Using a rope that will snap if the tension in it exceeds 356 N, you need to lower a bundle of old roofing material weighing 478 N from a point 7.50 m above the ground. (a) What magnitude of the bundle's acceleration will put the rope on the verge of snapping? (b) At that acceleration, with what speed would the bundle hit the ground?

Answer:

a) [tex]k_{avg}=6.22\times 10^{-21}[/tex]

b) [tex]k_{avg}=8.61\times 10^{-21}[/tex]

c)  [tex]k_{mol}=3.74\times 10^{3}J/mol[/tex]

d)   [tex]k_{mol}=5.1\times 10^{3}J/mol[/tex]

Explanation:

Average translation kinetic energy ([tex]k_{avg} [/tex]) is given as

[tex]k_{avg}=\frac{3}{2}\times kT[/tex]    ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

or

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

[tex]k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8[/tex]  

[tex]k_{avg}=6.22\times 10^{-21}J[/tex]

b) at T = 143° C

or

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

[tex]k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416[/tex]  

[tex]k_{avg}=8.61\times 10^{-21}J[/tex]

c ) The translational kinetic energy per mole of an ideal gas is given as:

       [tex]k_{mol}=A_{v}\times k_{avg}[/tex]

here   [tex]A_{v}[/tex] = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

        [tex]k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}[/tex]

          [tex]k_{mol}=3.74\times 10^{3}J/mol[/tex]

d) now at T = 143° C

        [tex]k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}[/tex]

          [tex]k_{mol}=5.1\times 10^{3}J/mol[/tex]

Answer:

2.63 Hz

Explanation:

m = mass of the object = 8.0 kg

x = stretch in the spring = 3.6 cm = 0.036 m

k = spring constant of the spring

using equilibrium of force

Spring force = weight of object

k x = m g

k (0.036) = (8) (9.8)

k = 2177.78 N/m

frequency of oscillation is given as

[tex]f = \frac{1}{2\pi }\sqrt{\frac{k}{m}}[/tex]

[tex]f = \frac{1}{2\pi }\sqrt{\frac{2177.78}{8}}[/tex]

[tex]f [/tex] =  2.63 Hz

The frequency of the spring is about 2.6 Hz

[tex]\texttt{ }[/tex]

Simple Harmonic Motion is a motion where the magnitude of acceleration is directly proportional to the magnitude of the displacement but in the opposite direction.

[tex]\texttt{ }[/tex]

The pulled and then released spring is one of the examples of Simple Harmonic Motion. We can use the following formula to find the period of this spring.

[tex]T = 2 \pi\sqrt{\frac{m}{k}}[/tex]

T = Periode of Spring ( second )

m = Load Mass ( kg )

k = Spring Constant ( N / m )

[tex]\texttt{ }[/tex]

The pendulum which moves back and forth is also an example of Simple Harmonic Motion. We can use the following formula to find the period of this pendulum.

[tex]T = 2 \pi\sqrt{\frac{L}{g}}[/tex]

T = Periode of Pendulum ( second )

L = Length of Pendulum ( kg )

g = Gravitational Acceleration ( m/s² )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

mass of the object = m = 8.0 kg

extension of the spring = x = 3.6 cm = 3.6 × 10⁻² m

Unknown:

frequency of the spring = f = ?

Solution:

Firstly, we will calculate the spring constant as follows:

[tex]F = kx[/tex]

[tex]mg = kx[/tex]

[tex]k = mg \div x[/tex]

[tex]k = \frac{mg}{x}[/tex]

[tex]\texttt{ }[/tex]

Next, we could calculate the frequency of the spring as follows:

[tex]f = \frac{1}{2 \pi}\sqrt{\frac{k}{m}}[/tex]

[tex]f = \frac{1}{2 \pi}\sqrt{\frac{mg / x}{m}}[/tex]

[tex]f = \frac{1}{2 \pi}\sqrt{\frac{g}{x}}[/tex]

[tex]f = \frac{1}{2 \pi}\sqrt{\frac{9.8}{3.6 \times 10^{-2}}}[/tex]

[tex]f = \frac{35\sqrt{2}}{6\pi} \texttt{ Hz}[/tex]

[tex]f \approx 2.6 \texttt{ Hz}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Simple Harmonic Motion

[tex]\texttt{ }[/tex]

Keywords: Simple , Harmonic , Motion , Pendulum , Spring , Period , Frequency

Solution:

Given:

height of geyser, h = 50 m

speed of water at ground can be given by third eqn of motion with v = 0 m/s:

[tex]u^{2} = v^{2} + 2gh[/tex]

putting v = 0 m/s in the above eqn, we get:

u = [tex]\sqrt{2gh}[/tex]                      (1)

Now, pressure is given by:

[tex]\Delta p = \frac{1}{2}u^{2}\rho \\[/tex]        (2)

where,

[tex]\Delta p[/tex] = density of water = 1000 kg/[tex]m^{3}[/tex]

g = 9.8 m/[tex]s^{2}[/tex]

Using eqn (1) and (2):

[tex]\Delta p = \frac{1}{2}(\sqrt{2gh})^{2}\rho[/tex]

⇒ [tex]\Delta p = \frac{1}{2}\rho gh[/tex]

[tex]\Delta p = \frac{1}{2}\times2\times 1000\times 9.8\times 50[/tex] = [tex]\Delta p[/tex] =   490000 Pa

[tex]p_{atm}[/tex] = 101325 Pa

For absolute pressure, p:

p = [tex]\Delta p[/tex] + [tex]p_{atm}[/tex]

p = 490000 - 101325

p = 388675 Pa

Therefore, pressure in hot springs for the water to attain the height of 50m is 388675 Pa

Answer:

2.82 × 10^6 watt

Explanation:

H = 120 m,

Mass per second = 2400 kg/s

Power = work / time

Power = m g H / t

Power = 2400 × 9.8 × 120 / 1

Power = 2.82 × 10^6 Watt

Answer:

The correct option is B. an alpha particle.

Explanation:

When ²³Na is bombarded with protons [tex]_{1}^{1}\textrm{H}[/tex], ²⁰Ne and one alpha particle [tex]_{2}^{4}\textrm{He}[/tex] is released.

The reaction is as follows:

[tex]_{11}^{23}\textrm{Na} + _{1}^{1}\textrm{H} \rightarrow _{10}^{20}\textrm{Ne} + _{2}^{4}\textrm{He}[/tex]

Therefore, an alpha particle and ²⁰Ne are released when ²³Na is bombarded with protons.

Answer:

42°C is equivalent to 315.15 K.

Explanation:

In this question, we need to convert the temperature from one scale to another. A glass of cold water has a temperature of 42 °C. The conversion from degree Celsius to kelvin is given by :

[tex]T_k=T_c+273.15[/tex]

Where

[tex]T_c\ and\ T_k[/tex] are temperatures in degree Celsius and kelvin respectively.

So, [tex]T_k=42+273.15[/tex]

[tex]T_k=315.15\ K[/tex]

So, 42°C is equivalent to 315.15 K. Hence, this is the required solution.

Answer : The height of the unit cell is, [tex]4.947\AA[/tex]

Explanation : Given,

Density of zinc = [tex]7.14g/cm^3[/tex]

Atomic radius = [tex]1.332\AA[/tex]

Atomic weight of zinc = 65.37 g/mole

As we know that, zinc has haxagonal close packed crystal structure. The number of atoms in unit cell of HCP is, 6.

Formula used for density :

[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]      .............(1)

where,

[tex]\rho[/tex] = density  of zinc

Z = number of atom in unit cell  = 6 atoms/unit cell (for HCP)

M = atomic mass

[tex](N_{A})[/tex] = Avogadro's number  = [tex]6.022\times 10^{23}atoms/mole[/tex]

a = edge length of unit cell

[tex]a^3[/tex] = volume of unit cell

Now put all the values in above formula (1), we get

[tex]7.14g/cm^3=\frac{(6\text{ atoms per unit cell})\times (65.37g/mole)}{(6.022\times 10^{23}atoms/mole)\times a^{3}}[/tex]

[tex]V=a^{3}=9.122\times 10^{-23}cm^3[/tex]

[tex]V=9.122\times 10^{-29}m^3[/tex]

Now we have to calculate the height of the unit cell.

Formula used :

[tex]V=6\sqrt {3}r^2h[/tex]

where,

V = volume of unit cell

r = atomic radius = [tex]1.332\AA=1.332\times 10^{-10}m[/tex]

conversion used : [tex](1\AA=10^{-10}m)[/tex]

h = height of the unit cell

Now put all the given values in this formula, we get:

[tex]9.122\times 10^{-29}m^3=6\sqrt {3}\times (1.332\times 10^{-10}m)^2h[/tex]

[tex]h=4.947\times 10^{-10}m=4.947\AA[/tex]

Therefore, the height of the unit cell is, [tex]4.947\AA[/tex]

To find the height of the zinc unit cell, we apply the geometric relationship of the hexagonal close-packed structure to the given atomic radius. After performing the calculation, we get the height of the unit cell. Density provides context for the metal's packing efficiency but is not directly used in the calculation.

To calculate the height of the unit cell of zinc, we must first determine the type of crystal structure zinc has. Zinc crystallizes in a hexagonal close-packed (hcp) structure. Given that its atomic radius is 1.332 Å (or 0.1332 nm), we can use this information along with the geometry of the hcp structure to calculate the height (c) of the unit cell.

In an hcp structure, the height (c) can be found using the relationship between the atomic radius (r) and the height, which is c = 2\sqrt{\frac{2}{3}}\cdot r. Substituting the given value for the atomic radius:

c = 2\sqrt{\frac{2}{3}}\cdot (0.1332 \text{ nm})

Calculating this gives us the height of the unit cell for zinc.

While this question does not require use of the density directly, it's worth noting that the density of zinc is 7.14  g/cm³, which indicates a relatively high atomic packing efficiency typical for metals. Using the atomic weight (65.37 grams/mole) and the Avogadro's number, we could further explore the relationship between the density and the unit cell parameters if needed.

Answer:

The diameter of the circle at the surface is 1.814 m.

Explanation:

Given that,

Depth = 80 cm

Let the critical angle be c.

From Snell's law

[tex]sin c=\dfrac{1}{n}[/tex]

[tex]c=sin^{-1}\dfrac{1}{n}[/tex]

The value of n for water,

[tex]n = \dfrac{4}{3}[/tex]

Put the value of n in the equation (I)

[tex]c =sin^{-1}\dfrac{3}{4}[/tex]

[tex]c=48.6^{\circ}[/tex]

We know that,

[tex]tan c=\dfrac{r}{h}[/tex]

We calculate the radius of the circle

[tex]r =h tan c[/tex]

[tex]r=80\times10^{-2}\times\tan48.6^{\circ}[/tex]

[tex]r =0.907\ m[/tex]

The diameter of the circle at the surface

[tex]d =2\times r[/tex]

[tex]d =2\times 0.907[/tex]

[tex]d =1.814\ m[/tex]

Hence, The diameter of the circle at the surface is 1.814 m.

Answer:

From lowest to highest acceleration:

3rd train

2nd train

1st train

Explanation:

The acceleration of an object can be found by using Newton's second law:

[tex]a=\frac{F}{m}[/tex]

where

a is the acceleration

F is the net force on the object

m is the mass of the object

We notice that for equal values of the forces F, the acceleration a is inversely proportional to the mass, m. Therefore, greater mass means lower acceleration, and viceversa.

So, the train with lowest acceleration is the one with largest mass, i.e. the 3rd train consisting of 50 equally loaded freight cars. Then, the 2nd train has larger acceleration, since it consists of 50 empty freight cars (so its mass is smaller). Finally, the 1st train (a single empty car) is the one with largest acceleration, since it is the train with smallest mass.

The acceleration of the three trains can be ranked as follows: the single empty freight car train, the 50 empty freight cars train, and the 50 equally loaded freight cars train.

The accelerations of the three trains can be ranked as follows:

When an equal force is applied to each train, the train with fewer cars will experience a greater acceleration. This is because the force is distributed over fewer cars, resulting in a larger net force acting on each car.

For example, consider two cars:

Therefore, the single empty freight car train will have the highest acceleration, followed by the 50 empty freight cars train, and finally the 50 equally loaded freight cars train.

Answer:

The minium static friction force to complete this turn is F= 29,690.94 N.

Explanation:

R= 95m

m= 610 kg

V= 68 m/s

an= V²/R

an= 48.67 m/s²

F= m * an

F= 29690.94 N

The approximate minimum static friction force to complete the turn is approximately 18806 N.

When a race car turns a corner, the wings on the car push it into the track, increasing the normal force. This increased normal force allows for larger friction forces, which help the car maintain traction and complete the turn.

For the given Formula One racetrack with a half-circle turn of diameter 190 m and a speed of 68 m/s, we can find the approximate minimum static friction force using the centripetal force equation. The centripetal force in this situation is equal to the product of the mass (610 kg), the speed squared (68 m/s) squared, and the reciprocal of the radius (half of the diameter = 95 m).

The approximate minimum static friction force to complete this turn is approximately 18806 N.

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Answer:

c) 500 N

Explanation:

If the block moves with constant speed there is no acceleration. We draw a free body diagram and define the forces on the body.

[tex]F_{f}=FrictionForce\\F_{e}=ExternalForce\\N=NormalForce\\W=Weight\\[/tex]

The equation for the frictional force is:

[tex]F_{f}=N*H_{k}[/tex]

Where [tex]H_{k}[/tex] is the kinetic friction coefficient. We write the equilibrium equation in the y direcction:

[tex]\sum F_{y}\rightarrow N-W=0\\N=W[/tex]

We replace this result in the equation for the frictional force:

[tex]F_{f}=W*H_{k}[/tex]

We replace the data given by the exercise and find the weight

[tex]W=\frac{F_{f}}{H_{k}}\\W=\frac{100}{0.2}=500\: N[/tex]

To calculate the force exerted by a stream of water on a moving plate, determine the change in momentum of the water as it hits the plate. By taking into account the relative velocity of the water to the plate, its cross-sectional area, and the density of water, we find the force exerted to be 17670 Newtons.

The subject of this question is related to physics, specifically fluid dynamics. It involves calculating the force exerted by a jet of water on a moving plate. The plate and the water jet are moving in the same direction, but at different speeds. To calculate this, we need to understand the change in momentum of the water as it hits the plate.

First, we calculate the velocity of water relative to the plate, which is 40 m/s - 10 m/s = 30 m/s. The water's diameter is 5 cm, which gives us a cross-sectional area (A) of 19.63 cm2. We can obtain the volume flow rate (Q) of the water from Q = Av, which gives 0.589 m3/s. Multiplying this by the density of water (ρ=1000 kg/m3) gives a mass flow rate (ṁ) of 589 kg/s.

The force exerted by the water on the plate is equal to the rate of change of momentum of the water. Since the water is coming to rest on the plate, the initial momentum of the water relative to the plate is -ṁv, and the final momentum is zero. Therefore, the force exerted on the plate is Δp/Δt = ṁv = 589 kg/s * 30 m/s = 17670 N.

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The force exerted by the water stream on the plate is [tex]\( 2356.2 \, \text{N} \).[/tex]

To determine the force exerted by the water stream on the moving plate, we need to apply the principles of fluid dynamics and the conservation of momentum.

Given Data:

Step-by-Step Solution:

1. Calculate the cross-sectional area of the jet:

[tex]\[ A = \pi \left(\frac{D}{2}\right)^2 = \pi \left(\frac{0.05}{2}\right)^2 = \pi \left(0.025\right)^2 = \pi \times 0.000625 = 0.001963495 \, \text{m}^2 \][/tex]

2. Determine the velocity of the water relative to the plate:

[tex]\[ V_{\text{rel}} = V_{\text{jet}} - V_{\text{plate}} = 40 \, \text{m/s} - 10 \, \text{m/s} = 30 \, \text{m/s} \][/tex]

3. Calculate the mass flow rate of the water jet:

[tex]\[ \dot{m} = \rho \cdot A \cdot V_{\text{jet}} \][/tex]

Assuming the density of water [tex]\( \rho \) is \( 1000 \, \text{kg/m}^3 \):[/tex]

[tex]\[ \dot{m} = 1000 \, \text{kg/m}^3 \times 0.001963495 \, \text{m}^2 \times 40 \, \text{m/s} \][/tex]

[tex]\[ \dot{m} = 78.54 \, \text{kg/s} \][/tex]

4. Apply the conservation of momentum to find the force:

The change in momentum of the water relative to the plate will result in the force exerted on the plate. Since the water splatters in all directions in the plane of the plate, we assume the relative velocity of the water to the plate goes to zero after hitting the plate (as it splatters and no longer moves in a single direction).

The force exerted by the water on the plate is given by the rate of change of momentum:

[tex]\[ F = \dot{m} \cdot V_{\text{rel}} \][/tex]

Substituting the values:

[tex]\[ F = 78.54 \, \text{kg/s} \times 30 \, \text{m/s} \][/tex]

[tex]\[ F = 2356.2 \, \text{N} \][/tex]

Final answer:

The branch falls from 98.0 m with a final speed of 44 m/s, and the boulder must be ejected with a velocity of 103.4 m/s to reach a height of 545 m, based on the work-energy theorem and neglecting air resistance.

Explanation:

Work Energy Theorem Examples

For a branch falling from the height of a tree with no air resistance, the work-energy theorem tells us that the kinetic energy gained by the branch when it hits the ground must equal the potential energy it had at the top.

Using the formula for gravitational potential energy (PE = mgh, where m is mass, g is the acceleration due to gravity, and h is height) and kinetic energy (KE = 0.5 * m * v^2), we can solve for the velocity when the branch reaches the ground.

For instance, using a height of 98.0 m for the redwood tree and the acceleration due to gravity g = 9.8 m/s^2, we find that the final velocity (v) can be calculated as:

PE_top = KE_bottom

mgh = 0.5 * m * v^2

v = sqrt(2gh) = sqrt(2 * 9.8 m/s^2 * 98.0 m) = 44 m/s

For the boulder ejected from a volcano, we're looking for the initial velocity required for the boulder to reach a maximum height of 545 m. The potential energy at the top (PE_top) will be equal to the kinetic energy at the bottom (KE_bottom) since we're negating air resistance.

PE_top will be m * g * h. KE_bottom = PE_top = 0.5 * m * v^2, so solving for the initial velocity (v) gives us:

v = sqrt(2gh) = sqrt(2 * 9.8 m/s^2 * 545 m) = 103.4 m/s

Answer:

Output Voltage = 0.96 volts

Explanation:

As we know by the equation of transformer we have

[tex]\frac{V_1}{V_2} = \frac{N_1}{N_2}[/tex]

here we know that

[tex]V_1[/tex] = voltage of primary coil = 120 Volts

[tex]V_2[/tex] = voltage of secondary coil

[tex]N_1[/tex] = number of turns in primary coil = 500

[tex]N_2[/tex] = number of turns in secondary coil = 4

now we will have

[tex]\frac{120}{V} = \frac{500}{4}[/tex]

[tex]V = 0.96 Volts[/tex]

Answer:

Part a)

decrease in internal energy is 2.23 k Cal

Part b)

Efficiency will be 4.9 %

Explanation:

Part A)

As per first law of thermodynamics we know that

[tex]Q = W + \Delta U[/tex]

here we have

Q = -8854 J

W = 463 J

now we have

[tex]-8854 J = 463 J + \Delta U[/tex]

so we have

[tex]\Delta U = -9317 J = - 2.23 kCal[/tex]

So decrease in internal energy is 2.23 k Cal

Part B)

Efficiency of the woman is given as

[tex]\eta = \frac{W}{\Delta U}[/tex]

here we have

[tex]\eta = \frac{463}{9317} = 0.049 [/tex]

So efficiency will be 4.9 %

Answer:

The work done by the steam is 213 kJ.

Explanation:

Given that,

Mass = 5 kg

Pressure = 150 kPa

Temperature = 200°C

We need to calculate the specific volume

Using formula of work done

[tex]W=Pm\DeltaV[/tex]

[tex]W=Pm(\dfrac{RT_{1}}{P_{atm}}-\dfrac{RT_{2}}{P_{atm}}[/tex]

[tex]W=\dfrac{PmR}{P_{atm}}(T_{2}-T_{1})[/tex]

Where,R = gas constant

T = temperature

P = pressure

[tex]P_{atm}[/tex]=Atmosphere pressure

m = mass

Put the value into the formula

[tex]W=\dfrac{150\times10^{3}\times5\times287.05}{1.01\times10^{5}}\times(473-373)[/tex]

[tex]W=213\ kJ[/tex]

Hence, The work done by the steam is 213 kJ.    

The work done by 5 kg of saturated water heated at constant pressure to 200°C can be calculated by applying the first law of thermodynamics, W = PΔV, where W is the work done, P is the pressure, and ΔV is the change in volume of the gas. However, without information on the change in volume in this scenario, the actual calculations cannot be performed.

The subject of this question is Physics, specifically thermodynamics, which deals with the transfer of heat and the work performed during this process. The given scenario involves a mass of 5 kg of saturated water vapor being heated at a constant pressure (150 kPa) until the temperature raises to 200°C.

We can apply the first law of thermodynamics to answer this question. The law states that the change in the internal energy of a system is equal to the heat supplied to the system minus the work done by the system. In this case, the work done by the system is the pressure times the change in volume (W = pΔV).

The scenario doesn't provide a direct change in volume, meaning the actual calculations can't be performed in this context. However, if the initial and final volumes were available, the calculations would follow the process described above. It is worth noting that for water vapor (or any other ideal gas), the volume would increase when heated if the pressure is held constant, so in theory, the system would perform positive work.

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Explanation:

Mass of the astronaut, m₁ = 170 kg

Speed of astronaut, v₁ = 2.25 m/s

mass of space capsule, m₂ = 2600 kg

Let v₂ is the speed of the space capsule. It can be calculated using the conservation of momentum as :

initial momentum = final momentum

Since, initial momentum is zero. So,

[tex]m_1v_1+m_2v_2=0[/tex]

[tex]170\ kg\times 2.25\ m/s+2600\ kg\times v_2=0[/tex]

[tex]v_2=-0.17\ m/s[/tex]

So, the change in speed of the space capsule is 0.17 m/s. Hence, this is the required solution.

Answer: The probability that the sample average sediment density is at most 3.00 = 0.9913

The probability that the sample average sediment density is between 2.61 and 3.00 = 0.4913

Explanation:

Given : Mean : [tex]\mu=2.61 [/tex]

Standard deviation : [tex]\sigma =0.82[/tex]

Sample size : [tex]n=25[/tex]

The value of z-score is given by :-

[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

a) For x= 3.00

[tex]z=\dfrac{3.00-2.61}{\dfrac{0.82}{\sqrt{25}}}=2.38[/tex]

The p-value : [tex]P(z\leq2.38)=0.9913[/tex]

b) For x= 2.61

[tex]z=\dfrac{2.61-2.61}{\dfrac{0.82}{\sqrt{25}}}=0[/tex]

The p-value : [tex]P(0<z\leq2.38)=P(2.38)-P(0)=0.9913-0.5=0.4913[/tex]

Answer:

15.7 m

Explanation:

m = mass of the sled = 125 kg

v₀ = initial speed of the sled = 8.1 m/s

v = final speed of sled = 0 m/s

F = force applied by the brakes in opposite direction of motion = 261

d = stopping distance for the sled

Using work-change in kinetic energy theorem

- F d = (0.5) m (v² - v₀²)

- (261) d = (0.5) (125) (0² - 8.1²)

d = 15.7 m

Answer:

Time required to completely fill this container = 78.43 seconds

Explanation:

Volume of container = 4 m³

Rate of filling of container = 0.051 m³/s

We have the equation

             [tex]\texttt{Time required}=\frac{\texttt{Volume of container}}{\texttt{Rate of filling of container}}[/tex]

Substituting

             [tex]\texttt{Time required}=\frac{4}{0.051}=\frac{4000}{51}=78.43 s[/tex]

Time required to completely fill this container = 78.43 seconds

Answer:

Charge, [tex]q=3.24\times 10^{-18}\ C[/tex]

Explanation:

A moving particle encounters an external electric field that decreases its kinetic energy from 9650 eV to 8900 eV as the particle moves from position A to position B The electric potential at A is 56.0 V and that at B is 19.0 V. We need to find the charge of the particle.

It can be calculated using conservation of energy as :

[tex]\Delta KE=-q(V_B-V_A)[/tex]

[tex]q=\dfrac{\Delta KE}{(V_B-V_A)}[/tex]

[tex]q=\dfrac{ 9650\ eV-8900\ eV}{(19\ V-56\ V)}[/tex]

q = -20.27 e

[tex]q =-20.27e\times \dfrac{1.6\times 10^{-19}\ C}{e}[/tex]

[tex]q=-3.24\times 10^{-18}\ C[/tex]

Hence, this is the required solution.

Answer:

50,000 N

Explanation:

d = 0.2 m, r = 0.1 m,  f = 500 N,

D = 2 m, R = 1 m

Let F be the weight lifted by the jack.

By use of Pascal's law

f/a = F / A

F = f x A / a

F = 500 x 3.14 x 1 x 1 / (3.14 x 0.1 x 0.1)

F = 50,000 N

Final answer:

The force between two charged particles will increase by a factor of 64 if their separation distance is reduced to one-eighth of the original distance, according to Coulomb's Law.

Explanation:

The student is asking about how the force between two charged particles changes when their separation distance changes. Coulomb's Law, which describes the electrostatic interaction between electrically charged particles, states that the force (F) between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (r) between them. Coulomb's Law is mathematically expressed as F = k * (|q¹*q|²/r²), where k is Coulomb's constant, and q1 and q2 are the charges.

If the distance between the particles is reduced to one-eighth of the original distance, the new force F' can be calculated using the relation F' = F * (1/(1/8)²), since the force varies inversely with the square of the distance. After performing the calculations, we find that the new force is 64 times the original force of 7.5x10² N.

Answer:

The S wave arrives 6 sec after the P wave.

Explanation:

Given that,

Distance of P = 45 km

Speed of p = 5000 m/s

Speed of S = 3000 m/s

We need to calculate the time by the P wave

Using formula of time

[tex]t = \dfrac{D}{v}[/tex]

Where, D = distance

v = speed

t = time

Put the value in to the formula

[tex]t_{p} =\dfrac{45\times1000}{5000}[/tex]

[tex]t_{p} = 9\ sec[/tex]

Now, time for s wave

[tex]t_{s}=\dfrac{45000}{3000}[/tex]

[tex]t =15\ sec[/tex]

The required time is

[tex]\Delta t=t_{s}-t_{p}[/tex]

[tex]\Delta t=15-9[/tex]

[tex]\Delta t =6\ sec[/tex]

Hence, The S wave arrives 6 sec after the P wave.

Answer:

The bullet will have been dropped vertically h= 0.54 meters by the time it hits the target.

Explanation:

d= 100m

V= 300 m/s

g= 9.8 m/s²

d= V*t

t= d/V

t= 0.33 s

h= g*t²/2

h=0.54 m

Answer:

Depth of the pool, h = 4.004 cm

Explanation:

Pressure at the bottom, P = 39240 N/m²

The density of water, d = 1000 kg/m³

The pressure at the bottom is given by :

P = dgh

We need to find the depth of pool. Let h is the depth of the pool. So,

[tex]h=\dfrac{P}{dg}[/tex]

[tex]h=\dfrac{39240\ N/m^2}{1000\ kg/m^3\times 9.8\ m/s^2}[/tex]

h = 4.004 m

So, the pool is 4.004 meters pool. Hence, this is the required solution.

Answer:

V = 48 Volts

Explanation:

Since we know that electric potential is a scalar quantity

So here total potential of a point is sum of potential due to each charge

It is given as

[tex]V = V_1 + V_2 + V_3[/tex]

here we have potential due to 50 nC placed at y = 6 m

[tex]V_1 = \frac{kQ}{r}[/tex]

[tex]V_1 = \frac{(9\times 10^9)(50 \times 10^{-9})}{\sqrt{6^2 + 8^2}}[/tex]

[tex]V_1 = 45 Volts[/tex]

Now potential due to -80 nC charge placed at x = -4

[tex]V_2 = \frac{kQ}{r}[/tex]

[tex]V_2 = \frac{(9\times 10^9)(-80 \times 10^{-9})}{12}[/tex]

[tex]V_2 = -60 Volts[/tex]

Now potential due to 70 nC placed at y = -6 m

[tex]V_3 = \frac{kQ}{r}[/tex]

[tex]V_3 = \frac{(9\times 10^9)(70 \times 10^{-9})}{\sqrt{6^2 + 8^2}}[/tex]

[tex]V_3 = 63 Volts[/tex]

Now total potential at this point is given as

[tex]V = 45 - 60 + 63 = 48 Volts[/tex]

Answer:

1.87 s

Explanation:

d = distance traveled by the water wave = 64 m

t = time taken to travel the distance = 14 s

[tex]v[/tex] = speed of water wave

Speed of water wave is given as

[tex]v=\frac{d}{t}[/tex]

[tex]v=\frac{64}{14}[/tex]

[tex]v[/tex] = 4.6 m/s

[tex]\lambda[/tex] = wavelength of the wave = 859 cm = 8.59 m

T = period of the wave

period of the wave is given as

[tex]T = \frac{\lambda }{v}[/tex]

[tex]T = \frac{8.59 }{4.6}[/tex]

T = 1.87 s

Answer:

10.99 m

Explanation:

m = mass of the block = 0.245 kg

k = spring constant of the vertical spring = 4975 N/m

x = compression of the spring = 0.103 m

h = height to which the block rise

Using conservation of energy

Potential energy gained by the block = Spring potential energy

mgh = (0.5) k x²

(0.245) (9.8) h = (0.5) (4975) (0.103)²

h = 10.99 m

Explanation:

It is given that,

Mass of the object, m = 0.8 g = 0.0008 kg

Electric field, E = 534 N/C

Distance, s = 12 m

Time, t = 1.2 s

We need to find the acceleration of the object. It can be solved as :

m a = q E.......(1)

m = mass of electron

a = acceleration

q = charge on electron

"a" can be calculated using second equation of motion as :

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

[tex]s=0+\dfrac{1}{2}at^2[/tex]

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

[tex]a=\dfrac{2\times 12\ m}{(1.2\ s)^2}[/tex]

a = 16.67 m/s²

Now put the value of a in equation (1) as :

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

[tex]q=\dfrac{0.0008\ kg\times 16.67\ m/s^2}{534\ N/C}[/tex]

q = 0.0000249 C

or

[tex]q=2.49\times 10^{-5}\ C[/tex]

Hence, this is the required solution.