A proton moves at 4.50 × 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.60 × 103 N/C. Ignoring any gravitational effects, find

(a) the time interval required for the proton to travel 5.00 cm horizontally,

(b) its vertical displacement during the time interval in which it travels 5.00 cm horizontally,

(c) the horizontal and vertical components of its velocity after it has traveled 5.00 cm horizontally.

Answer:

a)frequency of the oscillation = 1/(2*pi)*square root (k/x) =1/(2*pi)*square root (450/(15*10^-2))=8.72 cycle/second

b)spring potential energy = 0.5*k*(x)^2

=0.5*450*(15*10^-2)^2 =5.0625 joule

maximum kinetic energy =spring potential energy

c)

maximum kinetic energy=5.0625

kinetic energy=0.5*m*v^2

v=square toot ((5.0625/(0.5*2.5)) =2 m/s

Answer:

a)

85.05 N/m

b)

179.81 rad/s

Explanation:

a)

k = spring constant of the spring

m = mass of the block = 0.473 kg

x = stretch caused in the spring = 0.109 m

h = height dropped by the block = 0.109 m

Using conservation of energy

Spring potential energy gained by the spring = Potential energy lost by the block

(0.5) k x² = mgh

(0.5) k x² = mgx

(0.5) k x = mg

(0.5) k (0.109) = (0.473) (9.8)

k = 85.05 N/m

b)

angular frequency is given as

[tex]w = \sqrt{\frac{k}{m}}[/tex]

[tex]w = \sqrt{\frac{85.05}{0.473}}[/tex]

[tex]w [/tex] = 179.81 rad/s

The spring constant of the spring is 42.54 N/m, and the angular frequency of the block's vibrations is 4.88 rad/s.

To find the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is proportional to its displacement. In this case, the weight of the block is equal to the force provided by the spring at the equilibrium position.

Using the equation F = kx, where F is the force, k is the spring constant, and x is the displacement, we can solve for k. Since the block momentarily comes to rest after dropping 0.109 m, we can set the force provided by the spring equal to the weight of the block and solve for k.

Given:

Mass of the block (m) = 0.473 kg

Displacement of the block (x) = 0.109 m

Using the equation F = kx, we can rewrite it as k = F/x. The weight of the block is equal to its mass multiplied by the acceleration due to gravity (9.8 m/s^2), so the force provided by the spring is 0.473 kg * 9.8 m/s^2 = 4.6354 N. Substituting these values into the equation, we find the spring constant (k) to be:

k = 4.6354 N / 0.109 m = 42.54 N/m

To find the angular frequency of the block's vibrations, we can use the equation:

ω = sqrt(k/m)

Substituting the values of k and the mass of the block (m) = 0.473 kg into the equation, we can calculate the angular frequency (ω):

ω = sqrt(42.54 N/m / 0.473 kg) = 4.88 rad/s

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Answer:

The acceleration and charge are 17.122 m/s² and [tex]8.6\times10^{-5}\ C[/tex]

Explanation:

Given that,

Mass of object = 0.8 g

Electric field = 159 N/C

Distance = 72 m

Time = 2.9 s

We know that,

The electric force is

[tex]F = Eq[/tex]....(I)

The newton's second law

[tex]F=ma[/tex]

Put the value of F in the equation (I)

[tex]ma=Eq[/tex]...(II)

We calculate the acceleration

Using equation of motion

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

[tex]a =\dfrac{2s}{t^2}[/tex]

[tex]a=\dfrac{2\times72}{(2.9)^2}[/tex]

[tex]a=17.122\ m/s^2[/tex]

From equation (II)

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

[tex]q=\dfrac{0.8\times10^{-3}\times17.122}{159}[/tex]

[tex]q=0.000086148427673\ C[/tex]

[tex]q=8.6\times10^{-5}\ C[/tex]

Hence, The acceleration and charge are 17.122 m/s² and [tex]8.6\times10^{-5}\ C[/tex]

Answer:

(i) W = 8.918 N

(ii) [tex]V = 9.1 \times 10^{-4} m^3[/tex]

(iii) d = 9.1 cm

Explanation:

Part a)

As we know that weight of cube is given as

[tex]W = mg[/tex]

[tex]W = \rho V g[/tex]

here we know that

[tex]\rho = 0.91 g/cm^3[/tex]

[tex]Volume = L^3[/tex]

[tex]Volume = 10^3 = 1000 cm^3[/tex]

now the mass of the ice cube is given as

[tex]m = 0.91 \times 1000 = 910 g[/tex]

now weight is given as

[tex]W = 0.910 \times 9.8 = 8.918 N[/tex]

Part b)

Weight of the liquid displaced must be equal to weight of the ice cube

Because as we know that force of buoyancy = weight of the of the liquid displaced

[tex]W_{displaced} = 8.918 N[/tex]

So here volume displaced is given as

[tex]\rho_{water}Vg = 8.918[/tex]

[tex]1000(V)9.8 = 8.918[/tex]

[tex]V = 9.1 \times 10^{-4} m^3[/tex]

Part c)

Let the cube is submerged by distance "d" inside water

So here displaced water weight is given as

[tex]W = \rho_{water} (L^2 d) g[/tex]

[tex]8.918 = 1000(0.10^2 \times d) 9.8[/tex]

[tex]d = 0.091 m[/tex]

so it is submerged by d = 9.1 cm inside water

The ice cube weighs approximately 8.9 N. It would displace a volume of 910 cm³ of water, which means that 91% of the ice cube would be under the water surface.

The first part of the question asks what is the weight of the ice cube. To find the weight, we need to first calculate the volume of the cube, which is 10 cm × 10 cm × 10 cm = 1000 cm³.

Given the density of ice is 0.91 g/cm³, we can multiply this by the volume to find the mass of the cube: 0.91 g/cm³ × 1000 cm³ = 910 g. The weight is then calculated by multiplying the mass by gravity, roughly 9.8 m/s², which gives us 8.9 N.

For the second part, about what volume of liquid water is displaced, the amount of water displaced by the ice cube equals the volume of the ice cube that's beneath the water surface.

Because the cube's density is less than that of water (0.91 g/cm³ vs 1.00 g/cm³), it will displace an amount of water with equal mass. Given that 1 cm³ of water has a mass of 1 g, the ice cube will displace 910 cm³ of water.

In the last part, we need to find how much of the cube is under the water surface. This is the ratio of the mass of the water displaced to the total mass of the ice cube, which is 910 g / 1000 g = 0.91 or 91% of the cube.

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Answer:

  Collision will occur.

  Speed of red train when they collide = 0 m/s.

  Speed of green train when they collide = 10 m/s.

Explanation:

Speed of red train = 72 km/h = 20 m/s

Speed of green train = 144 km/h = 40 m/s.

Deceleration of trains = 1 m/s²

For red train:-

    Equation of motion v = u + at

              u = 20 m/s

              v = 0 m/s

              a = -1 m/s²

    Substituting

             0 = 20 - 1 x t

             t = 20 s.

    Equation of motion s = ut + 0.5at²

              u = 20 m/s

              t = 20 s

              a = -1 m/s²    

    Substituting

             s = 20 x 20 - 0.5 x 1 x 20² = 200 m

   So red train travel 200 m before coming to stop.

For green train:-

    Equation of motion v = u + at

              u = 40 m/s

              v = 0 m/s

              a = -1 m/s²

    Substituting

             0 = 40 - 1 x t

             t = 40 s.

    Equation of motion s = ut + 0.5at²

              u = 40 m/s

              t = 40 s

              a = -1 m/s²    

    Substituting

             s = 40 x 40 - 0.5 x 1 x 40² = 800 m

   So green train travel 800 m before coming to stop.

 Total distance traveled = 800 + 200 = 1000 m>950 m.

  So both trains collide.

  Distance traveled by green train when red train stops(t=20s)

     Equation of motion s = ut + 0.5at²

              u = 40 m/s

              t = 20 s

              a = -1 m/s²    

    Substituting

             s = 40 x 20 - 0.5 x 1 x 20² = 600 m

    Total distance after 20 s = 600 + 200 = 800 m< 950m . So they collide after red train stops.

  Speed of red train when they collide = 0 m/s.

  Distance traveled by green train when they collide = 950 - 200 = 750 m

  Equation of motion v² = u² + 2as

              u = 40 m/s

              s= 750 m

              a = -1 m/s²    

    Substituting  

              v² = 40² - 2 x 1 x 750 = 100

               v = 10 m/s

  Speed of green train when they collide = 10 m/s.

Final answer:

The red train traveling at 72 km/h and green train at 144 km/h will collide because their combined stopping distances exceed their initial separation. The speeds at impact are not provided, but the collision is inevitable due to their insufficient stopping distance.

Explanation:

To determine if a collision occurs between the red train traveling at 72 km/h and the green train traveling at 144 km/h, we need to convert their speeds into meters per second and calculate the stopping distance for both trains based on their deceleration.

The red train is traveling at 72 km/h, which is equivalent to 20 m/s (since 72 km/h / 3.6 = 20 m/s). The green train is traveling at 144 km/h, which is equivalent to 40 m/s (since 144 km/h / 3.6 = 40 m/s).

To calculate the stopping distance, use the equation d = v2 / (2a), where d is the stopping distance, v is the initial velocity, and a is the deceleration. So, for the red train, the stopping distance is (20 m/s)2 / (2 × 1.0 m/s2) = 200 m. For the green train, the stopping distance is (40 m/s)2 / (2 × 1.0 m/s2) = 800 m.

Adding both stopping distances, we get a total of 200 m + 800 m = 1000 m. Since the trains are only 950 m apart, their combined stopping distance exceeds the separation, meaning they will collide.

Just before the collision, both trains have been decelerating for the same amount of time. Given that the total stopping time can be found from v = at where v is the final velocity and t is time, we find that their time to stop (if unobstructed) would be t = v / a. Since the red train decelerates from 20 m/s, its stopping time is 20 m/s / 1 m/s2 = 20 s. For the green train, 40 m/s / 1 m/s2 = 40 s. Since they have not yet reached 20 s before collision, we know they will still be moving upon impact.

The collision occurs before either train can come to a complete stop, and thus we would use the physics of constant deceleration to determine their speeds at the moment of impact, but as the full calculation is not provided in this answer, it would require additional work to determine exact speeds.

The work done on the particle as it moves from x = b to x = a is k(1/a - 1/b).

To calculate the work done on a particle by the attractive force, we need to find the integral of the force function over the distance the particle moves. In this case, the force function is given by F(x) = k/x^2, where k is the constant. The work done when the particle moves from x = b to x = a is given by:

Work = ∫(k/x^2) dx from x = b to x = a

To evaluate this integral, we need to use the power rule of integration. The result will be:

Work = k(1/a - 1/b)

Therefore, the work done on the particle as it moves from x = b to x = a is k(1/a - 1/b).

Answer:

The time taken for ATP to diffuse across an average cell is 0.66 seconds

Explanation:

It is given that,

Diffusion constant of ATP is, [tex]D=3\times 10^{-10}\ m^2s^{-1}[/tex]

Distance to be diffused across is, [tex]x=20\ \mu m=20\times 10^{-6}\ m[/tex]

We need to find the time taken for ATP to diffuse across an average cell. It is given by :

[tex]t=\dfrac{x^2}{2D}[/tex]

[tex]t=\dfrac{(20\times 10^{-6}\ m)^2}{2\times 3\times 10^{-10}\ m^2s^{-1}}[/tex]

t = 0.66 seconds

So, the time taken for ATP to diffuse across an average cell is 0.66 seconds. Hence, this is the required solution.

Answer:

45.3°C

Explanation:

Heat gained = mass × specific heat × increase in temperature

q = mC (T − T₀)

Given C = 0.128 J/g/°C, m = 94.0 g, q = 305 J, and T₀ = 20.0°C:

305 J = (94.0 g) (0.128 J/g/°C) (T − 20.0°C)

T = 45.3°C

The final temperature of the metal, after adding 305 J of heat to 94.0 g of metal initially at 20.0 °C, is 45.34 °C. The calculation uses the specific heat capacity formula and involves solving for the change in temperature.

To find the final temperature, we can use the formula:

where:

First, solve for ΔT:

Next, find the final temperature:

Therefore, the final temperature of the metal is 45.34 °C.

Answer:

A primitive solid is a 'building block' that you can use to work with in 3D. Rather than extruding or revolving an object, AutoCAD has some basic 3D shape commands at your disposal.

Explanation:

Answer: The freezing point of solution is -3.34°C

Explanation:

Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of calcium nitrate follows:

[tex]Ca(NO_3)_2(aq.)\rightarrow Ca^{2+}(aq.)+2NO_3^-(aq.)[/tex]

The total number of ions present in the solution are 3.

[tex]Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}[/tex]

Where,

[tex]m_{solute}[/tex] = Given mass of solute [tex](Ca(NO_3)_2)[/tex] = 11.3 g

[tex]M_{solute}[/tex] = Molar mass of solute [tex](Ca(NO_3)_2)[/tex] = 164  g/mol

[tex]W_{solvent}[/tex] = Mass of solvent (water) = 115 g

Putting values in above equation, we get:

[tex]\text{Molality of }Ca(NO_3)_2=\frac{11.3\times 1000}{164\times 115}\\\\\text{Molality of }Ca(NO_3)_2=0.599m[/tex]

[tex]\Delta T=iK_fm[/tex]

where,

i = Vant hoff factor = 3

[tex]K_f[/tex] = molal freezing point depression constant = 1.86°C/m.g

m = molality of solution = 0.599 m

Putting values in above equation, we get:

[tex]\Delta T=3\times 1.86^oC/m.g\times 0.599m\\\\\Delta T=3.34^oC[/tex]

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

[tex]\Delta T=\text{freezing point of water}-\text{freezing point of solution}[/tex]

[tex]\Delta T[/tex] = 3.34 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

[tex]3.34^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-3.34^oC[/tex]

Hence, the freezing point of solution is -3.34°C

Final answer:

The freezing point of a solution made by dissolving 11.3 g of Ca(NO3)2 in 115 g of water is -3.34 °C. To find this, we calculate molality, account for the dissociation of ions and use the freezing point depression constant for water.

Explanation:

To calculate the freezing point depression of a solution of Ca(NO3)2 in water, we first determine the molality of the solution. With the provided mass of Ca(NO3)2 (11.3 g) and its formula weight (164 g/mol), we find there are 0.0689 moles of Ca(NO3)2. Since we only have 115 g of water, to convert to kilograms, we have 0.115 kg. The molality (m) is then 0.0689 moles / 0.115 kg = 0.599 m. Since Ca(NO3)2 dissociates into three ions (Ca2+, 2NO3-), the van't Hoff factor (i) is 3.

The depression of the freezing point is determined using the formula ΔTf = i * Kf * m, where Kf is the molal freezing point depression constant for water (1.86 °C/m). So the depression is ΔTf = 3 * 1.86 °C/m * 0.599 m = 3.34 °C.

The freezing point of the solution is then 0 °C - 3.34 °C = -3.34 °C, which is the answer.

Answer:

147,099,713.4 km

152,096,286.6 km

Explanation:

a = 149598000 km

e = 0.0167

The formula to find the perihelion

Rp = a ( 1 - e) = 149598000 ( 1 - 0.0167) = 147,099,713.4 km

The formula for aphelion

Ra = a ( 1 + e) = 149598000 ( 1 + 0.0167) = 152,096,286.6 km

To find the minimum and maximum distances from Earth to the Sun (perihelion and aphelion), we calculate them using Earth's semi-major axis of 149,598,000 kilometers and the eccentricity of 0.0167. The perihelion is 147,099,014 kilometers, and the aphelion is 152,096,986 kilometers.

The student's question revolves around finding the minimum (perihelion) and maximum (aphelion) distances from the Earth to the Sun, given the length of half of the major axis — also known as the semi-major axis — and the eccentricity of Earth's orbit. The semi-major axis (a) is 149,598,000 kilometers and the eccentricity (e) is 0.0167. The distance from the center of the ellipse, where Earth's orbit is, to the focus (c) is equal to the product of the semi-major axis and the eccentricity (c = ae).

The perihelion distance is the semi-major axis minus the distance c, resulting from the Earth being at the closest point to the Sun in its orbit. Conversely, the aphelion distance is the semi-major axis plus the distance c, when Earth is farthest from the Sun. Therefore, the perihelion (rp) can be calculated as rp = a - c, and the aphelion (ra) as ra = a + c.

Using the formula c = ae, we find that c is approximately 2,498,986 kilometers (149,598,000 km * 0.0167). Thus:

Answer:

a) 1/4 what it is now

Explanation:

As we know that force of gravitation between two planets at some distance "r" from each other is given as

[tex]F_g = \frac{Gm_1m_2}{r^2}[/tex]

now since we know that if the distance between Earth and Sun is changed

So the force of gravity will be given as

[tex]\frac{F_g'}{F_g} = \frac{r_1^2}{r_2^2}[/tex]

now we know that the distance between sun and earth is changed to twice the initial distance between them

so we have

[tex]r_2 = 2r_1[/tex]

so new gravitational force between sun and earth is given as

[tex]F_g' = \frac{r_1^2}{(2r_1)^2}F_g[/tex]

[tex]F_g' = \frac{1}{4}F_g[/tex]

Answer:

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Answer:

Heat energy needed = 3036.17 kJ

Explanation:

We have

     heat of fusion of water = 334 J/g

     heat of vaporization of water = 2257 J/g

     specific heat of ice = 2.09 J/g·°C

     specific heat of water = 4.18 J/g·°C

     specific heat of steam = 2.09 J/g·°C

Here wee need to convert 1 kg ice from -13°C to vapor at 100°C

First the ice changes to -13°C from 0°C , then it changes to water, then its temperature increases from 0°C to 100°C, then it changes to steam.

Mass of water = 1000 g

Heat energy required to change ice temperature from -13°C to 0°C

          H₁ = mcΔT = 1000 x 2.09 x 13 = 27.17 kJ

Heat energy required to change ice from 0°C to water at 0°C

          H₂ = mL = 1000 x 334 = 334 kJ

Heat energy required to change water temperature from 0°C to 100°C  

          H₃ = mcΔT = 1000 x 4.18 x 100 = 418 kJ    

Heat energy required to change water from 100°C to steam at 100°C  

          H₄ = mL = 1000 x 2257 = 2257 kJ    

Total heat energy required

          H = H₁ +  H₂ + H₃ + H₄ = 27.17 + 334 + 418 +2257 = 3036.17 kJ

Heat energy needed = 3036.17 kJ

Answer:

23.4 m/s

Explanation:

f = actual frequency of the wave = 6.2 x 10⁹ Hz

[tex]f_{app}[/tex] = frequency observed as the ball approach the radar

[tex]f_{rec}[/tex] = frequency observed as the ball recede away from the radar

V = speed of light

[tex]v[/tex] = speed of ball

B = beat frequency = 969 Hz

frequency observed as the ball approach the radar is given as

[tex]f_{app}=\frac{f(V+v)}{V}[/tex]                                 eq-1

frequency observed as the ball recede the radar is given as

[tex]f_{rec}=\frac{f(V-v)}{V}[/tex]                                  eq-2

Beat frequency is given as

[tex]B = f_{app} - f_{rec}[/tex]

Using eq-2 and eq-1

[tex]B = \frac{f(V+v)}{V}- \frac{f(V-v)}{V}[/tex]

inserting the values

[tex]969 = \frac{(6.2\times 10^{9})((3\times 10^{8})+v)}{(3\times 10^{8})}- \frac{(6.2\times 10^{9})((3\times 10^{8})-v)}{(3\times 10^{8})}[/tex]

[tex]v[/tex] = 23.4 m/s

To change the wave speed on a guitar string from 328 m/s to 363 m/s, the tension must be increased by 22.04%, as calculated through the wave speed formula for a string, taking into account the ratio of the squares of the wave speeds.

To find by what percent the tension in a guitar string must increase to change the wave speed on the string from 328 m/s to 363 m/s, we can use the formula for wave speed on a string, which is [tex]v = (\sqrt{T/mu}\)[/tex], where v is the wave speed in meters per second, T is the tension in Newtons, and [tex](mu)[/tex] is the linear mass density in kilograms per meter.

Starting with the initial tension (let's call it T1) and the initial wave speed (328 m/s), we have:

[tex]v1 = \(\sqrt{T1/\mu}\)[/tex]

For the final wave speed (363 m/s), we have:

[tex]v2 = \(\sqrt{T2/mu}\)[/tex]

Using the ratio of the squares of the wave speeds:

[tex]\((v2/v1)^2 = (T2/T1)[/tex]

[tex]\((363/328)^2 = (T2/T1)[/tex]

[tex]\(T2/T1 = 1.2204[/tex]

To find the percentage increase in tension:

Percent increase = [tex]\((T2/T1 - 1) \times 100\%)[/tex]

Percent increase = [tex]\((1.2204 - 1) \times 100\%\)[/tex]

Percent increase = 22.04%

Therefore, the tension in the guitar string must be increased by 22.04% to achieve a wave speed change from 328 m/s to 363 m/s.

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:

option (A)

Explanation:

Angle between the plane of coil and the magnetic field = 60 degree

Angle between the normal of coil and the magnetic field = 90 - 60 = 30 degree

N = 7, B = 0.4 T, T = 0.06 s , w = 2 pi / T, A = 900 cm^2 = 0.09 m^2

peak emf, e0 = N x B x A x (2 x 3.14 / T) x Sin 30

e0 = 7 x 0.4 x 0.09 x 6.28 x 0.5  / 0.06 = 13.18 Volt = 13 volt

Answer:

option d)

Explanation:

The amount of heat required to convert the 1 g of ice at 0 degree C to 1 g water at 0 degree C is called latent heta of fusion.

H = m Lf

where,, h is the heat required, m is the mass and Lf is the latent heat of fusion

1140 = m x 79.7

m = 14.3 g

Answer:

0.091 sec

Explanation:

f = frequency of oscillation of the mass attached to the spring = 11 Hz

T = Time period of oscillation of the mass attached to the spring = ?

Time period and frequency of oscillation of the mass attached to the end of spring are related as

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

Inserting the values

[tex]T = \frac{1}{11}[/tex]

T = 0.091 sec

Answer:

c

Explanation:

When a satellite is orbiting the earth , a constant force is being applied on it which means it must has acceleration. Also the direction of satellite is always being changed when it is orbitting to there is always change in the velocity vector which means acceleration.

You can view in the attached diagram to understand how the velocity is being changed.

Answer:

Force constant, K = 101.49 N/m

Explanation:

It is given that,

Mass, m = 30.6 kg

Time period of oscillation, T = 3.45 s

We need to find the force constant of the spring. The time period of the spring is given by :

[tex]T=2\pi\sqrt{\dfrac{m}{K}}[/tex]

[tex]K=\dfrac{4\pi^2m}{T^2}[/tex]

[tex]K=\dfrac{4\pi^2\times 30.6\ kg}{(3.45\ s)^2}[/tex]

K = 101.49 N/m

So, the force constant of the spring is 101.49 N/m. Hence, this is the required solution.

Answer:

[tex]intensity = 12.98 W/m^2[/tex]

[tex]Energy = 3.93 J[/tex]

Explanation:

As we know that the magnitude of electric field intensity is given as

[tex]E = 98.9 V/m[/tex]

now we know that intensity of the wave is given as the product of energy density and speed of the wave

[tex]intensity = \frac{1}{2}\epsilon_0 E^2 c[/tex]

[tex]intensity = \frac{1}{2}(8.85 \times 10^{-12})(98.9)^2(3\times 10^8)[/tex]

[tex]intensity = 12.98 W/m^2[/tex]

so intensity is the energy flow per unit area per unit of time

so the energy that flows through the area of 0.0259 m^2 in 11.7 s is given as

[tex]Energy = Area \times time \times intensity[/tex]

[tex]Energy = 0.0259(11.7)(12.98)[/tex]

[tex]Energy = 3.93 J[/tex]

Answer:

(a) 328 Nm

(b) 79.35 Nm

Explanation:

N = =150, side = 17.5 cm = 0.175 m, i = 42 A, B = 1.7 T

A = side^2 = 0.175^2 = 0.030625 m^2

(a) Torque = N x i x A x B x Sinθ

For maximum torque, θ = 90 degree

Torque = 150 x 42 x 0.030625 x 1.7 x Sin 90

Torque = 328 Nm

(b) θ = 14 degree

Torque =  150 x 42 x 0.030625 x 1.7 x Sin 14

Torque = 79.35 Nm

The torque on a current-carrying loop in a magnetic field is calculated using the formula T = NIAB sin ϴ. Part A finds the maximum torque by setting ϴ to 90 degrees, and Part B calculates the torque at an angle of 14 degrees by applying the sine of that angle.

To calculate the torque on a current-carrying loop in a magnetic field, we use the formula T = NIAB sin ϴ, where T is the torque, N is the number of turns in the coil, I is the current, A is the area, B is the magnetic field strength, and ϴ is the angle between the magnetic field and the normal to the plane of the loop.

Part A: Maximum Torque

To find the maximum torque, we set ϴ to 90°, as sin(90°) = 1. Thus, Tmax = NIAB. Using the given values: N = 150 turns, I = 42 A, A (area of the square loop) = (0.175 m)2 and B = 1.7 T, we calculate the maximum torque.

Part B: Torque at 14°

To find the torque when ϴ is 14°, sin(14°) is used in the formula. We thus get T = NIAB sin(14°).

Answer:

Momentum, p = 5 kg-m/s

Explanation:

The magnitude of the momentum of an object is the product of its mass m and speed v i.e.

p = m v

Mass, m = 3 kg

Velocity, v = 1.5 m/s

So, momentum of this object is given by :

[tex]p=3\ kg\times 1.5\ m/s[/tex]

p = 4.5 kg-m/s

or

p = 5 kg-m/s

So, the magnitude of momentum is 5 kg-m/s. Hence, this is the required solution.

The required tension to produce the transverse waves with a speed of 50.0 m/s on a 0.06 kg, 5.00 m long string is approximately 30.0 N. If the string has a tension of 8.00 N, the wave speed will be roughly 26.0 m/s.

This question is about the physics of waves on strings and involves the concepts of wave speed, tension, and linear mass density. Let's handle this in two parts.

(a) To determine the required tension to produce the transverse waves with a speed of 50.0 m/s, we first need to calculate the string's linear mass density, which is the mass of the string divided by its length. So, the linear mass density (μ) would be 0.06 kg / 5.00 m = 0.012 kg/m. Now there is a formula that determines the wave speed (v) on a string: v = sqrt(FT/μ) where FT is the tension in the string. Rearranging to solve for FT, we get FT = μv^2. Substituting the values we have,

(b) If the tension is 8.00 N, we can use the same formula to calculate the wave speed. This time, rearranging for v, we get v = sqrt(FT/μ). Substituting the values we have, v = sqrt((8.00 N)/(0.012 kg/m)) which gives us approximately 26.0 m/s. Therefore, the wave speed with a tension of 8.00 N is roughly 26.0 m/s.

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The linear mass density of the string is calculated first, which is 0.012 kg/m. Using this in the wave speed equation, the required tension to produce a wave speed of 50.0 m/s is 30.0 N. If the tension is 8.00 N, then the resulting wave speed would be around 25.82 m/s.

The student is attempting to produce transverse waves on a string. To determine the required tension to achieve a wave speed of 50.0 m/s, we should first calculate the linear mass density of the string, using the formula μ = m/L, where m is the mass of the string and L is its length. So, for a string 5.0 m long and a mass of 0.060 kg, μ = 0.060 kg / 5.00 m = 0.012 kg/m.

For (a), the formula for wave speed is v = √(FT/μ), where FT is the tension in the string and μ is the linear mass density. We need to rearrange it to calculate the required tension: FT = μ * v^2 = 0.012 kg/m * (50.0 m/s)^2 = 30.0 N.

For (b), using the same formula and given new tension value, the wave speed is v = √(FT/μ) = √(8.00 N / 0.012 kg/m) = approx. 25.82 m/s.

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Answer:

The velocity of the 4.00 kg object after the collision is 12 m/s.

Explanation:

Given that,

Mass of object [tex]m_{1} = 4.00\ kg[/tex]

Velocity of object [tex]v_{1} = 20.0\ m/s[/tex]

Mass of another object [tex]m_{2} = 6\ kg[/tex]

Velocity of another object [tex]v_{2}= 12.0\ m/s[/tex]

We need to calculate the relative velocity

[tex]v_{r}=v_{1}-v_{2}[/tex]

[tex]v_{r}=20-12=8\ m/s[/tex]

The relative velocity is 8 m/s in west before collision.

We know that,

In one dimensional elastic collision, the relative velocity before collision equals after collision but with opposite sign.

So, The relative velocity after collision must be 8 m/s in east.

So, The object of 6.00 kg is going 20 m/s and the object of 4.00 kg is slows down to 12 m/s.

Hence, The velocity of the 4.00 kg object after the collision is 12 m/s.

The 4.00-kg object is moving east at 2.00 m/s after the perfectly elastic collision with the 6.00-kg object.

This question is tackled using the concept of Conservation of Momentum. Since the collision is perfectly elastic, both the momentum and kinetic energy of the system are conserved.

In the east-west direction, momentum before is equal to momentum after the collision. Let's denote the velocity of the 4.00-kg object after the collision as v1. Thus, P(before) = P(after) gives us:
4.00kg * 20.00m/s (west) + 6.00kg * 12.00m/s (east) = 4.00kg * v1 (west) + 6.00kg * -12.00m/s (west).

Solving the above equation we find that v1 = -2.00 m/s, where the negative sign indicates it’s moving towards the east. Therefore, the 4.00-kg object is moving east at 2.00 m/s after the collision.

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The speed at which the satellite travels in its circular orbit around the Earth is 3,000 m/s.

In order to find the speed at which the satellite travels, we can use the formula:

v = 2πr / T

where v is the velocity of the satellite, r is the radius of the orbit, and T is the period of the satellite. Substituting the given values into the formula, we have:

v = (2π × 6,370,000) / (1.26 × 10^4) = 3,000 m/s

Therefore, the speed at which the satellite travels is 3,000 m/s.

Answer:

It takes to a spacecraft 100,837.13 years to cross the galaxy if could travels at 99% the speed of light.

Explanation:

d= 9.461 *10²⁰m

V= 297 *10⁶ m/s

t= d/V

t= 3.18 * 10¹² seconds = 100837.13 years

Answer:

[tex]a = 0.53 m/s^2[/tex]

Explanation:

initially the merry go round is at rest

after 6.73 s the merry go round will accelerates to 20 rpm

so final angular speed is given as

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 2\pi ( \frac{20}{60})[/tex]

[tex]\omega = 2.10 rad/s[/tex]

so final tangential speed is given as

[tex]v = r\omega[/tex]

[tex]v = 1.71 (2.10) = 3.58 m/s[/tex]

now average acceleration of the girl is given as

[tex]a = \frac{v_f - v_i}{\Delta t}[/tex]

[tex]a = \frac{3.58 - 0}{6.73}[/tex]

[tex]a = 0.53 m/s^2[/tex]