Suppose 1.4 mol of an ideal gas is taken from a volume of 2.5 m3 to a volume of 1.0 m3 via an isothermal compression at 27°C. (a) How much energy is transferred as heat during the compression, and (b) is the transfer to or from the gas?

Answer and Explanation:

We know by Newton's second law of motion that Force can be given by the rate of change of momentum i.e.,

F = m[tex]\frac{dp}{dt}[/tex]

where, p is momentum

Now, when EM wave falls on perfectly reflecting body , change in momentum is : -(p+p) = -2p

i.e., after reflection momentum is twice of its initial value

In case of absorption of radiation of EM wave as in perfectly black painted body, change in momentum  is half of that in reflection i.e., '-p'

Since, the force, F is equal to the change in momentum, the Force erxerted by the wave will also be half i.e., [tex]\frac{F}{2}[/tex] or 0.5F

Final answer:

When the wave is absorbed, the force is half compared to when the wave is reflected, that's means the force will be F/2

Explanation:

When an electromagnetic wave falls on a white, perfectly reflecting surface, it exerts a force F on that surface. According to Maxwell's predictions, electromagnetic waves carry momentum, and thus when these waves are absorbed by an object, they exert a force in the direction of the wave's propagation, known as radiation pressure. In the case of a perfectly reflecting surface,

the force is twice as great as that on a perfectly absorbing surface because the change in momentum when the wave is reflected is twice as much as when it is absorbed, due to the wave's momentum being reversed while conserving the system's total momentum.

When the same electromagnetic wave falls on a surface that has been painted a perfectly absorbing black, the force exerted on the surface will be half compared to the force on a perfectly reflecting surface.

This is because the absorbing surface does not reflect the electromagnetic waves, and thus, the change in momentum is only due to the wave being absorbed rather than being both absorbed and reflected back.

The property of the surface, whether it is absorbing or reflecting, significantly changes the amount of force exerted by the electromagnetic wave on the surface, that's means the force will be F/2 for the absorbing surface.

Answer:

1764 kilowatt

Explanation:

h = 120 m, mass per second = 1500 kg/s

Power = m g h / t

Power = 1500 x 9.8 x 120 / 1

Power = 1764000 Watt

Power = 1764 kilowatt

Answer:

d = 506.25 ft

Explanation:

As we know by kinematics that

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

here we know that initially the stone is dropped from rest from the edge of the roof

so here initial speed will be zero

now we have

[tex]v_i = 0[/tex]

also the acceleration of the stone is due to gravity which is given as

[tex]g = 32 ft/s^2[/tex]

now we have

[tex]v_f = 180 ft/s[/tex]

so from above equation

[tex]180^2 - 0 = 2(32)d[/tex]

[tex]d = 506.25 ft[/tex]

Using the equation of motion, we find that the height of the roof from which the stone was dropped is approximately 506.25 feet.

The subject question involves a physics concept relating to the motion of objects under the influence of gravity, in particular, the calculation of displacement during free fall. We can solve this problem using one of the fundamental equations of motion: final velocity2 = initial velocity2 + 2*a*d. In this case, initial velocity (u) = 0 (as the stone was dropped), final velocity (v) = -180 feet per second (negative as the stone is moving downwards), and acceleration (a) due to gravity is -32 feet/second2 (negative as gravity acts downward).

Substituting these values, we get:
(-180)2 = (0)2 + 2*(-32)*d.

This simplifies to:
32400 = -64d.

Solving for d (which represents the height of the roof), we get:
d = -32400 / -64 = 506.25 feet. Hence, the roof is approximately 506.25 feet high.

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Answer:

option c) 0.88c is the correct answer

Explanation:

using the Lorrentz equation we have

[tex]t=\frac{d}{v}\sqrt{1-(\frac{v}{c})^2}[/tex]

where,

t = time taken to cover the distance

d = Distance

v = velocity

c = speed of light

given

d = 15 light years

Now,

[tex]8years=\frac{15years\times c}{v}\sqrt{1-(\frac{v}{c})^2}[/tex]

or

[tex](\frac{v}{c})^2\times (\frac{8}{15})^2=1-(\frac{v}{c})^2[/tex]

or

[tex](\frac{v}{c})^2\times 0.284=1-(\frac{v}{c})^2[/tex]

or

[tex](\frac{v}{c})^2\times 0.284 + (\frac{v}{c})^2 =1[/tex]

or

[tex](1.284\times \frac{v}{c})^2 =1[/tex]

or

[tex] (\frac{v}{c}) =\sqrt{\frac{1}{0.284}}[/tex]

or

[tex]v=0.88c[/tex]

Answer:

c) 359%

Explanation:

k = spring constant of the spring

x₁ = initial stretch of the spring = 7 cm = 0.07 m

x₂ = final stretch of the spring = 15 cm = 0.15 m

Percentage increase in energy is given as

[tex]\frac{\Delta U \times 100}{U_{1}}=\frac{((0.5)k{x_{2}}^{2} - {(0.5)k x_{1}}^{2})(100)}{(0.5)k{x_{1}}^{2}}[/tex]

[tex]\frac{\Delta U \times 100}{U_{1}}=\frac{({x_{2}}^{2} - {x_{1}}^{2})(100)}{{x_{1}}^{2}}[/tex]

[tex]\frac{\Delta U \times 100}{U_{1}}=\frac{({0.15}^{2} - {0.07}^{2})(100)}{{0.07}^{2}}[/tex]

[tex]\frac{\Delta U \times 100}{U_{1}}[/tex] = 359%

Answer:

A yard is shorter than a meter.

Explanation:

>>>1 yard is 0.914 m, so a yard is shorter than a meter.

>>>1 mile is 1.609 km, so a mile is longer than a kilometer

>>>1 foot is 30.48cm, so a foot longer than a centimeter

>>> 1 inch is 2.54cm, so an inch is longer than a centimeter

From the above relationships, only a yard is shorter than a meter is true. Others are wrong.

Answer:

Intensity of a wave is Power per unit area.

Explanation:

The intensity of wave is defined as the power per unit area in the direction of motion of wave i.e.

[tex]I=\dfrac{P}{A}[/tex]

Also, the intensity of a wave is directly proportional to the square of its amplitude. Let a is the amplitude of a wave. So,

[tex]I\propto a^2[/tex]

The SI unit of power is watt and the SI unit of area is m². So, the SI unit of intensity is W/m². Hence, the correct option is (a) "Intensity of a wave is Power per unit area".

Answer:

a)  Tangential speed = 3.57 m/s

b) Centripetal acceleration = 17.79 m/s²

c) Maximum tangential speed the ball can have = 5.41 m/s

Explanation:

a) Tangential speed, v = rω

   Radius, r = 0.86 m

   Angular speed, ω = 0.660 rev/s = 0.66 x 2 x π = 4.15 rad/s

  Tangential speed, v = rω = 0.86 x 4.15 = 3.57 m/s

b) Centripetal acceleration,

          [tex]a=\frac{v^2}{r}=\frac{3.57^2}{0.86}=14.79m/s^2[/tex]

c) Maximum tension in horizontal circular motion

           [tex]T=\frac{mv^2}{r}\\\\136=\frac{4\times v^2}{0.86}\\\\v=5.41m/s[/tex]

  Maximum tangential speed the ball can have = 5.41 m/s

The questions involve concepts of circular motion: tangential speed, centripetal acceleration, and the tension in the rope maintaining the motion. Tangential speed is derived from the radius of the circle and the angular velocity, centripetal acceleration is derived from the square of tangential speed divided by the radius, and the maximum tangibility speed before the rope breaks can be calculated from its tension.

This is a problem of circular motion, and it's crucial to understand the concepts of tangential speed, centripetal acceleration, and the tension in the rope.

The tangential speed of a body moving along a circular path is its normal linear speed for that moment. It can be calculated by using the formula v = rω, where v is the tangential speed, r is the radius of the circle, and ω is the angular velocity. Make sure to convert the angular speed from rev/s to rad/s (ω = 0.660 rev/s * 2π rad/rev).

Centripetal acceleration is the rate of change of tangential velocity, and it always points toward the center of the circle. It can be calculated using the formula ac = v²/r.

The tension in the rope is the force exerting on the ball to keep it moving circularly. If it exceeds its maximum limit (136N in this case), the rope will break. The maximum tangential speed can be derived from the equation T = m * v² / r.

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Answer:

16.67 years

Explanation:

It moves a distance of 30 m in a year.

The time required tp move 500 m = 500 / 30 = 16.67 years

Answer:

15.7 N

Explanation:

Draw a free body diagram.  The block has four forces acting on it.  Gravity pulling down, normal force pushing perpendicular to the plane, friction pointing up the plane, and applied force F pushing up the plane.

Sum of the forces normal to the plane:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces parallel to the plane:

∑F = ma

Nμ + F − mg sin θ = 0

F = mg sin θ − Nμ

Substituting:

F = mg sin θ − mgμ cos θ

F = mg (sin θ − μ cos θ)

Given mg = 87.0 N, θ = 24.1°, and μ = 0.25 (because the block is not moving):

F = 87.0 N (sin 24.1° − 0.25 cos 24.1°)

F = 15.7 N

The minimum magnitude of the force F required to prevent the block from slipping is approximately 14.114 N.

The minimum magnitude of the force F required to prevent the block from slipping is given by the equation:

[tex]\[ F = w \sin(\theta) - \mu_s w \cos(\theta) \][/tex]

where w is the weight of the block, [tex]\( \theta \)[/tex] is the angle of inclination, and [tex]\mu_s \)[/tex] is the coefficient of static friction.

Given:

-[tex]\( w = 87.0 \, \text{N} \)[/tex]

- [tex]\( \theta = 24.1^\circ \)[/tex]

- [tex]\( \mu_s = 0.25 \)[/tex]

First, convert the angle from degrees to radians because the sine and cosine functions in most calculators require the input to be in radians:

[tex]\[ \theta_{\text{rad}} = \theta \times \frac{\pi}{180^\circ} \][/tex]

[tex]\[ \theta_{\text{rad}} = 24.1^\circ \times \frac{\pi}{180^\circ} \approx 0.420 \text{ radians} \][/tex]

Now, calculate the minimum force F using the given formula:

[tex]\[ F = w \sin(\theta_{\text{rad}}) - \mu_s w \cos(\theta_{\text{rad}}) \][/tex]

[tex]\[ F = 87.0 \, \text{N} \times \sin(0.420) - 0.25 \times 87.0 \, \text{N} \times \cos(0.420) \][/tex]

[tex]\[ F \ =87.0 \, \text{N} \times 0.412 - 0.25 \times 87.0 \, \text{N} \times 0.910 \][/tex]

[tex]\[ F \ = 35.864 \, \text{N} - 21.75 \, \text{N} \][/tex]

[tex]\[ F \ =14.114 \, \text{N} \][/tex]

Final answer:

The question revolves around the final temperature of a mixture of ice and water in a thermally insulated system. The specific heats and enthalpy of fusion are required to solve, but the proper method involves using the conservation of energy.

Explanation:

The student is asking about the final temperature of a system consisting of a 0.0811 kg ice cube at 0°C and 0.397 kg of water at 14.8°C. To solve this, we can use the principle of conservation of energy, stating that the heat lost by the water will equal the heat gained by the ice cube as it warms up, melts, and then possibly continues to heat up as water. However, without knowing the specific heat capacities and the enthalpy of fusion, we cannot provide a numerical answer. Given the specific heats and the heat of fusion, one would set up an equation where the heat lost by the water equals the heat gained by the ice, and solve for the final temperature. The specific heat of water, the specific heat of ice, and the enthalpy of fusion of ice would be necessary information to perform actual calculations.

Answer:

Velocity of throwing = 34.335 m/s

Explanation:

Time taken by the tennis ball to reach maximum height, t = 0.5 x 7 = 3.5 seconds.

Let the initial velocity be u, we have acceleration due to gravity, a = -9.81 m/s² and final velocity = 0 m/s

Equation of motion result we have v = u + at

Substituting

             0 = u - 9.81 x 3.5

             u = 34.335 m/s

Velocity of throwing = 34.335 m/s

Hey there!:

q1 = 4.25 * 10^-6 C is at x1 = 0 m

q2 = - 5.75 * 10^-6 C is at x2 = 0.23 m

q3 = - 9 uC = -9 * 10^-6 C

for the net force on q1 to be in -x direction

q3 must be placed to the left of q1

the net force on q1 , F = k * q1 * q3 /( x3^2) - k * q1 * q2 /( x2 - x3)^2

6 = 9 * 10^9 * 10^-6 * 4.25 *  10^-6 * (9 /x3^2 - 5.75 /( 0.25 - x3 )^2 )

solving for x3

x3 = - 0.22 m

the charge q3 is placed at x3 =  - 0.22 m

Hope this helps!

Answer:

5.062 x 10^-9 mg

Explanation:

Charge on glass rod = 89 uC

Charge on silk cloth = - 89 uC

mass of scarf, M = 100 g

Number of excess electrons on silk cloth = charge / charge of one electron

n = (89 x 10^-6) / ( 1.6 x 10^-19)

n = 5.5625 x 10^14

mass of one electron = 9.1 x 10^-31 kg

Mass of  5.5625 x 10^14 electrons =  5.5625 x 10^14 x 9.1 x 10^-31 kg

ΔM = 5.062 x 10^-16 kg

Δ M / M = (5.062 x 10^-16) / 0.1 = 5.062 x 10^-15 kg = 5.062 x 10^-9 mg

It is not noticeable.

Answer:

h₍₁₎ = 495,1 meters

h₍₂₎ = 480,4 m

h₍₃₎ = 455,9 m

...

..

Explanation:

The exercise is "free fall". t = [tex]\sqrt{\frac{2h}{g} }[/tex]

Solving with this formula you find the time it takes for the stone to reach the ground (T) = 102,04 s

The heights (h) according to his time (t) are found according to the formula:

h(t) = 500 - 1/2 * g * t²

Remplacing "t" with the desired time.

Answer:

Option (D)

Explanation:

In the projectile motion

Horizontal distance = horizontal velocity × time

3.3 = u × 0.66

u = 5 m/s

Muzzle velocity is 5 m/s

Answer:

YES THERE ARE VERY MANY BOOKS THAT YOU CAN BUY OR BORROW FROM YOUR LOCAL LIBRARY, HOPE THIS HELPS! HAVE A GREAT DAY!

Explanation:

Answer:

- 100 Nm²/C

Explanation:

E = magnitude of electric field along x-direction = 10 N/C

θ = angle made by the direction of electric field with rectangular surface = 30

φ = angle made by the direction of electric field with normal to the rectangular surface = 90 + 30 = 120

A = area of the rectangular surface = 2 x 10 = 20 m²

Φ = Electric flux crossing the surface

Electric flux is given as

Φ = E A Cosφ

Φ = (10) (20) Cos120

Φ = - 100 Nm²/C

Final answer:

The student must use Coulomb's law to calculate the net electrostatic force between two charged red blood cells, and then use energy conservation to determine the initial speed needed for the cells to overcome this force and just touch. So, the initial speed of each cell is approximately [tex]\( 662.8 \, \text{m/s} \)[/tex].

Explanation:

Coulomb's Law and Initial Speed Calculation

The student's problem involving two red blood cells carrying negative charges requires the use of Coulomb's law to determine the electrostatic force of repulsion between them. Since the cells are modeled as spheres, the charges are assumed to be uniformly distributed. To find the initial speed needed for the cells to just touch, one must consider energy conservation principles, where the initial kinetic energy of the cells should be equivalent to the electrostatic potential energy at the point of closest approach (when they just touch).

The formula for Coulomb's law is
F = k * |q1 * q2| / r^2, where F is the force between the charges, q1 and q2 are the charges, r is the separation distance, and k is Coulomb's constant (8.9875 x 10^9 Nm^2/C^2). The potential energy (U) due to electrostatic force is given by U = k * |q1 * q2| / r. By setting the kinetic energy (1/2 * m * v^2) for both cells equal to U, and solving for v, we can find the initial speed of each cell.

To find the initial speed of each cell using the potential energy and kinetic energy, you can follow these steps:

Given:

[tex]- Charges: \( q_1 = -2.50 \, \text{pC} \), \( q_2 = -3.30 \, \text{pC} \)\\- Separation distance: \( r = 3.75 \times 10^{-6} \, \text{m} \)\\- Coulomb's constant: \( k = 8.9875 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)\\- Mass: \( m = 9.05 \times 10^{-14} \, \text{kg} \)\\[/tex]

1. Calculate the potential energy \( U \) using Coulomb's law:

  [tex]\[ U = \frac{k \cdot |q_1 \cdot q_2|}{r} \] \[ U = \frac{8.9875 \times 10^9 \, \text{Nm}^2/\text{C}^2 \cdot |(-2.50 \times 10^{-12} \, \text{C}) \cdot (-3.30 \times 10^{-12} \, \text{C})|}{3.75 \times 10^{-6} \, \text{m}} \][/tex]

2. Calculate the potential energy \( U \):

  [tex]\[ U = \frac{8.9875 \times 10^9 \, \text{Nm}^2/\text{C}^2 \cdot 8.25 \times 10^{-24} \, \text{C}^2}{3.75 \times 10^{-6} \, \text{m}} \] \[ U = \frac{7.465625 \times 10^{-14} \, \text{Nm}}{3.75 \times 10^{-6} \, \text{m}} \] \[ U = 1.989 \times 10^{-8} \, \text{J} \][/tex]

3. Set the kinetic energy equal to the potential energy:

 [tex]\[ \frac{1}{2} m v^2 = U \][/tex]

4. Solve for \( v \):

  [tex]\[ v = \sqrt{\frac{2U}{m}} \] \[ v = \sqrt{\frac{2 \times 1.989 \times 10^{-8} \, \text{J}}{9.05 \times 10^{-14} \, \text{kg}}} \][/tex]

5. Calculate \( v \):

 [tex]\[ v \approx \sqrt{\frac{2 \times 1.989 \times 10^{-8} \, \text{J}}{9.05 \times 10^{-14} \, \text{kg}}} \] \[ v \approx \sqrt{4.391 \times 10^5 \, \text{m}^2/\text{s}^2} \] \[ v \approx 662.8 \, \text{m/s} \][/tex]

So, the initial speed of each cell is approximately [tex]\( 662.8 \, \text{m/s} \)[/tex].

Since there is no viscous drag, the conservation of energy provides the necessary framework to equate the initial kinetic energy to the electrostatic potential energy. Accounting for the fact that each blood cell has a different charge, the calculation would involve determining the net repulsive force and then the corresponding speed for each cell to overcome this force and just barely touch.

The initial speed required for each red blood cell to just barely touch is approximately [tex]\( 1.479 \times 10^{3} \)[/tex]m/s.

Given:

- Mass of each red blood cell, [tex]\( m = 9.05 \times 10^{-14} \)[/tex] kg

- Charge on the first cell, [tex]\( q_1 = -2.50 \) pC \( = -2.50 \times 10^{-12} \) C[/tex]

- Charge on the second cell, [tex]\( q_2 = -3.30 \) pC \( = -3.30 \times 10^{-12} \) C[/tex]

- Radius of each cell, [tex]\( r = 3.75 \times 10^{-6} \) m[/tex]

 When the cells just barely touch, the distance between their centers will be 2r , since each cell has a radius r.

The electrostatic force F between the two charges can be calculated using Coulomb's law:

[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]

where \( k \) is Coulomb's constant, [tex]\( k = 8.9875 \times 10^9 \)[/tex] N m²/C².

Substituting the values, we get:

[tex]\[ F = (8.9875 \times 10^9) \frac{|(-2.50 \times 10^{-12})(-3.30 \times 10^{-12})|}{(2 \times 3.75 \times 10^{-6})^2} \][/tex]

Now, we need to find the initial kinetic energy ( KE ) of each cell, which must be equal to the work done by the electrostatic force when the cells move from infinity to a distance ( 2r ) apart. The work done by the electrostatic force is equal to the potential energy U at a distance 2r:

[tex]\[ U = k \frac{|q_1 q_2|}{2r} \][/tex]

The initial kinetic energy of each cell is given by:

[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]

where v is the initial speed we want to find.

Since there are no other forces doing work (no viscous drag), the total mechanical energy is conserved, and the initial kinetic energy must equal the potential energy at the point of touching:

[tex]\[ \frac{1}{2} m v^2 = k \frac{|q_1 q_2|}{2r} \][/tex]

Solving for v, we get:

[tex]\[ v = \sqrt{\frac{2k |q_1 q_2|}{m (2r)}} \][/tex]

Substituting the values, we have:

[tex]\[ v = \sqrt{\frac{2 \times (8.9875 \times 10^9) \times |(-2.50 \times 10^{-12})(-3.30 \times 10^{-12})|}{9.05 \times 10^{-14} \times (2 \times 3.75 \times 10^{-6})}} \][/tex]

Now, let's calculate the value of \( v \):

[tex]\[ v = \sqrt{\frac{2 \times (8.9875 \times 10^9) \times (2.50 \times 10^{-12} \times 3.30 \times 10^{-12})}{9.05 \times 10^{-14} \times (2 \times 3.75 \times 10^{-6})}} \] \[ v = \sqrt{\frac{2 \times (8.9875 \times 10^9) \times (8.25 \times 10^{-24})}{9.05 \times 10^{-14} \times (7.5 \times 10^{-6})}} \] \[ v = \sqrt{\frac{2 \times (7.4234375 \times 10^{-14})}{6.79875 \times 10^{-20}}} \][/tex]

[tex]\[ v = \sqrt{\frac{14.846875 \times 10^{-14}}{6.79875 \times 10^{-20}}} \] \[ v = \sqrt{2.18375 \times 10^{6}} \] \[ v \approx 1.479 \times 10^{3} \text{ m/s} \][/tex]

Answer:

The magnitude of the angular momentum of the Mars in a circular orbit around the sun is [tex]3.53\times10^{39}\ kg-m^2/s[/tex]

Explanation:

Given that,

Mass of mars [tex]M=6.42\times10^{23}\ kg[/tex]

Radius [tex]r'= 3.40\times10^{6}[/tex]

Orbit radius [tex]r=2.28\times10^{11}\ m[/tex]

Time period = 687 days

We need to calculate the magnitude of the angular momentum of the mars

Using formula of angular momentum

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

Here, [tex] I = mr^2[/tex]

[tex] L=mr^2\omega[/tex]

[tex] L=mr^2\times\dfrac{2\pi}{T}[/tex]

Where,

L = angular momentum

I = moment of inertia

r = radius

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

Put the value into the formula

[tex]L=6.42\times10^{23}\times(2.28\times10^{11})^2\times\dfrac{2\pi}{687\times24\times3600}[/tex]

[tex]L=3.53\times10^{39}\ kg-m^2/s[/tex]

Hence, The magnitude of the angular momentum of the Mars in a circular orbit around the sun is [tex]3.53\times10^{39}\ kg-m^2/s[/tex]

The angular momentum of Mars in its circular orbit around the Sun can be calculated by using the formula L = mvr, where m is the mass of Mars, v is its orbital velocity, and r is the radius of its orbit. The orbital velocity is calculated based on the time Mars takes to complete one orbit.

The question is asking for the calculation of the angular momentum of Mars in its circular orbit around the sun. The angular momentum of any body moving in a circular path is given by the product of its mass (m), its orbital velocity (v), and the radius of its circular path (r). The orbital velocity can be calculated as the circumference of the circular path (2πr) over the period of orbit.

Given that the mass of Mars is 6.42×1023 kg, orbit radius is 2.28×1011 m, and that Mars completes one orbit in 687 days, we can convert these days into seconds as this is the standard SI unit for time, then calculate the orbital speed.

Angular momentum can then be calculated using the formula L = mvr

The full set of calculations is often more complicated, because they have to account for the gravitational influence of other celestial bodies, but for our purposes, this basic calculation should suffice.

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Answer:

Time period of oscillations is 0.62 s

Explanation:

Due to suspension of weight the change in the length of the spring is given as

[tex]\Delta L = L_f - L_i[/tex]

[tex]\Delta L = 21.5 - 11.9 = 9.6 cm[/tex]

now we know that spring is stretched due to its weight so at equilibrium the force due to weight is counter balanced by the spring force

[tex]mg = kx[/tex]

[tex]0.037 (9.81) = k(0.096)[/tex]

[tex]k = 3.78 N/m[/tex]

Now the period of oscillation of spring is given as

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

Now plug in all values in it

[tex]T = 2\pi \sqrt{\frac{0.037}{3.78}}[/tex]

[tex]T = 0.62 s[/tex]

The period of the oscillation for the spring with a spring constant of 3.78 N/m will be 0.7316 sec.

We know that the formula of the spring force is written as,

[tex]F = k \times \delta[/tex]

Given to us,

weight on the spring, F = 37 g = (9.81 x 0.037) = 0.36297 N

Deflection in spring, δ = (21.5 - 11.9) = 9.6 cm = 0.096 m

Substitute the value,

[tex]0.36297 = k \times 0.096\\k= 3.78\rm\ N/m[/tex]

We know that the weight is been pulled down therefore, an external force is been applied to the spring to extend it further to 25.2 cm.

We know that the formula of the spring force is written as,

[tex]F = k \times \delta[/tex]

Given to us

Deflection in the spring, δ = (25.2 - 11.9) = 13.3 cm = 0.133 m

Spring constant, k = 3.78 N/m

Substitute the values,

[tex]F = 3.78 \times 0.133\\\\F = 0.5\\\\m \times g = 0.5\\\\m = 0.05125\rm\ kg[/tex]

We know the formula for the period of oscillation is given as,

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

Substitute the values,

m = 0.05125 kg

k = 3.78 N/m

[tex]T = 2\pi \sqrt{\dfrac{0.05125}{3.78}}\\T = 0.7316\rm\ sec[/tex]

Hence, the period of the oscillation for the spring with a spring constant of 3.78 N/m will be 0.7316 sec.

Learn more about Oscillation:

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Answer:

Force, F = 77 N

Explanation:

A child in a wagon seem to fall backward when you give the wagon a sharp pull forward. It is due to Newton's third law of motion. The forward pull on wagon is called action force and the backward force is called reaction force. These two forces are equal in magnitude but they acts in opposite direction.

We need to calculate the force is needed to accelerate a sled. It can be calculated using the formula as :

F = m × a

Where

m = mass = 55 kg

a = acceleration = 1.4 m/s²

[tex]F=55\ kg\times 1.4\ m/s^2[/tex]

F = 77 N

So, the force needed to accelerate a sled is 77 N. Hence, this is the required solution.

The child in the wagon appears to fall backward due to inertia, which is the property that resists changes in motion. To accelerate a 55 kg sled at 1.4 m/s2, a force of 77 N is required.

A child in a wagon seems to fall backward when you give the wagon a sharp pull forward due to inertia. Inertia is the resistance of any physical object to a change in its state of motion or rest. The child's body tends to remain in a state of rest while the wagon accelerates forward, resulting in the child appearing to fall backward.

Now, to address the second part of the question:

To find this, we use Newton's second law, F = ma where F is force, m is mass, and a is acceleration.

Therefore, F = 55 kg * 1.4 m/s2 = 77 N.

The correct answer is 77 N.

Answer:

1.2826 x 10^-13 m

Explanation:

[tex]\lambda  = \frac{h}{\sqrt{2 m K}}[/tex]

Here, k be the kinetic energy and m be the mass

K = 50 KeV = 50 x 1.6 x 10^-16 J = 80 x 10^-16 J

m = 1.67 x 10^-27 kg

[tex]\lambda  = \frac{6.63 \times  10^{-34}}{\sqrt{2 \times 1.67\times 10^{-27}\times 80\times 10^{-16}}}[/tex]

λ = 1.2826 x 10^-13 m

Answer:

The minimum coefficient of friction must exist between the road and tires to prevent the car from slipping is μ= 0.28

Explanation:

m= 1500 kg

r= 52m

Vt= 12m/s

g= 9.8 m/s²

Vt= ω * r

ω= Vt/r

ω= 0.23 rad/s

ac= ω²* r

ac= 2.77 m/s²

Fr= m* ac

Fr= 4153.84 N

W= m*g

W= 14700 N

Fr= μ * W

μ= Fr/W

μ=0.28

Answer:

Vector angular momentum about this axis of the sphere is:

L= 3.76[tex]\hat k[/tex] kg-m²/sec

Explanation:

The formula for the moment of inertia of a sphere is:

[tex]I=\frac{2}{5}\times MR^2[/tex]

Given:

Mass of the sphere = 13.5 kg

Radius of the sphere = 0.490 m

Thus, moment of inertia :

[tex]I=\frac{2}{5}\times 13.5\times (0.490)^2 kg\ m^2[/tex]

[tex]I=1.29654 kg\ m^2[/tex]

The expression for the angular momentum is:

L=I×ω

Given:

Angular speed(ω) = 2.9 rad/s

I, above calculated = 1.29654 kgm⁻²

Thus, angular momentum is:

L= 1.29654×2.9 kg-m²/sec

L= 3.76 kg-m²/sec

Given, the sphere is turning counterclockwise about the vertical axis. Thus, the direction of the angular momentum will be on the upper side of the plane. ( [tex]+\hat k[/tex] ).

Thus, angular momentum with direction is:

L= 3.76[tex]\hat k[/tex] kg-m²/sec

Answer: The initial temperature of the system comes out to be 147 °C

Explanation:

To calculate the initial temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

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

where,

[tex]V_1\text{ and }T_1[/tex] are the initial volume and temperature of the gas.

[tex]V_2\text{ and }T_2[/tex] are the final volume and temperature of the gas.

We are given:

[tex]V_1=2.50L\\T_1=?K\\V_2=1.75L\\T_2=25^oC=(25+273)K=294K[/tex]

Putting values in above equation, we get:

[tex]\frac{2.50L}{T_1}=\frac{1.75L}{294K}\\\\T_1=420K[/tex]

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

[tex]T(K)=T(^oC)+273[/tex]

[tex]420=T(^oC)+273\\T(^oC)=147^oC[/tex]

Hence, the initial temperature of the system comes out to be 147 °C

Answer:

The nuclear binding energy = [tex]1.466\times 10^{-13} KJ[/tex]

The binding energy per nucleon =1.37×10⁻¹⁵ KJ/nucleon

Explanation:

Given:

Number of protons = 47

Number of neutrons = 107-47 = 60

Now,

the mass defect (m)= Theoretical mass - actual mass

m = [tex]47 (1.007825) + 60(1.008665) - 106.9051[/tex]

since,

mass of proton = 1.007825 amu

Mass of neutron = 1.008665 amu

thus,

m = [tex]47.367775 + 60.5199 - 106.9051[/tex]

or

m = [tex]0.982575 amu[/tex]

also

1 amu = [tex]1.66\times 10^{-27} kg[/tex]

therefore,

m =  [tex]0.982575 amu \times 1.66\times 10^{-27} kg[/tex]

or

m = [tex]1.6310745\times 10^{-27} kg[/tex]

now,

Energy = mass × (speed of light)²

thus,

Energy = [tex]1.6310745\times 10^{27} kg ( 3\times 10^8 ms^{-1} )[/tex]

or

Energy = [tex]14.66\times 10^{-11} kgm^2/s^2[/tex]

or

Energy = [tex]1.466\times 10^{-10} J[/tex] = [tex]1.466\times 10^{-13} KJ[/tex]

Therefore the nuclear binding energy = [tex]1.466\times 10^{-13} KJ[/tex]

Now,

the binding energy per nucleon = [tex]\frac{1.466\times 10^{-13} KJ}{107}[/tex] = 1.37×10⁻¹⁵ KJ/nucleon

Explanation:

It is given that, a stone travel over level ground if it is thrown upward at an angle of 30.0° with respect to the horizontal.

Speed with which it is thrown, v = 11 m/s

We need to find the maximum height above the ground level. The maximum height attained by the stone is given by :

[tex]h=\dfrac{v^2sin^2\theta}{2g}[/tex]

[tex]h=\dfrac{(11\ m/s)^2sin^2(30)}{2\times 9.8\ m/s^2}[/tex]

h = 1.54 m

So, the stone will travel a distance of 1.54 meters.

Range of a projectile is given by :

[tex]R=\dfrac{v^2sin\ 2\theta}{g}[/tex]

[tex]R=\dfrac{(11\ m/s)^2sin(60)}{9.8\ m/s^2}[/tex]

R = 10.69 m

So, the maximum range achieved with the same initial speed is 10.69 meters. Hence, this is the required solution.

A stone thrown upward at a 30.0° angle with a speed of 11.0 m/s will travel approximately 10.74 meters over level ground. For maximum range, the same stone thrown at a 45° angle would travel approximately 12.34 meters.

We need to use the equations of projectile motion. The horizontal range (R) of a projectile is given by the formula R = (v^2 * sin(2θ)) / g, where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity (9.81 m/s² on Earth). For maximum range, the ideal launch angle is 45° as it provides the optimal balance between vertical and horizontal components of velocity. Thus, for the same initial speed, the maximum range is achieved at this angle.

To calculate the range at 30°: R = (11.0^2 * sin(60°)) / 9.81 = (121 * 0.866) / 9.81 ≈ 10.74 meters.

For maximum range at v = 11.0 m/s and θ = 45°: R = (11.0^2 * sin(90°)) / 9.81 = (121 * 1) / 9.81 ≈ 12.34 meters.

Answer:

[tex]a = 2.7 \times 10^{-3} m/s^2[/tex]

Explanation:

As we know that acceleration due to gravity is given as

[tex]a = \frac{GM}{r^2}[/tex]

here we know that

[tex]M = 5.97 \times 10^{24} kg[/tex] (mass of earth)

r = distance at which acceleration is to be find out

[tex]r = 3.84 \times 10^8 m[/tex]

now from the formula of acceleration due to gravity we know

[tex]a = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(3.84 \times 10^8)^2}[/tex]

[tex]a = 2.7 \times 10^{-3} m/s^2[/tex]

Answer:

(a) 9.65 mH

(b) doubled

Explanation:

A = 6 cm^2 = 6 x 10^-4 m^2, l = 20 cm = 0.2 m, i = 2.8 A, N = 1600

(a) The formula for the self inductance of a solenoid is given by

L = μ0 N^2 A / l

L = 4 x 3.14 x 10^-7 x 1600 x 1600 x 6 x 10^-4 / 0.2

L = 9.65 x 10^-3 H

L = 9.65 mH

(b) As we observe that the inductance is proportional to the area so if the area is doubled, then inductance is also doubled.