Consider water flowing through a cylindrical pipe with a variable cross-section. The velocity is v at a point where the pipe diameter is 1.0 cm. At a point where the pipe diameter is three times larger, the velocity is:

nine times larger than the initial velocity
one ninth the initial velocity
three times larger than the initial velocity
the same as the initial velocity
one third the initial velocity

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:

[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:

Radius between electron and proton[tex] = 6.804\times 10^{-10}m[/tex]

Explanation:

The motion of the electron is carried out in the orbit due to the balancing of the electrostatic force between the proton and the electron and the centripetal force acting on the electron.

The electrostatic force is given as = [tex]\frac{kq_1q_2}{r^2}[/tex]

Where,

k = coulomb's law constant (9×10⁹ N-m²/C²)

q₁ and q₂  = charges = 1.6 × 10⁻¹⁹ C

r = radius between the proton and the electron

Also,

Centripetal force on the moving electron is given as:

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

where,

[tex]m_e[/tex] = mass of the electron (9.1 ×10⁻³¹ kg)

V = velocity of the moving electron (given: 6.1 ×10⁵ m/s)

Now equating both the formulas, we have

[tex]\frac{kq_1q_2}{r^2}[/tex] = [tex]\frac{m_eV^2}{r}[/tex]

⇒[tex]r = \frac{kq_1q_2}{m_eV^2}[/tex]

substituting the values in the above equation we get,

[tex]r = \frac{9\times 10^{9}\times (1.6\times 10^{-19})^2}{9.1\times 10^{-31}\times (6.1\times 10^5)^2}[/tex]

⇒[tex]r = 6.804\times 10^{-10}m[/tex]

Answer:

Electric field, E = 45.19 N/C

Explanation:

It is given that,

Surface charge density of first surface, [tex]\sigma_1=0.2\ nC/m^2=0.2\times 10^{-9}\ C/m^2[/tex]

Surface charge density of second surface, [tex]\sigma=-0.6\ nC/m^2=-0.6\times 10^{-9}\ C/m^2[/tex]

The electric field at a point between the two surfaces is given by :

[tex]E=\dfrac{\sigma}{2\epsilon_o}[/tex]

[tex]E=\dfrac{\sigma1-\sigma_2}{2\epsilon_o}[/tex]

[tex]E=\dfrac{0.2\times 10^{-9}-(-0.6\times 10^{-9})}{2\times 8.85\times 10^{-12}}[/tex]

E = 45.19 N/C

So, the magnitude of the electric field at a point between the two surfaces is 45.19 N/C. Hence, this is the required solution.

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:

- 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

Answer:

Factor by which the intrnal energy changes is 66.63

Explanation:

Given:

When the is motor operated normally, the current through the motor,I₁ =0.750A

Motor winding resistance, R = 19.6Ω

Thus, the Power dissipated at Normal condition,(P₁):

P₁=I₁²×R

P₁=(0.750)²×19.6= 11.025 W

Also,

Power = Rate of change of energy = [tex]\frac{dE}{dt}[/tex]

Now,

when the motor is connected to 120 V power supply, the current through the  motor (I₂) =[tex]\frac{V}{R}[/tex]

or

I₂ =[tex]\frac{120}{19.6}A[/tex]

or

I₂ =6.122 A

Thus, the power dissipated due the current I₂ will be

P₂ = I₂² × R

⇒ P₂ = (6.122)² × 19.6 = 734.693 W

Hence, the factor by which the internal energy rate in resistor winding increases is = [tex]\frac{P_2}{P_1}=\frac{734.693}{11.025}=[/tex]66.63

The solution involves using Ohm's Law and the formula for power dissipation to determine the increase in the rate of change of internal energy in the motor windings when the rotor seizes and compare it to normal operating conditions.

The student's question pertains to a direct current motor and involves calculating the rate of change of internal energy, back electromotive force (emf), and current under various scenarios. To address the student's question, one would need to apply Ohm's Law (V = IR, where V is voltage, I is current, R is resistance) and the power dissipation formula (P = I²R, where P is power).

Explanation of Solution

Initially, the motor's normal operation can be described by the values given: I = 0.750 A, V = 120 V, R = 19.6 Ω. The power dissipated (P) by the motor windings when the motor is operating normally would be P = I²R = (0.750 A)²(19.6 Ω).

However, if the rotor seizes, the motor will essentially become a static resistor and the current through it will increase. This is because without the back emf generated by the rotating rotor, the voltage across the motor windings will be the full supply voltage. The current in this case (let's call it I') can be calculated using Ohm's Law: I' = V/R = 120 V / 19.6 Ω. The new power dissipation (P') when the rotor seizes would be P' = I'²R. To find by what factor the rate of change of internal energy increases, we would need to compare P' to the original power dissipation P by calculating P'/P.

Answer:

2 N

Explanation:

Wight of object = mg = 8 N

Coefficient of friction, u = 0.25

The normal reaction acting on the object = N = mg = 8 N

According to the laws of limiting friction,

The friction force, f = uN

f = 0.25 x 8 = 2 N

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

Explanation:

The number of charges flowing per second inside the conductor is called the current through the conductor. Mathematically, it is given by :

[tex]I=\dfrac{Q}{t}[/tex]

Where

Q is the charge

t is time

Since, Q = ne

n is number of electrons

e is the charge on one electron

So, to find the current through the conductor we must know the number of electrons moving per second through it. Hence, this is the required solution.

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

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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Final answer:

The longer wire will have twice the resistance of the shorter wire.

Explanation:

The resistance of a wire depends on both its length and cross-sectional area. In this case, the longer wire has twice the length and twice the cross-sectional area of the shorter wire. The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. Therefore, the longer wire will have twice the resistance of the shorter wire.

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:

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:

111 N

Explanation:

weight (downwards) = 600 N

reaction force (upwards) = 900 N

t = 0.37 s

Net force

F = reaction force - weight = 900 - 600 = 300 N

Impulse = force x time = 300 x 0.37 = 111 N

Answer:

The observer experienced a change I'm frequency of 6971.42Hertz

Explanation:

In waves, the frequency of a wave is directly proportional to its velocity i.e F ∝ V

F = kV

k = F/V

F1/V1 = F2/V2 = k... 1

Given F1 = 800Hz at V1 = 35m/s

F2 = ? when V2 = 340m/s

Substituting this given datas into equation 1 to get the new frequency F2:

800/35 = F2/340

35F2 = 800×340

F2 = 800×340/35

F2 = 7771.42Hertz

Change of frequency = F2-F1

Change of frequency = 7771.42-800

Change of frequency = 6971.42Hertzz

The change of frequency does the observer hear due to police car siren is 6971.42 Hz.

Frequency of wave is the number of waves, which is passed thorough a particular point at a unit time.

For the cars approaching to the observer, the Doppler formula can be given as

[tex]\dfrac{f_1}{v_1}=\dfrac{f_2}{v_s}[/tex]

Here, [tex](v_s)[/tex] is the speed of the sound.

A police car has an 800-Hz siren. It is traveling at 35.0 m/s on a day when the speed of sound through air is 340 m/s.

As, the car approaches and passes an observer. Thus by the above formula,

[tex]\dfrac{800}{35}=\dfrac{f_2}{340}\\f_2=7771.41\rm Hz[/tex]

As the frequency of police car siren is 800-Hz. Thus, the change of the frequency observed by the observer is,

[tex]\Delta f=7771.42-8000\\\Delta f=6971.42\rm Hz[/tex]

Thus, change of frequency does the observer hear due to police car siren is 6971.42 Hz.

Learn more about the frequency here;

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Answer:

2954.6 N/C, 46.36 degree from positive  axis

Explanation:

E1 = 1300 N/C, θ1 = 35 degree

E2 = 1700 N/C, θ2 = 55 degree

Now write the electric fields in vector form

E1 = 1300 ( Cos 35 i + Sin 35 j) = 1064.9 i + 745.6 j

E2 = 1700 ( Cos 55 i + Sin 55 j) = 975.08 i + 1392.6 j

Resultant electric field

E = E1 + E2

E =  1064.9 i + 745.6 j + 975.08 i + 1392.6 j

E = 2039.08 i + 2138.2 j

Magnitude of E

E = sqrt (2039.08^2 + 2138.2^2)

E = 2954.6 N/C

Let it makes an angle Φ from X axis

tan Φ = 2138.2 / 2039.08 = 1.049

Φ = 46.36 degree from positive X axis.

This question is about calculating the overall electric field from two individual fields using vector addition. This involves resolving each field into its components, adding the respective components, and then using Pythagoras' theorem and the tangent inverse relation to determine the overall magnitude and direction of the resultant field.

The total electric field can be determined by using vector addition to add together the individual electric fields due to each of the two components mentioned in your question. Firstly, we resolve each field into its x and y components. The x-component and y-component for E1 are E1*cos(θ1) and E1*sin(θ1), respectively. Similarly, the x and y components for E2 are E2*cos(θ2) and E2*sin(θ2), respectively.

We then add together the respective x and y components: E(x total) = E1*cos(θ1) + E2*cos(θ2) and E(y total) = E1*sin(θ1) + E2*sin(θ2).

The overall magnitude of the resultant electric field can be calculated using Pythagoras' theorem, √[(E(x total))^2 + (E(y total))^2]. The direction of the total electric field can be evaluated using the tangent inverse relation θ total = tan^-1 [E(y total)/E(x total)].

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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:

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.

The angular momentum of a rotation object is the product of its moment of inertia and its angular velocity:

L = Iω

L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Apply the conservation of angular momentum. The total angular momentum before disks A and B are joined is:

L_{before} = (3.3)(6.6) + B(-9.3)

L_{before} = -9.3B+21.78

where B is the moment of inertia of disk B.

The total angular momentum after the disks are joined is:

L_{after} = (3.3+B)(-2.1)

L_{after} = -2.1B-6.93

L_{before} = L_{after}

-9.3B + 21.78 = -2.1B - 6.93

B = 4.0kg·m²

The moment of inertia of disk B is 4.0kg·m²

Answer:

Part a)

E = 4812.5 N/C

Part b)

Two field must be perpendicular to each other as well as perpendicular to the velocity of charge so that net force on the moving charge will be zero

Explanation:

For uniform electric and magnetic field we will have a charge particle moving through it such that net force on it is zero

so here we have

[tex]qE = qvB[/tex]

magnetic force on moving charge will balanced by electrostatic force on moving charge

[tex]v = \frac{E}{B}[/tex]

now we have

[tex]v = 8.75 km/s[/tex]

B = 0.550 T

now we have

[tex]E = (8.75 \times 10^3)(0.550)[/tex]

[tex]E = 4812.5 N/C[/tex]

Part b)

Two field must be perpendicular to each other as well as perpendicular to the velocity of charge so that net force on the moving charge will be zero

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:

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.

Final answer:

The angular magnification produced by a +4.00 cm focal length lens with an object at 3.75 cm distance is approximately 7.25 times.

Explanation:

The question presented is a problem in optics, more specifically in the subtopic of lens magnification, which is a part of Physics. To calculate the angular magnification for a lens, we can use the thin lens formula and the magnification formula. Given that the focal length of the lens (f) is +4.00 cm and the object distance (do) is 3.75 cm, we first determine if an image is formed by using the lens equation 1/f = 1/do + 1/di. However, since this equation will not yield a real value for di (image distance) because do is smaller than f, we can't directly use it to find the angular magnification.

Instead, the angular magnification (M) of a simple lens while viewing a close object is approximately given by M = 1 + (25 cm / f), where 25 cm is the near point of a normal human eye. Thus, for this lens:

M = 1 + (25 cm / 4.00 cm)

M = 1 + 6.25

M = 7.25

This means the lens would produce an angular magnification of approximately 7.25 times when the object is placed at 3.75 cm from the lens.

Answer:

(a) 1.042 J

b) 0.108 m

Explanation:

(a)

W = work done to stretch the spring = 2 J

k = spring constant

L₀ = initial length = 30 cm = 0.30 m

L = final length = 42 cm = 0.42 m

x = stretch of the spring = L - L₀ = 0.42 - 0.30 = 0.12 m

Work done to stretch the spring is given as

W = (0.5) k x²

2 = (0.5) k (0.12)²

k = 277.78 N/m

x₀ = initial stretch of the spring = 35 - 30 = 5 cm = 0.05 m

x = final stretch of the spring = 40 - 30 = 10 cm = 0.10 m

W' = work needed to stretch the spring from 35 cm to 40 cm

Work needed to stretch the spring from 35 cm to 40 cm is given as

W' = (0.5) k (x² - x₀²)

W' = (0.5) (277.78) (0.10² - 0.05²)

W' = 1.042 J

b)

x = Stretch of the spring beyond natural length

F = force = 30 N

Spring force is given as

F = k x

30 = (277.78) x

x = 0.108 m

(a) The work is needed to stretch the spring from 35 cm to 40 cm is 1.042 J.

(b) Beyond the length of 0.108 m  force of 30 N keep the spring stretched.

In order to stretch or compress the spring some amount of work has to be done. Which is given as,

          [tex]\rm{W=\frac{1}{2} Kx^{2} }[/tex]  

where k = spring constant of the spring

          x = compression of spring

          W= Work required or spring work

Initial length is given by 30 cm = 0.30 m

final length is given by  42 cm = 0.42 m

When spring is stretched change in the length occurs denoted by x

x = final length - initial length =0.42 - 0.30 = 0.12 m

[tex]\rm{W=\frac{1}{2} Kx^{2} }[/tex]

[tex]\rm{2=\frac{1}{2} K(0.12)^{2} }[/tex]

K= 277.78 N/m

(a) work is needed to stretch the spring from 35 cm to 40 cm

stretching length from30cm to 35 is 5 cm which is 0.05 m

stretching length from cm to 30 is 40 cm is 10 cm which is 0.1m

x = final stretch of the spring = 40 - 30 = 10 cm = 0.10 m

Work needed to stretch the spring from 35 cm to 40 cm is given

W = (0.5)K([0.1]² - [0.5]²)

W = (0.5) (277.78) (0.10² - 0.05²)

W = 1.042 J

x is stretch of the spring beyond natural length

F = force = 30 N

Spring force is given as

F = k x

30 = (277.78) x

x = 0.108 m

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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

The maximum speed at which the truck can briefly move around the curve without making the toolbox slide can be found by setting the sliding acceleration equal to the centripetal acceleration (v²/r) and solving for v. Factors like tire friction and flat terrain are also taken into account.

To determine how fast the truck can go before the toolbox starts to slide, we need to calculate the maximum static friction (the force that prevents sliding).

The formula to identify the maximum speed that didn't result in sliding, we would use centripetal acceleration, which is the product of tangential speed squared divided by the curve's radius.

The acceleration of the crate on the truck bed must equal centripetal acceleration to prevent sliding. Given that the sliding acceleration is 2.06 m/s², setting this value to equal centripetal acceleration (v²/r where r is 22 m) and solving for v (velocity/speed), would provide the maximum speed at which the truck can move without the toolbox sliding.Tire friction also plays a role, as it allows vehicles to move at higher speeds without sliding off the road.

This also assumes that the road is relatively flat and the truck moves uniformly, similar to the situation illustrated in Figure 24.7 with a motorist moving in a straight direction.

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The speed at which a toolbox in a truck will start to slide when the truck is turning around a curve depends on the balance of the centripetal force and static friction force. These depend on the truck's speed, the curve's radius, the toolbox's mass and the friction coefficient. Without specific values, a numerical answer can't be given.

The question is asking about the speed at which the toolbox in the truck would start sliding due to the forces acting upon it when the truck turns around a curve. This is related to centripetal force and frictional force.

When the truck turns, the toolbox experiences a centripetal force which pushes it towards the center of the curve. At the same time, friction between the toolbox and the truck bed is resisting this pushing force, keeping the box in place. At some point, if the truck is going too fast, the centripetal force will overcome friction, and the toolbox will start to slide.

We can use the formula for centripetal force, which is F = mv²/r, where m is the mass of the toolbox, v is the velocity of the truck and r is the radius of the curve. And we know that the maximum static friction force (F_max) is f_s * m * g, where f_s is the static friction coefficient and g is the acceleration due to gravity.

The toolbox starts to slide when F = F_max, so we can find the maximum speed (v_max) before sliding happens by making these two equations equal to each other and solving for v. Without the specific values for the mass of the toolbox and the static friction coefficient, we cannot calculate a numerical answer.

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