(80×106 W/m2 / (3.0×108 m/s × 8.85×10−12 C2/N·m2))

After performing the calculations:

B = 1.69×10−3 T

Therefore, the amplitude of the magnetic field for an electromagnetic wave with an intensity of 80 MW/m2 is approximately 1.69 mT (milliteslas).

Answer:

Velocity of the particle at time t = a

        [tex]v(a)=-\frac{12}{a^3}[/tex]

Velocity of the particle at time t = 1

         [tex]v(1)=-12m/s[/tex]

Velocity of the particle at time t = 2

         [tex]v(2)=-1.5m/s[/tex]

Velocity of the particle at time t = 3

          [tex]v(3)=-0.44m/s[/tex]

Explanation:

Displacement,

          [tex]s(t)=\frac{6}{t^2}[/tex]

Velocity is given by

          [tex]v(t)=\frac{ds}{dt}=\frac{d}{dt}\left ( \frac{6}{t^2}\right )=-\frac{12}{t^3}[/tex]

Velocity of the particle at time t = a

        [tex]v(a)=-\frac{12}{a^3}[/tex]

Velocity of the particle at time t = 1

         [tex]v(1)=-\frac{12}{1^3}=-12m/s[/tex]

Velocity of the particle at time t = 2

         [tex]v(2)=-\frac{12}{2^3}=-1.5m/s[/tex]

Velocity of the particle at time t = 3

          [tex]v(3)=-\frac{12}{3^3}=-0.44m/s[/tex]

The velocity of the particle for different times (t = a, 1, 2, and 3 seconds) is found by differentiating the displacement equation s = 6/t^2 and applying the values of t to get the velocities: v(a) = -12/a^3, v(1) = -12 m/s, v(2) = -1.5 m/s, and v(3) ≈ -0.44 m/s.

The question you've asked is about finding the velocity of a particle moving in a straight line where its displacement is given by s = 6/t2, and t is the time in seconds. To find the velocity (v) at any given time, we need to take the derivative of the displacement with respect to time. So the derivative of s with respect to t gives us v = -12/t3. Let's apply this formula for t = a, 1, 2, and 3 seconds.

Note that the negative sign indicates the direction of velocity is opposite to the direction assumed as positive in the displacement equation.

Answer:

Explanation:

To estimate the focal length of a convex lens follow the following steps.

1. take a convex lens.

2. Stand near a window which is just opposite to a wall.

3. Look at a tree which is far away from the window by the convex lens.

4. focus the image of the tree on the wall which is opposite to the window.

5. You wll observe that by changing the position of convex lens a sharp and inverted and small image is seen on the wall.

5. Now measure the distance between the lens and the wall.

7. This distance is the rough focal length of the convex lens.

To estimate the focal length of an unmarked convex lens, you can hold the lens in front of a bright object and project the image onto a blank wall until it's clear and at its smallest size. The distance between the lens and wall is the focal length. Refraction of light in materials like water changes their lens properties.

To estimate the focal length of an unmarked convex lens quickly, you can employ a simple method using readily available materials. Here's a step-by-step explanation of this experimental process:

This method is based on the fact that when an object is placed at a great distance from a converging lens (much greater than the focal length), the image is formed at the focal point on the other side of the lens. By finding this point of clear image formation, you effectively measure the focal length of the convex lens.

Impact of Refraction

When you fill a glass or plastic bottle with water, it can act as a converging lens due to the refraction of light. The water inside the bottle has a different index of refraction compared to the air, which allows the bottle to focus light and form images like a lens. The curvature of the bottle and the water's index of refraction contribute to the lens properties and focal length of the water bottle lens.